Resonance Kinematics

(1)

RESONANCE

TM Page # 1Page # 1Page # 1Page # 1Page # 1

MECHANICS

Mechanics is the branch of physics which deals with the cause and effects of motion of a particle, rigid objects and deformable bodies etc. Mechanics is classified under two streams namely Kinematics and Dynamics.

Mechanics

Kinematics Dynamics (or Kinetics) The word kinematics means It is branch of mechanics which is ‘science of motion’.branch of concerned about the causes (i.e.

the

mechanics which deals with force , torque) that cause motion study of motion without going of bodies.

into the cause of motion, i.e. force, torque etc.

1.

1. MOTION AND REST

MOTION AND REST

Motion is a combined property of the object and the observer. There is no meaning of rest or motion without the observer. Nothing is in absolute rest or in absolute motion.

An object is said to be in motion with respect to a observer, if its position changes with respect to that observer. It may happen by

both ways either observer moves or object moves.

2.

2. RECTILINEAR MOTION

RECTILINEAR MOTION

Rectilinear motion is motion, along a straight line or in one dimension. It deals with the kinematics of a particle in one dimension.

2.1

2.1 Position

Position

The position of a particle refers to its location in the space at a certain moment of time. It is concerned with the question − “ where is the

particle at a particular moment of time ? ”

2.2

2.2 Displacement

Displacement

The change in the position of a moving object is known as displacement. It is the vector joining the initial position of the particle to its final position during an interval of time.

2.3

2.3 Distance

Distance

The length of the actual path travelled by a particle during a given time interval is called as distance. The distance travelled is a scalar quantity which is quite different from displacement. In general, the distance travelled between two points may not be equal to the magnitude of the displacement between the same points.

“E pur, si muove.” (Translated: “Even so, it does move.”)

When Galileo was declared a heretic by the Catholic Church and chose to recant his beliefs in order to spare his life, he is known to have said:

“I must altogether abandon the false opinion that the Sun is the center of the world and immovable, and that the Earth is not the center of the world, and moves, and that I must not hold, defend, or teach... the said doctrine.”

But afterwards, as he rose from his kneeling position, his body weary from seventy years of age, legend has it that he murmured, in Italian the above statement.

RECTILINEAR MOTION

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RESONANCE

Example 1. Ram takes path 1 (straight line) to go from P to Q and Shyam takes path 2 (semicircle). (a) Find the distance travelled by Ram and Shyam?

P Q

1

2

100 m (b) Find the displacement of Ram and Shyam?

Sol. (a) Distance travelled by Ram = 100 m

Distance travelled by Shyam = π(50 m) = 50π m (b) Displacement of Ram = 100 m

Displacement of Shyam = 100 m

2.4

2.4 Average Velocity

Average Velocity

(in an interval) :

The average velocity of a moving particle over a certain time interval is defined as the displacement divided by the lapsed time.

Average Velocity = erval int time nt displaceme for straight line motion, along x−axis, we have

v_av =

v

= <v> =

t

x

Δ

= i f i f

t

x

−

The average velocity is a vector in the direction of displacement. For motion in a straight line, directional aspect of a vector can be taken care of by +ve and -ve sign of the quantity.

2.5

2.5 Average Speed

Average Speed

(in an interval)(in an interval)(in an interval)(in an interval)(in an interval)

Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place. It helps in describing the motion along the actual path.

Average Speed = interval time travelled distance NOTE :

(a) Average speed is always positive in contrast to average velocity which being a vector, can be positive or negative.

(b) If the motion of a particle is along a straight line and in same direction then, average velocity = average speed.

(c) Average speed is, in general, greater than the magnitude of average velocity.

The dimension of velocity and speed is [LT-1_{] and their SI unit is meters per second (m/s)}

Example 2. In the example 1, if Ram takes 4 seconds and Shyam takes 5 seconds to go from P to Q, find (a) Average speed of Ram and Shyam?

(b) Average velocity of Ram and Shyam?

Sol. (a) Average speed of Ram =

4 100

m/s = 25 m/s

Average speed of Shyam =

5

50 π

m/s = 10π m/s

(b) Average velocity of Ram =

4

100

m/s = 25 m/s Average velocity of Shyam =

5 100

m/s = 20 m/s

Example 3. A particle travels half of total distance with speed v₁ and next half with speed v₂ along a straight line. Find out the average speed of the particle?

Sol. Let total distance travelled by the particle be 2s. Time taken to travel first half =

1

v

s

Time taken to travel next half =

2

v

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RESONANCE

TM Page # 3Page # 3Page # 3Page # 3Page # 3 Average speed = taken time Total covered distance Total = 2 1 v s v s s 2 + = 1 2 2 1 v v v v 2 + Q.1 A particle covers 4 3

of total distance with speed v₁ and next 4 1

with v₂. Find the average speed of the particle? Ans. 2 1 2 1 3v v v 4v +

Q.2. A car is moving with speed 60 Km/h and a bird is moving with speed 90 km/h along the same direction as shown in figure. Find the distance trav-elled by the bird till the time car reaches the tree?

240 m

Ans. 360 m

2.6

2.6 Instantaneous Velocity

Instantaneous Velocity

(at an instant) :

The velocity at a particular instant of time is known as instantaneous velocity. The term “velocity” usually means instantaneous velocity.

V_inst. = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ Δ → Δ t x lim 0 t =

_dt

dx

In other words, the instantaneous velocity at a given moment (say , t) is the limiting value of the average velocity as we let Δt approach zero. The limit as Δt → 0 is written in calculus notation as dx/dt and is called the derivative of x with respect to t.

2.7

2.7 Average acceleration (in an interval):

Average acceleration (in an interval):

The average acceleration for a finite time interval is defined as : Average acceleration =

erval

int

time

velocity

in

change

Average acceleration is a vector quantity whose direction is same as that of the change in velocity.

av

a

r

=

t

v

Δ

Δr

=

t

v

_f _i

Δ

− r

r

Since for a straight line motion the velocities are along a line, therefore a_av=

t

v

Δ

= i f i f t t v v − −

(where one has to substitute v_f and v_i with proper signs in one dimensional motion)

2.8

2.8 Instantaneous Acceleration (at an instant):

Instantaneous Acceleration (at an instant):

The instantaneous acceleration of a particle is its acceleration at a particular instant of time. It is defined as the derivative (rate of change) of velocity with respect to time. We usually mean instantaneous acceleration when we say “ acceleration”. For straight motion we define instantaneous acceleration as :

a = dt dv = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ Δ → Δ t v lim 0 t and in general

_a

r

= dt v dr = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ Δ → Δ t v lim 0 t r

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RESONANCE

Example 4. Position of a particle as a function of time is given as x = 5t2_{+ 4t + 3. Find the velocity and}

acceleration of the particle at t = 2 s?

Sol. Velocity; v = dt dx = 10t + 4 At t = 2 s v = 10(2) + 4 ⇒ v = 24 m/s Acceleration; a = ₂ 2 dt x d = 10 Acceleration is constant, so at t = 2 s a = z10 m/s2

Q.3 The position of a particle moving on X-axis is given by x = At3_{+ Bt}2_{+ Ct + D.}

The numerical values of A, B, C, D are 1, 4, -2 and 5 respectively and SI units are used. Find (a) the dimensions of A, B, C and D, (b) the velocity of the particle at t = 4 s, (c) the acceleration of the particle at t = 4s, (d) the average velocity during the interval t =0 to t = 4s, (e) the average acceleration during the interval t = 0 to t = 4 s.

Ans. [(a) [A] = [LT -3_{], [B] = [LT} -2_{], [C] = [LT} -1_{] and [D] = [L] ; (b) 78 m/s ; (c) 32 m/s}2 _{; (d) 30 m/s ; (e) 20 m/s}2_]

3.

3. MO

MO

MOTION WITH UNIF

TION WITH UNIFORM VEL

TION WITH UNIF

ORM VEL

ORM VELOCIT

OCIT

OCITYYYYY

Consider a particle moving along x−axis with uniform velocity u starting from the point x = x_i at t = 0. Equations of x, v, a are : x (t) = x_i + ut ; v (t) = u ; a (t) = 0

• x− t graph is a straight line of slope u through x_i.

• as velocity is constant, v − t graph is a horizontal line.

• a−t graph coincides with time axis because a = 0 at all time instants.

x O xi t u is negative slo pe =_u v O u t positive velocity v O u t negative velocity

4.

4. UNIF

UNIF

UNIFORML

ORML

ORMLY ACCELERA

Y ACCELERA

Y ACCELERATED MO

TED MO

TED MOTION

TION

If a particle is accelerated with constant acceleration in an interval of time, then the motion is termed as uniformly accelerated motion in that interval of time.

For uniformly accelerated motion along a straight line (x−axis) during a time interval of t seconds, the following important results can be used.

(a) v = u + at (b) s = ut + 1/2 at2 s = vt − 1/2 at2 x_f = x_i + ut + 1/2 at2 (c) v2_{= u}2_{+ 2as} (d) s = 1/2 (u + v) t (e) s_n = u + a/2 (2n − 1)

u = initial velocity (at the beginning of interval) a = acceleration

v = final velocity (at the end of interval) s = displacement (x_f− x_i)

x_f = final coordinate (position) x_i = initial coordinate (position) s_n = displacement during the nth_sec

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RESONANCE

Example 5. A particle moving rectilinearly with constant acceleration is having initial velocity of 10 m/s. After some time, its velocity becomes 30 m/s. Find out velocity of the particle at the mid point of its path?

Sol. Let the total distance be 2x.

∴ distance upto midpoint = x Let the velocity at the mid point be v and acceleration be a.

From equations of motion

v2_{= 10}2_{+ 2ax} _{____ (1)} 302_{= v}2_{+ 2ax} _{____ (2)} (2) - (1) gives v2_{- 30}2_{= 10}2_{- v}2 ⇒ _v2_{= 500} ⇒ _{v =} 5 10 m/s

Example 6. A police inspector in a jeep is chasing a pickpocket an a straight road. The jeep is going at its maximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance d away, and the motorcycle starts with a constant accelera-tion a. Show that the pick pocket will be caught if _v_≥ _2ad .

Sol. Suppose the pickpocket is caught at a time t after motorcycle starts. The distance travelled by the motor-cycle during this interval is

2

at

2

1 s

=

____ (1) During this interval the jeep travels a distance

vt d s+ = ____ (2) By (1) and (2),

at

d

vt

2

1

2

₊

₌

or, a ad 2 v v t 2 ₋ ± =

The pickpocket will be caught if t is real and positive. This will be possible if _v2 _≥_2ad_or, _v_≥ _2ad

Q.4 A car deccelerates from a speed of 20 m/s to rest in a distance of 100 m. What was its acceleration, assumed constant?

Ans. [- 2 m/s2_]

Q. 5 A 150 m long train accelerates uniformly from rest. If the front of the train passes a railway worker 50 m away from the station at a speed of 25 m/s, what will be the speed of the back part of the train as it

passes the worker?

Ans. [50 m/s]

5 .

GRAPHS IN UNIF

GRAPHS IN UNIFORML

ORML

ORMLY ACCELERA

Y ACCELERATED MO

Y ACCELERA

TED MO

TED MOTION

TION

TION (a

≠ 0)

• x is a quadratic polynomial in terms of t. Hence x − t graph is a parabola.

xi x a > 0 t 0 xi x a < 0 t 0 x-t graph

• v is a linear polynomial in terms of t. Hence v−t graph is a straight line of slope a. v u a is positive slop e= a t 0 v u a is negative slop_{e =} a t 0

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RESONANCE

v-t graph

• a−t graph is a horizontal line because a is constant.

a a positive acceleration t 0 a a negative acceleration 0 a-t graph

6.

6. REACTION TIME

REACTION TIME

When a situation demands our immediate action. It takes some time before we really respond. Reaction time is the time a person takes to observe, think and act.

7

77

7 7...

DIRECTIONS OF VECTORS IN STRAIGHT LINE MOTION

In straight line motion, all the vectors (position, displacement, velocity & acceleration) will have only one component (along the line of motion) and there will be only two possible directions for each vector.

• For example, if a particle is moving in a horizontal line (x−axis), the two directions are right and left. Any vector directed towards right can be represented by a positive number and towards left can be represented by a negative number.

• For vertical or inclined motion, upward direction can be taken +ve and downward as −ve

line of motion l in e of m o ti on lin e o f m otio n + + ₊ --

-• For objects moving vertically near the surface of the earth, the only force acting on the particle is its weight (mg) i.e. the gravitational pull of the earth. Hence acceleration for this type of motion will always be a = −g i.e. a = − 9.8 m/s2₍−_{ve sign, because the force and acceleration are directed}

downwards, If we select upward direction as positive).

NOTE :

(a) If acceleration is in same direction as velocity, then speed of the particle increases.

(b) If acceleration is in opposite direction to the velocity then speed decreases i.e. the particle slows down. This situation is known as retardation.

Example 7. Mr. Sharma brake his car with constant acceleration from a velocity of 25 m/s to 15 m/s over a distance of 200m.

(a) How much time elapses during this interval? (b) What is the acceleration?

(c) If he has to continue braking with the same constant acceleration, how much longer would it take for him to stop and how much additional distance would he cover?

Sol. (a) We select positive direction for our coordinate system to be the direction of the velocity and choose the origin so that x_i = 0 when the braking begins. Then the initial velocity is u_x = +25 m/s at t = 0, and the final velocity and position are v_x = +15 m/s and x = 200 m at time t.

Since the acceleration is constant, the average velocity in the interval can be found from the average of the initial and final velocities.

∴ v_{av, x} =

2

1

(u_x + v_x) =

2

1

(15 + 25) = 20 m/s.

The average velocity can also be expressed as v_{av, x} =

t

x

Δ

. With

_Δ

_x

= 200 m and

Δ

_t

= t − 0, we can solve for t:

t = x av,

v

Δx

=

20

200

= 10 s.

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RESONANCE

(b) We can now find the acceleration using v_x = u_x+ a_xt a_x = t u v_x− _x = 10 25 5 1 − = − 1 m/s2_.

The acceleration is negative, which means that the positive velocity is becoming smaller as brakes are applied (as expected).

(c) Now with known acceleration, we can find the total time for the car to go from velocity u_x = 25 m/s to v_x = 0. Solving for t, we find

t = x x x a u v − =

1

25

0 −

−

= 25 s. The total distance covered is

x = x_i + u_xt +

2

1

a_xt2_{= 0 + (25)(25) +}

2

1

(−1)(25)2_{= 625}−_{312.5 = 312.5 m.}

Additional distance covered = 312.5 – 200 = 112.5 m.

Example 8. A particle is dropped from a tower. It is found that it travels 45 m in the last second of its journey. Find out the height of the tower?

Sol. Let the total time of journey be n seconds.

Using;

(

2 n

1 )

2 a

u

s

_n

=

+

−

45 = 0 +

2

10 )

1

2 (

n

−

n = 5 sec Height of tower; h =

2

1

gt2 ₌

2

1 ×

10

×

52_{= 125 m}

Example 9. A particle is dropped from height 100 m and another particle is projected vertically up with velocity 50 m/s from the ground along the same line. Find out the position where two particle will meet?

Sol. Let the upward direction as positive.

Let the particles meet at a distance y from the ground.

y=0m y=100m u=0 m/s u=50 m/s A B For particle A, y₀ = + 100 m u = 0 m/s a = − 10 m/s2 y = 100 + 0(t) − 2 1

×

10

×

t2 _{[y = y} 0 + ut + ₂ 1 at2_] = 100 - 5t2 _{---- (1)} For particle B, y₀ = 0 m ⇒ u = + 50 m/s ⇒ a = − 10 m/s2 y = 50(t) − 2 1

×

10

×

t2 = 50t − 5t2 _{---- (2)}

According to the problem; 50t − 5t2_{= 100}−_5t2

t = 2 sec

Putting t = 2 sec in eqn. (1), y = 100 − 20 = 80 m

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RESONANCE

Q. 6 A particle is thrown vertically with velocity 20 m/s . Find (a) the distance travelled by the particle in first 3 seconds, (b) displacement of the particle in 3 seconds.

Ans. [25m, 15m]

Q.7 A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 50 m high when

the stone was dropped, find its height when the stone hits the ground. Take g = 10 m/s2_. Ans. [68.5 m]

NOTE:- As the particle is detached from the balloon it is having the same velocity as that of balloon, but its

acceleration is only due to gravity and is equal to g.

8.

8. GRAPHIC

GRAPHIC

GRAPHICAL INTERPRET

AL INTERPRET

AL INTERPRETA

A

ATION OF SOME QU

TION OF SOME QU

TION OF SOME QUANTITIES

ANTITIES

8.1

8.1 Average Velocity

Average Velocity

If a particle passes a point P (x_i) at time t = t_i and reaches Q (x_f) at a later time instant t = t_f, its average velocity in the interval PQ is V_av =

t x Δ Δ = i f i f t t x x − −

This expression suggests that the average velocity is equal to the slope of the line (chord) joining the points corresponding to P and Q on the x−t graph.

x Q P t O xf xi ti tf

8.2

8.2 Instantaneous Velocity

Instantaneous Velocity

Consider the motion of the particle between the two points P and Q on the x−t graph shown. As the point Q is brought closer and closer to the point P, the time interval between PQ (Δt, Δt

′

, Δt

′′

,...) get progressively smaller. The average velocity for each time interval is the slope of the appropriate dotted line (PQ, PQ

′

, PQ

′′

...). As the point Q approaches P, the time interval approaches zero, but at the same time the slope of the dotted line approaches that of the tangent to the curve at the point P. As Δt → 0, V_av (=Δx/Δt) → V_inst.

Geometrically, as Δt → 0, chord PQ → tangent at P.

Hence the instantaneous velocity at P is the slope of the tangent at P in the x − t graph. When the slope of the x − t graph is positive, v is positive (as at the point A in figure). At C, v is negative because the tangent has negative slope. The instantaneous velocity at point B (turning point) is zero as the slope is zero.

8.3

8.3 Instantaneous Acceleration

Instantaneous Acceleration

The derivative of velocity with respect to time is the slope of the tangent in velocity time (v−t) graph.

v t a<0 a=0 0 a>0 a=0 a=0 8 . 4 8 . 48 . 4

8 . 48 . 4

Displanncement from v - t graph

Displacement = Δx = area under v−t graph.

Since a negative velocity causes a negative displacement, areas below the time axis are taken negative. In similar way, can see that Δ v = a Δt leads to the conclusion that area under a −−−−− t graph gives the change in velocity ΔΔΔΔΔv during that interval.

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RESONANCE

Example 10. Describe the motion shown by the following velocity-time graphs.

(a) (b)

Sol. (a) During interval AB: velocity is +ve so the particle is moving in +ve direction, but it is slowing down

as acceleration (slope of v-t curve) is negative. During interval BC: particle remains at rest as velocity is zero. Acceleration is also zero. During interval CD: velocity is -ve so the particle is moving in -ve direction and is speeding up as acceleration is also negative.

(b) During interval AB: particle is moving in +ve direction with constant velocity and acceleration is

zero. During interval BC: particle is moving in +ve direction as velocity is +ve, but it slows down until it comes to rest as acceleration is negative. During interval CD: velocity is -ve so the particle is moving in -ve direction and is speeding up as acceleration is also negative.

Important Points to Remember

• For uniformly accelerated motion (a ≠ 0), x−t graph is a parabola (opening upwards if a > 0 and opening downwards if a < 0). The slope of tangent at any point of the parabola gives the velocity at that instant.

• For uniformly accelerated motion (a ≠ 0), v−t graph is a straight line whose slope gives the acceleration of the particle.

• In general, the slope of tangent in x−t graph is velocity and the slope of tangent in v−t graph is the acceleration.

• The area under a−t graph gives the change in velocity.

• The area between the v−t graph gives the distance travelled by the particle, if we take all areas as positive.

• Area under v−t graph gives displacement, if areas below the t−axis are taken negative.

Example 11. For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the particle?

Sol. Distance travelled = Area under v-t graph (taking all areas as +ve.)

∴ Distance travelled = Area of trapezium + Area of triangle =

(

2

6 )

8

2

1 ×

+

4

5

2

1 ×

×

= 32 + 10 = 42 m

Displacement = Area under v-t graph (taking areas below time axis as -ive.)

∴ Displacement = Area of trapezium − Area of triangle =

(

2

6 )

8

2

1 ×

+

−

4

5

2

1 ×

×

= 32 − 10 = 22 m

Hence, distance travelled = 42 m and displacement = 22 m.

Q.8 For a particle moving along x-axis, following graphs are given. Find the distance travelled by the particle in 10 s in each case?

(10)

RESONANCE

9.

9. Interpretation of some more Graphs

Interpretation of some more Graphs

9.1

9.1 Position vs Time graph

Position vs Time graph

9.1.1 Zero Velocity

As position of particle is fix at all the time, so the body is at rest. Slope;

dt

dx

= tanθ = tan 0º = 0 Velocity of particle is zero

9.1.2 9.1.29.1.2

9.1.29.1.2 Uniform VelocityUniform VelocityUniform VelocityUniform VelocityUniform Velocity

Here tanθ is constant tanθ =

dt

dx

∴

dt

dx

is constant.

∴ velocity of particle is constant.

9.1.3

9.1.3 9.1.3 Non uniform velocity

Non uniform velocity

Non uniform velocity (increasing with time)

In this case;

As time is increasing, θ is also increasing.

∴

dt

dx

= tanθ is also increasing

Hence, velocity of particle is increasing.

9.1.4

9.1.4 9.1.4 Non uniform velocity (decreasing with time)

Non uniform velocity (decreasing with time)

In this case;

As time increases, θ decreases.

∴

dt

dx

= tanθ also decreases.

Hence, velocity of particle is decreasing.

9.2

9.2 Velocity vs time graph

Velocity vs time graph

9.2.1

9.2.1 9.2.1 Zero acceleration

Zero acceleration

Velocity is constant. tanθ = 0 ∴

dt

dv

= 0 Hence, acceleration is zero.

9.2.2 9.2.29.2.2

9.2.29.2.2 Uniform accelerationUniform accelerationUniform accelerationUniform accelerationUniform acceleration

tanθ is constant.

∴

dt

dv

= constant

Hence, it shows constant acceleration.

9.2.3 9.2.39.2.3

9.2.39.2.3 Uniform retardationUniform retardationUniform retardationUniform retardationUniform retardation

Since θ > 90º

∴ tanθ is constant and negative.

∴

dt

dv

= negative constant Hence, it shows constant retardation.

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RESONANCE

9.3

9.3 Acceleration vs time graph

Acceleration vs time graph

9.3.1 Constant acceleration 9.3.1 Constant acceleration9.3.1 Constant acceleration 9.3.1 Constant acceleration9.3.1 Constant acceleration

tanθ = 0

∴

dt

da

= 0

Hence, acceleration is constant.

9.3.2 Uniformly increasing acceleration 9.3.2 Uniformly increasing acceleration9.3.2 Uniformly increasing acceleration 9.3.2 Uniformly increasing acceleration9.3.2 Uniformly increasing acceleration

θ is constant. 0º < θ < 90º

⇒

tanθ > 0 ∴

dt

da

= tanθ = constant > 0

Hence, acceleration is uniformly increasing with time.

9.3.3 Uniformly decreasing acceleration 9.3.3 Uniformly decreasing acceleration9.3.3 Uniformly decreasing acceleration 9.3.3 Uniformly decreasing acceleration9.3.3 Uniformly decreasing acceleration

Since θ > 90º

∴ tanθ is constant and negative.

∴

dt

da

= negative constant

Hence, acceleration is uniformly decreasing with time

Example 12. The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw velocity vs time and acceleration vs time graph.

Sol: x = 4t2 ⇒ _{v =}

dt dx

= 8t

Hence, velocity-time graph is a straight line having slope i.e. tanθ = 8.

a =

dt dv

= 8

Hence, acceleration is constant throughout and is equal to 8.

Problem 10. The displacement vs time graph of a particle moving along a straight line is shown in the figure.

(12)

RESONANCE

11.

11. MO

MO

MOTION WITH NON-UNIF

TION WITH NON-UNIF

TION WITH NON-UNIFORM ACCELERA

ORM ACCELERA

ORM ACCELERATION

ORM ACCELERA

TION

TION (use of def

(use of def

(use of definit

inite int

init

e int

e integrals)

egrals)

Δx =

∫

f i t t

dt

)

t

(

v

(displacement in time interval t = t_i to t_f)

The expression on the right hand side is called the definite integral of v(t) between t = t_i and t = t_f. . Similarly change in velocity

Δv = v_f − v_i =

∫

f i t t dt ) t ( a

Table 2 : Some quantities defined as derivatives and integrals. v(t) =

dt

dx

v = slope of x−t graph a (t) =

dt

dv

a = slope of v−t graphs F (t) =

dt

dp

F = slope of p−t graph (p = linear momentum)

Δx =

∫

dx

=

∫

f i t t

dt

)

t

(

v

_Δx = area under v−t graph

Δv =

∫

dv

=

∫

f i t t

dt

)

t

(

a

_Δv = area under a−t graph

Δp =

∫

dp

=

∫

f i t t

dt

)

t

(

F

_Δp = area under F−t graph

W =

∫

dW

=

∫

f i x x

dx

)

x

(

F

_{W = area under F}₋_{x graph}

12.

12. SOL

SOL

SOLVING PR

VING PR

VING PROBLEMS WHICH INV

OBLEMS WHICH INV

OBLEMS WHICH INVOL

OBLEMS WHICH INV

OL

OLVES NONUNIF

VES NONUNIF

VES NONUNIFORM ACCELERA

VES NONUNIF

ORM ACCELERA

ORM ACCELERATION

TION

12.1

12.1 12.1 Acceleration depending on velocity v or time t

Acceleration depending on velocity v or time t

By definition of acceleration, we have a =

dt dv . If a is in terms of t,

∫

v v0 dv₌

∫

t 0 dt ) t ( a _{. If a is in} terms of v,

∫

=

∫

t 0 v v dt ) v ( a dv 0

. On integrating, we get a relation between v and t, and then

using

∫

x x₀

dx

₌

_∫

t 0

dt

)

t

(

v

, x and t can also be related.

1 2 . 2

1 2 . 21 2 . 2

Acceleration depending on velocity v or position x

a =

dt

dv

⇒ a =

dx

dv

dt

dx

⇒ a =

dt

dx

dv

⇒ a = v

dx

dv

This is another important expression for acceleration.

If a is in terms of x,

∫

v v₀

dv

v

=

∫

x x₀

dx

)

x

(

a

.

(13)

RESONANCE

TM Page # 13Page # 13Page # 13Page # 13Page # 13 If a is in terms of v,

∫

=

∫

x x v v0 0 dx ) v ( a dv v

On integrating, we get a relation between x and v. Using

∫

x x₀

v

(

x

)

dx

=

∫

t 0

dt

_{, we can relate x and t.}

Example 13. An object starts from rest at t = 0 and accelerates at a rate given by a = 6t. What is (a) its velocity and

(b) its displacement at any time t?

Sol. As acceleration is given as a function of time,

∴

∫

=

∫

t t v(t) ) v(t₀ ₀

a(t)dt

dv

Here t₀ = 0 and v(t₀) = 0 ∴ v(t) =

∫

t 0

6tdt

₌

0 t

2 t

6

2

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

= 6 (

2 t

2 - 0) = 3t2 So, v(t) = 3t2 As

Δ

=

∫

t t₀

v(t)dt

x

∴

Δ

=

∫

t 0 2

dt

3t

x

₌

0 t

3 t

3

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−0

3 t

3

3 = t3

Hence, velocity v(t) = 3t2_{and displacement} 3

t

=

Δx

Q.9. For a particle moving along x-axis, acceleration is given as

a

=

2v

2. If the speed of the particle is v₀ at x = 0, find speed as a function of x.

Ans. [

v

=

v

₀

e

2x]

Q. 10. For a particle moving along x-axis, velocity is given as a function of time as v = 2t2 _{+ sint. At t = 0, particle is}

at origin. Find the position as a function of time?

Ans. [

t

cos(t)

1

3

2

(14)

RESONANCE

S U M M A R Y

Rectilinear Motion Rectilinear MotionRectilinear Motion

Rectilinear MotionRectilinear Motion : Rectilinear motion is motion, along a straight line or in one dimension.

Displacement : Displacement : Displacement :

Displacement : Displacement : The vector joining the initial position of the particle to its final position during an interval of time.

Distance : Distance : Distance :

Distance : Distance : The length of the actual path travelled by a particle during a given time interval

Average Velocity Average Velocity Average Velocity Average Velocity Average Velocity =

erval int time nt displaceme = i f i f

t

x

−

Average Speed Average SpeedAverage Speed Average SpeedAverage Speed =

interval time

travelled distance

Instantaneous Velocity : Instantaneous Velocity : Instantaneous Velocity :

Instantaneous Velocity : Instantaneous Velocity : V_inst. = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ Δ → Δ t x lim 0 t =

_dt

dx

Average Acceleration Average AccelerationAverage Acceleration Average AccelerationAverage Acceleration

= erval int time velocity in change = i f i f t t v v − − Instantaneous Acceleration :

Instantaneous Acceleration :Instantaneous Acceleration : Instantaneous Acceleration :Instantaneous Acceleration :

a =

dt

dv

=

⎟

⎠

⎞

⎜

⎝

⎛

Δ

→ Δ

t

v

lim

0 t Equations of Motion Equations of MotionEquations of Motion Equations of MotionEquations of Motion

(a) v = u + at (b) s = ut + 1/2 at2 s = vt − 1/2 at2 x_f = x_i + ut + 1/2 at2 (c) v2_{= u}2_{+ 2as} (d) s = 1/2 (u + v) t (e) s_n = u + a/2 (2n − 1)

Important Points to Remember Important Points to RememberImportant Points to Remember Important Points to RememberImportant Points to Remember

• For uniformly accelerated motion (a ≠ 0), x−t graph is a parabola (opening upwards if a > 0 and opening downwards if a < 0). The slope of tangent at any point of the parabola gives the velocity at that instant.

• For uniformly accelerated motion (a ≠ 0), v−t graph is a straight line whose slope gives the acceleration of the particle.

• In general, the slope of tangent in x−t graph is velocity and the slope of tangent in v−t graph is the acceleration.

• The area under a−t graph gives the change in velocity.

• The area between the v−t graph gives the distance travelled by the particle, if we take all areas as positive.

• Area under v−t graph gives displacement, if areas below the t−axis are taken negative.

Maxima and Minima Maxima and MinimaMaxima and Minima Maxima and Minima Maxima and Minima Conditions for maxima are:−−−−−

dx

dy

= 0 (b) ₂ 2 dx y d < 0

Conditions for minima are:−−−−−

dx

dy

= 0(b) ₂ 2 dx y d > 0

Motion with Non-Uniform Acceleration Motion with Non-Uniform AccelerationMotion with Non-Uniform Acceleration Motion with Non-Uniform Acceleration Motion with Non-Uniform Acceleration

Δx =

∫

f i t t

dt

)

t

(

v

Δv = v_f − v_i =

∫

f i t t dt ) t ( a

Solving Problems which Involves Nonuniform Solving Problems which Involves NonuniformSolving Problems which Involves Nonuniform Solving Problems which Involves Nonuniform Solving Problems which Involves Nonuniform Acceleration AccelerationAcceleration Acceleration Acceleration If a is in terms of t,

∫

v v₀

dv

₌

∫

t 0

dt

)

t

(

a

If a is in terms of v,

∫

=

∫

t 0 v v dt ) v ( a dv 0 If a is in terms of x,

∫

v v₀

dv

v

=

∫

x x₀

dx

)

x

(

a

. If a is in terms of v,

∫

=

∫

x x v v0 0 dx ) v ( a dv v

(15)

RESONANCE

Exercise 1

OBJECTIVE PROBLEMS

(RECTILINEAR MOTIN)

BA

BATTTTTCH - CC

CH - CC

1. A motor car is going due north at a speed of 50 km/h. It makes a 90º left turn without changing the speed. The change in the velocity of the car is about

(A) 50 km/h towards west (B) 50 2 km/h towards south-west

(C) 50 2 km/h towards north-west (D) zero

2. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let x_A and x_B be the magnitude of displacements in the first 10 seconds and the next 10 seconds.

(A) x_A < x_B (B) x_A = x_B (C) x_A > x_B (D) the information is insufficient to decide the relation of x_A with x_B.

3. A ball takes t seconds to fall from a height h₁ and 2t seconds to fall from a height h₂. Then h₁/h₂ is

(A) 2 (B) 4 (C) 0.5 (D) 0.25

4. A body starts from rest and is uniformly acclerated for 30 s. The distance travelled in the first 10 s is x₁, next 10 s is x₂ and the last 10 s is x₃. Then x₁ : x₂ : x₃ is the same as

(A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 9

5. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is

(A) a upward (B) (g-a) upward (C) (g-a) downward (D) g downward

6. A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically downward and the ball B is thrown vertically upward with the same speed. The ball A hits the ground with a speed

υA and the ball B hits the ground with a speed υB. We have (A) υ _A > υ _B (B) υ _A < υ _B (C) υ _A = υ _B

(D) the relation between A and B depends on height of the building above the ground.

7. Figure shows position-time graph of two cars A and B. (A) Car A is faster than car B.

(B) Car B is faster than car A.

(C) Both cars are moving with same velocity.

A

0 5 x(m)

B

(D) Both cars have positive acceleration.

8. The displacement–time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point :

(A) C (B) D t x C D E F (C) E (D) F

9. A particle starts from rest and moves along a straight line with constant acceleration. The variation of velocity v with displacement S is : (A) S v (B) v S C) v S (D) v S

10. The displacement time graphs of two particles A and B are straight lines making angles of respectively 300_{and 60}0

with the time axis. If the velocity of A is v_A and that of B is v_B, then the value of

B A v v is (A) 1/2 (B) 1/ 3 (C) 3 (D) 1/3

11. The initial velocity of a particle is u (at t=0) and the acceleration f is given by (f = at). Which of the following relations is valid?

(A) v = u + at2 _{(B) v = u +}

2 at2

(C) v = u + at (D) v = u

12. A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by

(A) T = 2h/v (B) v h g h 2 T= + _(C) v 2 h g h 2 T= + _(D) v h 2 g 2 h T= +

(16)

RESONANCE

13. A body is released from the top of a tower of height h metre. It takes T seconds to reach the ground. Where is the ball at the time T/2 seconds ?

(A) at h/4 metre from the ground (B) at h/2 metre from the ground

(C) at 3h/4 metre from the ground (D) depend upon the mass of the ball

14. A stone is thrown vertically upward with an initial velocity u from the top of a tower, reaches the ground with a velocity 3u. The height of the tower is:

(A) g 3u2 (B) g 4u2 (C) g 6u2 (D) g 9u2

15. A particle starts from rest with uniform acceleration a. Its velocity after n seconds is v. The displacement of the body in the last two seconds is :

(A) n 1) -2v(n (B) n 1) -v(n (C) n 1) v(n+ (D) n 1) 2v(2n+

16. Consider the motion of the tip of the minute hand of a clock. In one hour

(A) the displacement is zero (B) the distance covered is zero

(C) the average speed is zero (D) the average velocity is zero

17. A particle moves along the X-axis as x = u(t – 2) + a(t – 2)2

(A) the initial velocity of the particle is u (B) the acceleration of the particle is a (C) the acceleration of the particle is 2a (D) at t =2s particle is at the origin.

18. Mark the correct statements for a particle going on a straight line:

(A) If the velocity and acceleration have opposite sign, the object is slowing down.

(B) If the position and velocity have opposite sign, the particle is moving towards the origin. (C) If the velocity is zero at an instant, the acceleration should also be zero at that instant.

(D) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

19. The velocity of a particle is zero at t = 0 (A) The acceleration at t = 0 must be zero (B) The acceleration at t = 0 may be zero.

(C) If the acceleration is zero from t = 0 to t = 10 s, the speed is also zero in this interval. (D) If the speed is zero from t = 0 to t = 10 s the acceleration is also in the interval.

20. Mark the correct statements:

(A) The magnitude of the velocity of a particle is equal to its speed.

(B) The magnitude of average velocity in an interval is equal to its average speed in that interval.

(C) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero (D) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

21. The velocity-time plot for a particle moving on a straight line is shown in fig. (A) The particle has constant acceleration

(B) The particle has never turned around. (C) The particle has zero displacement

(D) The average speed in the interval 0 to 10s is the same as the 10 20 30

t(s) -10 -20 10 0 v(m/s)

average speed in the interval 10s to 20s.

SUBJECTIVE PROBLEMS

1. A particle covers each 1/3 of the total distance with speed v₁, v₂ and v₃ respectively. Find the average speed of the particle ?

2. The position of a particle moving on x-axis is given by x = 4t3_{+ 3t}2_{+ 6t + 4. Find} (a) The velocity and acceleration of particle at t = 5 s.

(b) The average velocity and average acceleration during the interval t = 0 to t = 5 s, x = 4t3_{+ 3t}2_{+ 6t + 4} (a) The velocity and acceleration of particle at t = 5 s.

(b) The average velocity and average acceleration during the interval t = 0 to t = 5 s.

3. A train starts from rest and moves with a constant acceleration of 2.0 m/s2_{for half a minute. The brakes are then} applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

4. A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. Find (a) the acceleration and (b) the distance travelled during the (t+1)th second.

5. For a particle moving along x-axis, following graphs are given. Find the distance travelled by

the particle in 10 s in each case.

t(s) x(m) 10 10 (A) (B) t(s) v(m/s) 10 0 10 2

(17)

RESONANCE

6. For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the

particle? Also find the average velocity of the particle?

7. The acceleration of a cart started at t = 0, varies with time as shown in figure . Find the distance travelled in 30 seconds and draw the

position-time graph.

8. For a particle moving rectilinearly, acceleration as a function of speed is given as a = 8v2_{. Find the speed as a} function of x if the particle is having a speed of v₀ at x = 0?

9. Under what conditions does the magnitude of the average velocity equal to the average speed.

10. Can an object have increasing speed but its acceleration decreases? If yes, give an example; if not, explain why?

11. Figure shows four paths along which obejcts move from a starting point to a final point (particle is moving along the same straight line), all in the same time. The paths pass over a grid of equally spaced straight lines. Rank the paths according to

(a) the average velocity of the objects and

(b) the average speed of the objects, greatest first.

12. A man walking with a speed 'v' constant in magnitude and direction passes under a lantern hanging at a height H above the ground. Find the velocity with which the edge of the shadow of the man's head moves over the ground, if his height is 'h'.

13. An elevator is descending with uniform acceleration. To measure the acceleration , a person in the elevator drops a coin at the moment the elavator starts. The coin is 6 ft above the floor of the elevator at time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator. [Take g = 32 ft/s2_]

14. When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket moves upward for a while and then begins to fall. A parachute opens shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.

(a) How fast was the rocket climbing when the engine stopped?

(b) For how many seconds did the engine burn? (c) When did the rocket reach its

highest point? What was its velocity then?

(d) When did the parachute open up ? How fast was the rocket falling then? (e) How long did the rocket fall before the parachute opened?

(f) When was the rocket's acceleration greatest?

(g) When was the acceleration constant? What was its value then (to the nearest interger)?

1. A particle of mass 10−2_{kg is moving along the positive x}−_{axis under the influence of a force}

F(x) = ₂

x 2

K

− where K = 10−2_{N m}2_{. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. Find} (i) its velocity when it reaches x = 0.50 m

(ii) the time at which it reaches x = 0.25 m. [ JEE '98, 8 ]

2. In 1.0 sec. a particle goes from point A to point B moving in a semicircle of radius 1.0 m. The magnitude of average velocity is: [JEE '99, 2]

(A) 3.14 m/sec (B) 2.0 m/sec

(18)

RESONANCE

3. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the

ground as [ JEE '2000, 3 ]

(A) (B) (C) (D)

4. A block is moving down a smooth inclined plane starting from rest at time t = 0. Let S_n be the distance travelled by

the block in the interval t = n – 1 to t = n. The ratio

1 n n S S + is [JEE Scr. 2004, 3] (A) n 2 1 n 2 − (B) 1 n 2 1 n 2 + − (C) 1 n 2 1 n 2 − + (D) 1 n 2 n 2 −

5. A particle is initially at rest, It is subjected to a linear acceleration a , as shown in the figure. The maximum speed attained by the particle is

[JEE Scr. 2004; 3]

(A) 605 m/s (B) 110 m/s (C) 55 m/s (D) 550 m/s

6. The velocity displacement graph of a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be [JEE Scr. 2005; 3]

(A) (B) (C) (D)

EXERCISE # 1

OBJECTIVE PROBLEMS

OBJECTIVE PROBLEMSOBJECTIVE PROBLEMS OBJECTIVE PROBLEMSOBJECTIVE PROBLEMS

1. B 2. D 3. D 4. C 5. D 6. C 7. C 8. C 9. B 10. D 11. B 12. B 13. C 14. B 15. A 16. A,D 17. C,D 18. A,B,D 19. B,C,D 20. A 21.A,D SUBJECTIVE PROBLEMS SUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMS SUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMS

1. 3 1 3 2 2 1 3 2 1 v v v v v v v v 3v + +

2. (a) v_{t = 5 s} = 336 units, a_{t = 5 s} = 126 units ; (b) <v> = 121 units, <a> = 66 units

3. (a) 2.7 km , (b) 60 m/s, (c) 225 m and 2.25 km

4. (a) 50 m/s2_{; (b) 125 m} _{5. (A) 0; (B) 60m}

6. distance travelled = 10 m; displacement = 6 m; average velocity = 1.2 m/s 7. 1000 ft. , straight line (l j y j s[kk) parabolic curve (i j oy h; oØ)

parabolic curve (i j oy h; oØ)

8. v = v₀ e8x 9. <vr> = <v> ∴ time t displacmen = time travelled ce tan dis

When the distance travelled is equal to the displacement of the particle, i.e. particle moves along the straight line in the same direction without reversing the direction of motion.

10. Yes. Speed of object will increase if acceleration is in

the direction of velocity, yet its magnitude may decrease.

a =

dt dv

= slope of the curve.

acceleration (slope of the curve) is decreasing with time yet the speed is increasing.

11. (a) <vr₁> = <vr₂ > = <vr₃> = <vr₄ >

(b) v_{4, avg}> v_{1, avg} = v_{2, avg} > v_{3, avg}

12. _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −h H H v 13. 20 ft/s2 14. (a) 190 ft/s (b) 2 s (c) 8 s, 0 ft/s (d) 10.8 s, 90 ft/s (e) 2.8 s

(f) greatest acceleration happens 2 s after launch (g) constant acceleration between 2 and 10.8 s, – 32 ft/s2_.

EXERCISE # 2

1. (i) _Vr = −−−−−1 _iˆ m/s (ii) t =

4 3 3+

π

(19)

RESONANCE

RELA

RELATIVE MO

TIVE MO

TIVE MOTION

TION

Motion is a combined property of the object under study and the observer. Motion is always relative, there is no such term like absolute motion or absolute rest. Motion is always defined with respect to an observer or reference frame.

Reference frame :

Reference frame is an axis system from which motion is observed. A clock is attached to measure time. Reference frame can be stationary or moving. There are two types of reference frame:

(i) Inertial reference frame : A frame of reference in which Newton’s

first law is valid is called as inertial reference frame.

(ii) Non-inertial reference frame : A frame of reference in which

Newton’s first law is not valid is called as non-inertial reference frame.

Note : Earth is by definition a non-inertial reference frame because of its

cen-tripetal acceleration towards sun. But, for small practical applications earth is assumed stationary hence, it behaves as an inertial reference frame.

RELA

RELATIVE VEL

TIVE VEL

TIVE VELOCIT

OCIT

OCITYYYYY

Definition : Relative velocity of a particle (object) A with respect to B

is defined as the velocity with which A appears to move is B if consid-ered to be at rest. In other words, it is the velocity with which A appears to move as seen by the B considering itself to be at rest.

Relative motion along straight line

-v_A = dt dx_A , v_B = dt dx_B x_BA = x_B – x_A v_BA = dt dxB – dt dxA v_BA = v_B – v_A

⇒ v_AA = v_A – v_A = 0 (velocity of A with respect to A)

Note : velocity of an object w.r.t. itself is always zero. • Impossibility is a relative concept. • If my Theory of relativity is proven successful, Germany will claim me as a German and France will declare that I am a citizen of the world. Should my theory prove untrue, France will say that I am a German and Germany will declare that I am a Jew.

-Einstein

RELA

(20)

RESONANCE

Ex.1 An object A is moving with 5 m/s and B is moving with 20 m/s in the same direction. (Positive x-axis) (i) Find velocity of B with respect to A.

Sol. v_B = 20iˆ m/s ⇒ v_A = 5 iˆ m/s ⇒ v_B – v_A = 15 iˆ m/s (ii) Find velocity of A with respect to B

Sol. v_B = 20iˆ m/s, v_A = 15iˆ m/s ⇒ v_AB = v_A – v_B = – 15iˆ m/s

Note : v_BA = – v_AB

Ex.2 Two objects A and B are moving towards each other with velocities 10 m/s and 12 m/s respectively a shown.

(i) Find out velocity of A with respect to B.

Sol. v_AB = v_A – v_B = (10) – (–12) = 22 m/s towards right. (ii) Find out velocity of B with respect to A

v_BA = v_B – v_A = (–12) – (10) = –22 m/s towards left.

Velocity of Approach

It is the rate at which a separation between two moving particles decreases. If separation decreases velocity of approach is positive,

Velocity of approach = 22 m/s

Velocity of approach = 15 m/s

If separation increases, velocity of approach is negative. It is mainly called velocity of separation.

Velocity of separation

-It is the rate with which separation between two moving object increases. Velocity of separation = 2 m/s

Velocity of separation = 15 m/s

Illustration :

Two balls A and B are moving in the same direction with equal velocities, find out their relative velocity.

(21)

RESONANCE

Illustration :

A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively (g = 10 m/s2_{. Find}

separa-tion between them after one second

Sol. S_A = ut – 2 1 gt2 _{= 5t –} 2 1 × 10 × t2 = 5 × 1 – 5 × 12 _{= 5 – 5} _{= 0} S_B = ut – 2 1 gt2_. _{= 10 × 1 –} 2 1 × 10 × 12 = 10 – 5 = 5 ∴ S_B – S_A = separation = 5m. Alter : By relative ar_BA = ar_B – ar_A = (–10) – (–10) = 0 Also vr_BA= vr_B – vr_A = 10 – 5 = 5 m/s ∴ sBA r (in 1 sec) = vr_BA × t = 5 × 1 = 5 m

∴ Distance between A and B after 1 sec = 5 m.

Illustration :

A ball is thrown downwards with a speed of 20 m/s from top of a building 150 m high and simultaneously another ball is thrown vertically upwards with a speed of 30 m/s from the foot of the building. Find the time when both the balls will meet. (g = 10 m/s2₎ Sol. (I) S₁ = 20 t + 5 t2 + S₂ = 30 t – 5 t2 ____________________ 150 = 50 t ⇒ t = 3 s.

(II) Relative acceleration of both is zero since both have acceleration in downward direction

AB ar = ar_A – ar_B = g – g = 0 BA vr = 30 – (–20) = 50 s_BA = v_BA × t t = BA BA v s = 50 150 = 3 s

(22)

RESONANCE

Ex.5 Two cars C₁ and C₂ moving in the same direction on a straight road with velocities 12 m/s and 10 m/s respectively. When the separation between the two was 200 m C₂ started accelerating to avoid collision. What is the minimum acceleration of car C₂ so that they don’t collide.

Sol. By relative aC₁C₂ r = aC₁ r – aC₂ r = 0 – a = (–a) vC₁C₂ r = vC₁ r – vC₂ r = 12 – 10 = 2 m/s. So by relativity we want the car to stop.

∴ v2_{– u}2_{= 2as.} ⇒ 0 – 22_{= – 2 × a × 200} ⇒_{a =} 100 1 m/s2 = 0.1 m/s2 _{= 1 cm/s}2_.

∴ Minimum acceleration needed by car C₂ = 1 cm/s2

RELA

RELATIVE MO

TIVE MO

TIVE MOTION IN LIFT

TION IN LIFT

Illustration :

A lift is moving up with acceleration a. A person inside the lift throws the ball upwards with a velocity u relative to hand.

(a) What is the time of flight of the ball?

(b) What is the maximum height reached by the ball in the lift?

Sol. (a) ar_BL= ar_B – ar_L = (g + a) downwards s r = urt + 2 1 BL ar t2 _{0 = uT –} 2 1 (g + a)T2 ∴ T = ₍_g2₊u_a₎ (b) v2_{– u}2_{= 2 as} _{0 – u}2_{= –2(g + a) H} H = ) a g ( 2 u2 +

Projectile motion in a lift moving with acceleration a upwards

(1) Initial velocity = u

(2) Velocity at maximum height = u cos θ (3) T = 2_gusin₊_aθ (4) Maximum height (H) = ) a g ( 2 sin u2 2 + θ (5) Range = a g 2 sin u2 + θ

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RESONANCE

RELA

RELATIVE MO

TIVE MO

TIVE MOTION IN T

TION IN T

TION IN TW

TION IN T

W

WO DIMENSION

O DIMENSION

A

r

r _{= position of A with respect to O}

B

r r

= position of B with respect to O

AB

r r

= position of A with respect to B.

B A AB r r r r r r ₌ ₋ ∴ d(_dtrAB) r = dt ) r ( d r_A – dt ) r ( d r_B . ⇒ vr_AB =vr_A−vr_B dt ) v ( d r_AB = dt ) v ( d r_A – dt ) v ( d r_B ⇒ aAB aA aB r r r ₌ ₋

Note : These formulae are not applicable for light.

Illustration :

Object A and B has velocities 10 m/s. A is moving along East while B is moving towards North from the same point as shown. Find velocity of A relative to B (vr_AB)

Sol. vr_AB = vr_A – vr_B ∴ vAB r = 10√2 Note : vA vB r r ₋ = v2_A₊v_B2₋2v_Av_Bcos_θ

Illustration :

Two particles A and B are projected in air. A is thrown horizontally, B is thrown vertically up. What is the separation between them after 1 sec.

Sol. ar_AB = ar_A– ar_B = 0

∴ vr_AB = ₁₀2₊₁₀2 ₌₁₀ ₂

∴ s_AB = v_ABt = (10 2) t = 10 2 m Consider the situation, shown in figure

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RESONANCE

Ex.23 (i) Find out velocity of B with respect to A

A B

BA v v

vr =r −r = 20jˆ – 20iˆ (ii) Find out velocity of A with respect to B

B A

AB v v

vr =r −r = 20iˆ – 20jˆ

Ex.24

(1) Find out motion of tree, bird and old man as seen by boy. (2) Find out motion of tree, bird, boy as seen by old man (3) Find out motion of tree, boy and old man as seen by bird.

Sol. (1) With respect to boy : v_tree = 16 m/s (←) v_bird = 12 m/s (↑) v_{old man} = 18 m/s (←) (2) With respect to old man :

v_Boy = 18 m/s (→) v_Tree = 2 m/s (→)

v_Bird = 18 m/s (→) and 12 m/s (↑) (3) With respect to Bird :

v_Tree = 12 m/s (↓) and 16 m/s (←) v_{old man} = 18 m/s (←) and 12 m/s (↓). v_Boy = 12 m/s (↓).

MO

MOTION OF A TRAIN MO

TION OF A TRAIN MO

TION OF A TRAIN MOVING ON EQU

VING ON EQU

VING ON EQUA

VING ON EQU

A

ATTTTTOR :

OR :

If a train is moving at equator on the earth’s surface with a velocity v_TE relative to earth’s surface and a point on the surface of earth with velocity v_E relative to its centre, then

E T

TE v v

vr =r −r or vr_T =vr_TE +vr_E

So, if the train moves from west to east and if the train moves from east to west (the direction of motion of earth on its axis) (i.e. opposite to the motion of earth)