Kinematics

Kinematics

2

Chapter 3 : Kinematics

3.1 Motion in One Dimension 04

3.2 Motion in Two & Three Dimension 26

3.3 Projectile Motion 31

3.4 Relative Motion in One and Two Dimension 42

3.5 Circular Motion 47

Solved Examples 52

Exercises 60

Previous Years’ IITJEE Questions 67

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Interesting Fact

“Our nature

consist in

motion;

complete rest

is death.”

Blaise Pascal

Physicist

Galileo Galilei

Galileo Galilei (15 February 1564– 8 January 1642) was an Italian physicist, mathematician, astronomer and philosopher who played a major role in the Scientific Revolution. Galileo's theoretical and experimental work on the motions of bodies, was a precursor of the classical mechanics developed by Sir Isaac Newton. He also concluded that objects retain their velocity unless a force— often friction—acts upon them. According to Stephen Hawking, Galileo probably bears more of the responsibility for the birth of modern science than anybody else, and Albert Einstein called him the father of modern science.

Galileo Galilei

Galileo had dropped balls of the same material, but different masses, from the Leaning Tower of Pisa to demonstrate that their time of descent was independent of their mass.

Introduction

Our aim is to study the motion of a rigid body. A rigid body can also rotate as it moves. However, we shall restrict ourselves to translatory motion for the time being , in which every portion of the body moves in the same direction at the same rate. As all particles in the body move in the same way, we may as well consider the motion of a single particle in the body. Moreover, the cause of the motion will not be dealt with until we come to Newton’s laws of motion. In this lesson we study the motion itself.

Section - 3.1

Motion in One Dimension

To describe the motion of a particle, we introduce four important quantities, namely position, displacement, velocity and acceleration. In the general motion of a particle in two or three dimensions, these quantities are vectors, which have directions as well as magnitude. In this section we confine ourselves to motion in one dimension. For such restricted motion, the particle is confined to move in a straight line and there are only two directions, distinguished by designating them as positive and negative.

3.1.1 Displacement, Velocity and Speed

To understand the concept of displacement, let us set up a coordinate system by choosing some reference point on a line for the origin O. Any other point on the line can be assigned a number x which indicates how far the point is from the origin. The value of x will depend on the unit chosen as the measure of distance. If the point is towards right of origin, x is positive and if it is towards left of origin then x is negative.

Suppose that the particle is at x₁ at time t₁ and at x₂ at time t₂ then the displacement of the particle is x₂ – x₁. The change in position (displacement) is denoted by ∆x.

∆x

= x

₂

– x

₁

The average velocity of the particle is defined to be the ratio of displacement ∆x

and time interval ∆t = t₂ – t₁:

2 1 2 1 . avg x x x V t t t − ∆ = = ∆ −

x1 t = t1 t = t2 x = 0 x2 x-axis

The displacement and average velocity may be either positive or negative, depending on whether x₂ is greater or less than x₁. A positive value would indicate the motion towards the right and the negative value indicates the motion towards left. If we plot position (x) of the particle as a function of time (t), then in the graph

V_avg = 2 1 2 1 x x t t − − = BC AC = tanθ θ A x1 x2 x ∆t = – t2 t1 C B ∆x = – x2 x1 t1 t2

= Slope of line joining points A and B on x – t graph.

The distance travelled by the particle is not necessarily equal to the displacement. If the particle moves only in one direction then the displacement is equal to distance travelled, otherwise distance travelled is more than displacement.

x = x1 t = t1 t = t2 x = 0 x = x2 x-axis x = x1 x = 0 x = x2 x = x3 x-axis A C B

Distance travelled = Displacement Distance travelled = AB + BC

= (x₃ – x₁) + (x₃ – x₂) Displacement = x₂ – x₁

In general Distance travelled ≥ Displacement

The average speed of the particle is defined to be ratio of distance travelled and time interval.

Average speed = total distance

time taken

Note that while velocity is a vector quantity and has direction, speed has no direction. The SI units are meters per second.

Illustration 1:

If a particle is at x₁ = 10 m at t₁ = 2s and at x₂ = 6 m at t₂ = 4s find its displacement and average velocity for this time interval.

Solution :

By definition, displacement is ∆x = x₂ – x₁ = 6 – 10 = – 4 m and the average velocity is

V_avg = x t ∆ ∆ = 4 4 2 − − = – 2 m/s Illustration 2:

A man runs 200 m in first 25 seconds and then turns back and runs 100m in next 15 seconds towards the starting point. Find out his

(a) average velocity (b) average speed.

Solution :

(a) Let us take origin at the starting point. Then x₁ = 0, x₂ = 200 – 100 = 100 m ∆x = x₁ – x₁ = 100 – 0 = 100 m ∆t = 25 + 15 = 40 sec (0, 0) (100, 0) (200, 0) t = 0 t = 40 sec t = 25 sec V_avg = 100 40 = 2.5 m/s (b) Distance travelled = 200 + 100 = 300 m time taken = 25 + 15 = 40 sec

Average speed = 300

40 = 7.5 m/s

Instantaneous Velocity (Velocity at an instant)

The velocity at a particular instant of time is known as instantaneous velocity.

Avg. velocity for time interval ∆t is V_avg = x t ∆

∆ = slope of line AB.

∆x t1 t2''' t2'' t2' t2 t x1 x2''' x2'' x2' x2 A B B2 B1 B3 Bn x ∆t

As we reduce the interval ∆t, point B moves nearer and nearer to point A. (B

→B₁ → B₂ → B₃ --- → B_n) and (t₂ →t₂' → t₂'' → t₂''')

As ∆t →0, (∆t tends to zero), point B approaches point A and we can find

instantaneous velocity at point A.

T H E O R Y

∴ V_{instantaneous} = 0 lim t x t ∆ → ∆ ∆ = dx dt

= Slope of tangent at point A.'

∴ Slope of x – t graph at any point gives velocity at that point Magnitude of instantaneous velocity = Instantaneous speed

3.1.2 Average and Instantaneous Acceleration

When the velocity of the body changes continuously as the motion proceeds, the body is said to have accelerated motion. Suppose a particle moving along x-axis have velocity v₁ at point P₁ and velocity v₂ at point P₂. Figure below shows the graph of instantaneous velocity v plotted as a function of time t.

t1 t2 P1 P2 v1 v2 t v

Slope of tangent at P = Instantaneous acc at P1 1 n

Slope of tangent at P = Instantaneous acc at P Slope of line P P = Avg. acc during interval [ , ]

2 2

1 2 1 2

n

n

t t

The average velocity of the particle as it moves from P₁ to P₂ is defined as the ratio of change in velocity to the elapsed time.

a_avg = 2 1 2 1 v v v t t t − ₌∆ − ∆

In the figure, the average acceleration is represented by slope of the line joining points P₁ and P₂.

The instantaneous acceleration of a body, that is acceleration at some time t is defined in the same way as instantaneous velocity. Let the second point P₂ in figure be taken closer and closer to point P₁ and let average acceleration be calculated over shorter and shorter time intervals. The instantaneous acceleration at the first point is defined as the limiting value of the average acceleration when the second point is taken closer and closer to the first point.

a_{instantaneous} = lim_{t 0} v t ∆ → ∆ ∆ = dv dt

In the figure shown, the instantaneous acceleration at any point is equal to the slope of the tangent in the v – t graph at that point.

Illustration 3:

The position of a particle is given by x = (3t – 5t2 _{+ t}3_{) m where t is in seconds. Find the velocity and acceleration as a} function of time.

Solution :

To find the velocity of the particle we compute the derivative of x with respect to t.

v = dx

dt = d dt(3t – 5t

2 _{+ t}3_{) = (3 – 10t + 3t}2_{) m/sec} To find acceleration, we differentiate v with respect to time t.

a = dv

dt = – 10 + 6t = (6t – 10) m/s 2

In-Chapter Exercise - 1

1. The velocity of a car travelling on a straight road is given by the equation v = 6 + 8t – t2 _{where v is in meters per second and}

t in seconds. The instantaneous acceleration when t = 4.5 s is :

(a) 0.1 m/s2 _{(b) 1 m/s}2 _{(c) – 1 m/s}2 _{(d) – 0.1 m/s}2

2. The displacement x of a body varies with time t as x = – 2

3t

2 _{+ 16t + 2. The body will come to rest after time t = ... .}

3. An athlete takes 4 sec to reach his maximum speed of 36 km/hr. What is the magnitude of this avg. acceleration.

4. When a person leaves his home for shopping by his bike, the milometer reads 1032 miles. When he returns home after 1.5 hours the reading is 1122 miles.

(a) What is the average speed of the car during this period? (b) What is the average velocity? 5. The position of a particle is given by x = 6t – 3t2 _{where t is expressed in seconds and x in meter.}

(a) The acceleration of the particle is

(i) 6 m/sec2 _{(ii) – 6 m/sec}2 _{(iii) – 3 m/sec}2 _{(iv) None}

(b) The maximum value of position co-ordinate of particle on positive x-axis is

(i) 3 m (ii) 1 m (iii) 2 m (iv) 4 m

(c) The total distance travelled by the particle between t = 0 to t = 2 sec is

(i) 0 (ii) 3 m (iii) 4 m (iv) 6 m

3.1.3 Motion with Constant Acceleration

If a particle is accelerated with constant acceleration for a time interval, then following important results can be used.

v = u + at Here u = initial velocity s = ut + 1 2at 2 _{v = final velocity} = vt – 1 2at 2 _{a = acceleration} x_f = x_i + ut + 1 2at 2 _x

f = final position coordinate

v2 _{= u}2 _{+ 2as x}

i = initial position coordinate

s_n = u + 1

2a(2n – 1) sn = Displacement during n

th _second

If acceleration is in same direction as velocity, then speed of the particle increases and if in opposite direction then speed decreases.

Illustration 4:

A person travels half of the total distance with speed 20m/sec and next half with speed 30m/sec along a straight line. Find out the average speed of the particle?

Solution :

Let the total distance travelled by the particle be 2D. Time taken for 1st _{half =} D

20 Time taken for 2nd _{half =} D 30

Average speed = Total Distance Covered

Time take = 2D D D 20+30 = 24 m/sec T H E O R Y

Illustration 5:

Position of a particle moving along x-axis is a function of time as x = 3t2 _{+ 9t + 6 meters. Find} (a) Average velocity for time interval t : [2, 6] sec. (b) Instantaneous velocity at t = 3 sec. (c) Average acceleration for time interval t : [0, 6] sec. (d) Instantaneous acceleration at t = 4 sec.

Solution :

(a) V_avg. = Total Displacement Time interval =

(

6

) (

2

)

6 2 x t= −x t= − =

(

) (

)

2 2 3 6 9 6 6 3 2 9 2 6 4 × + × + − × + × + = 33 m/sec. (b) Instantaneous velocity = dx dt = d dt(3t 2 _{+ 9t + 6) = (6t + 9) m/sec} ∴ V at t = 3 sec = 6 × 3 + 9 = 27 m/sec.

(c) a_avg. = Change in Velocity Time interval = V( 6) V( 0) 6 t= − t= = (6 6 9) (6 0 9) 6 × + − × + = 6 m/sec2 (d) Instantaneous acceleration = dv dt = d dt(6t + 9) = 6 m/sec 2_. Illustration 6:

A car starts from rest at a constant acceleration of 8 m/s2_{. Find}

(a) How fast is it going at t = 10 sec (b) How far has it gone till t = 10 sec (c) Its average velocity for the interval t : [0, 8] sec.

Solution :

Using Equation of motion (Q acceleration = const.) (a) v = u + at

= 0 + 8 × 10 = 80 m/sec ∴ v(t = 10 sec) = 80 m/sec (b) s = ut + 1 2at 2 = 0 × 10 + 1 2 × 8 × 10 2 _{= 400 m} (c) V_avg = Displacement Time Interval = ( 8 sec) ( 0) 8 sec x t= −x t= = 2 1 0 8 8 8 0 2 8 × + × × − = 32 m/sec Illustration 7:

A particle moves in a straight line with constant acceleration. If it covers 10 m in the first second and 20 m in the next second, find its initial velocity and acceleration.

Solution :

Applying s = ut + 1

2at

2 _{for 1}st _{sec we get}

10 = v₀ × 1 + 1 2a × 1 2 _{⇒ 10 = v} 0 + 9 2 ––––––– (1) Applying s = ut + 1 2at

2 _{for 1}st _{two seconds we get}

30 = v₀ × 2 + 1 2a × 2

2 _{⇒ 30 = 2v}

0 + 2a ––––––– (2)

Solving we get v₀ = 5 m/sec and a = 10 m/s2_.

Illustration 8:

A particle with an initial velocity of 10 m/sec moves along a straight line with a constant acceleration. When the velocity of the particle is 50 m/sec, the direction of acceleration is reversed. Find the velocity of the particle when it reaches the starting points again.

Solution :

Let the magnitude of acceleration be a₀. For AB : v2 _{= u}2 _{+ 2as} (50)2 _{= (10)}2 _{+ 2a} 0 × (AB) ∴ AB = 0 1200 a A A B C vi = 10 m/sec vf = 50 m/sec v = 0 a0 a0 a0 For BC : v2 _{= u}2 _{+ 2as} 02 _{= (50)}2 _{+ 2(– a} 0)BC ∴ BC = 0 1250 a Now, CA = AB + BC = 0 2450 a ∴ v_f2 _{= 0}2 _{+ 2 × a} 0 × CA = 2 × a0 × 0 2450 a = 4900

∴ v_f = 70 m/sec. (In opposite direction to initial velocity)

Illustration 9:

A bus starts from a bus stop with a constant acceleration of 0.4 m/s2_{. A passenger arrives at the bus stop 6 sec after the} bus left. Then he immediately starts running towards the bus with a constant speed 5 m/sec. At what distance from the starting point will he catch the bus.

Solution : 5 m/sec

t = 6 sec t = T

0.4 m/s2

Let us assume that the bus starts at t = 0 from x = 0 and the man catches the bus at t = T.

Bus travelled for time interval = T and man ran for time interval (T – 6). Finally both of them reached at x = x_f . ∴ For Bus x_f = 0 × T + 1 2 × 0.4 T 2 _{–––––––– (1)} For Man x_f = 5(T – 6) –––––––– (2) T H E O R Y

Equating (1) and (2) we get 0.2 T2 _{= 5(T – 6)} On solving we get T = 10 sec and T = 15 sec So, the man will catch the bus at t = 10 sec. Distance travelled = 5(10 – 6) = 20 m.

Illustration 10:

A car moving rectilinearly with constant acceleration is having initial velocity v₀. After some time its velocity becomes

7v₀. Find out velocity of the particle at the midpoint of its path.

Solution :

From the figure

D D v0 v = ?? 7v0

( )

2 0 7v = v₀2 _{+ 2 × a} 0 × 2D or 3 2v0 2 _{= a} 0D –––––––– (1)

Let velocity at midpoints be v. Then

v2 _{= v} 0 2 _{+ 2 × a} 0 × D –––––––– (2) Putting (1) in (2) we get v2 _{= v} 0 2 _{+ 2} 2 0 3 2v       = 4v0 2 ∴ v at midpoint = 2v₀

3.1.4 Motion Under Gravity

When a particle moves under the influence of gravitational pull of Earth alone then its acceleration is g = 9.8 m/s2 _{(approx = 10} m/s2_{) downwards.}

Illustration 11:

A stone is thrown vertical upwards from ground level with u = 20 m/s. (a) Find the maximum height attained by the stone

(b) time interval t after which it returns to the point of projection (c) The velocity with which it strikes the ground (Take g = 10 m/s2₎

Solution :

20 m/sec Hmax

v = 0

Let us choose our origin O at the point of projection with +ve x-axis pointing in the vertical upwards direction.

Note that in this coordinate system, acceleration due to gravity is negative because it points in the downward direction.

Thus a = – 10 m/s2_{, u = 20 m/s}

(a) At the highest point, velocity of the particle will become zero. Let h be the maximum height. Thus S = h.

v2 _{– u}2 _{= 2a S}

we get0 – 202 _{= 2 ×(– 10) × h} ⇒ _{h =} 400

20 = 20 m

(b) Let time of flight be T. As the particle returns to the same position i.e. S = 0 after time interval t = T from the time of projection, using S = ut + 1 2at 2 _{we have} 2 1 0 20 10 2 t t = − × ⇒ t = 40 10 = 4 sec

(c) To find the velocity when it returns, we use v = u + at

⇒ v = 20 – 10 × 4

= – 20 m/s

Here, minus sign indicates that particle moves in the downward direction.

Note : It returns with same speed with which it was thrown.

In general, if a particle is thrown upwards with velocity v₀, then

Time of flight = 2v0

g

Max Height reached = 2 0 2

v g

Speed just before hitting ground = v₀ (downwards)

Illustration 12:

If another stone was thrown from the same point after one second in the last example, with initial velocity of 25 m/s , find (a) The coordinates of the point where they collide

(b) The velocities of the two stones when they collide. (Take g = 10 m/s2₎

Solution :

(a) Let x₁ and x₂ to be the coordinates of first and second stone after time t.

S₁ = x₁ – 0 = 20t – 1 210t 2 _{—————— (1)} S₂ = x₂ – 0 = 25(t – 1) – 1 210(t –1) 2 _{—————— (2)}

Please note that if the first stone is in flight for t second then second stone will be in flight for (t – 1) seconds as it was thrown after 1 sec. When they collide, they are at the same place, hence

x₁ = x₂

⇒ 20t – 5t2 _{= 25(t – 1) – 5(t – 1)}2 ⇒ t = 2 sec

Substituting t = 2 in eqn. (1) or eqn (2) , we find

h = x₁ = x₂ = 20 m

(b) The velocities of the stones after time t are

v₁ = 20 – 10t —————— (3)

v₂ = 25 – 10(t – 1) —————— (4)

Substituting t = 2 in eqns (3) and (4), we find

v₁ = 0 m/sec (i.e. stone is at its highest point.) v₂ = 15 m/sec

T H E O R Y

Illustration 13:

A body is thrown down from the top of a tower of height h with velocity 10 m/s. Simultaneously, another body is projected upward from bottom. They meet at a height 2h/3 from the ground level. If h = 60 m, find the initial velocity of the lower body. Solution : h v = 0 B v0 A A B 2 3 h

Let us choose the origin at the foot of the tower and y-axis pointing upwards. Let us refer upper and lower ball as A and B respectively

For Ball A : y_initial = h, y_final = 2

3 h u_initial = – 10 m/sec, a = – 10 m/s2 Applying y_f = y_i + ut + 1 2at 2 _{we have} 2 3 h = h – 10t – 1 2 × 10t 2 _{————— (1)}

For Ball B : y_initial = 0, y_final = 2

3 h , u_initial = v₀, a = – 10m/s2 Applying y = y_i + ut + 1 2at 2 _{we have} 2 3 h = 0 + v₀t – 1 2 × 10t 2 _{————— (2)}

Subtrating (1) from (2) we have 0 = – h + (v₀ + 10)t ⇒ t = 0 10

h v +

Putting this value in eq. (1) we have

0 = 2 0 0 10 3 ( 10) 10 h h h v v   × − −  +  +  2 0 0 2 4 10 10 h h v v     + −     + +     = 0

On solving we have v₀ = 15 5+ 5 = 38.5 m/sec (∴ h = 60 m)

Illustration 14:

A rocket is fired vertically and ascends with constant vertical acceleration of 19.6 m/s2 _{for 30 seconds. Its fuel is then all} used up and it continues as a free particle.

(a) What is the maximum altitude reached? (b) What is total time after which it strikes the ground again?

Solution :

Let us choose the vertical upward to be the positive direction. Let A be the point at which fuel is exhausted and let point B represent the maximum altitude.

From O → AA O A a = 19.6 m/s2 a = 9.8 m/s2 v = 0 B u = 0, a = 19.6 m/s2_{, t = 30s} OA = ut + 1 2at 2 _{= 0 + 9.8 × 900 = 8820 m} V_A = u + at = 0 + 19.6 × 30 = 588 m/s Form A →B u = V_A = 588 m/s, a = – 9.8 m/s2 v = V_B = 0

using v2 _{– u}2 _{= 2as we get} 0 – (588)2 _{= 2 × – 9.8 × AB} ⇒ AB = 17640 m

maximum altitude = OA + AB = 26.46 km

To find time, let us consider the path A →B →O

a = – 9.8 m/s2_{, s = – OA = – 8820, u = 588 m/s} s = ut + 1 2at 2 ⇒ – 8820 = 588 × t – 4.9t2 ⇒ t2 _{– 120t – 1800= 0} t = 120 14400 7200 2 + + = 133.5 sec So, the total time = 30 + 133.5 = 163.5 sec

Illustration 15:

A balloon is rising with constant acceleration 2m/sec2_{. Two stones are released from the balloon at the interval of 2 sec.} Find out the distance between the two stones 1 sec after the release of second stone. (g = 10 m/s2₎

Solution :

Acceleration of balloon = 2 m/sec2_{. Let v}

0 be the velocity of the balloon at the time of release of 1st stone and y0 be its attitude w.r.t ground.

After 2 seconds, balloon’s new velocity is

v = v₀ + 2 × 2 = (v₀ + 4) m/sec upwards v = + 4v0 v = v0 a = g a = g 2nd stone released 1st stone released v0 + 4 v0 y0 Its new altitude is y = y₀ + v₀ × 2 + 1

2 × 2 × 2 2 = y₀ + 2v₀ + 4

Altitude of 1st _{stone 3 seconds after its release}

y₁= y₀ + v₀ × 3 + 1

2(– 10) × 3 2 = y₀ + 3v₀ – 45

Altitude of 2nd _{stone 1 second after its release}

y₂= (y₀ + 2v₀ + 4) + (v₀ + 4) × 1 + 1

2(– 10) × 1 2 = y₀ + 3v₀ + 3

Distance between the stones = |y₂ – y₁|

= |(y₀ + 3v₀ + 3) – (y₀ + 3v₀ – 45)| = 48 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.

In-Chapter Exercise - 2

1. A ball is dropped from a height. If it takes 0.200s to cross the last 6.00 m before hitting the ground, the height from which it was dropped is ... (Take g = 10 m/s2₎

2. A parachutist drops freely from an aeroplane for 10 s before the parachute opens out. Then he descends with a net retardation of 2.5 m/s2_{. If he bails out the plane at a height of 2495 m and g = 10 m/s}2_{, his velocity on reaching the ground} will be v = ... . T H E O R Y

3. A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s2 _{due west. The distance covered by} the particle in the fifth second of its motion is ... .

4. In a car race, car A takes a time of t sec. Less than car B at the finish and presses the finishing point with a velocity v more than the car B. Assuming that the cars start from rest and travel with constant accelerations a₁ and a₂ respectively. Which of the following relation is correct :

(a) v = a a1 2 t (b) v = 2 a a1 2 t (c) v =

1 2 2

a a

t (d) None of these

5. A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time to the driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s2_{, the distance travelled by the car after he sees the} need to put the brakes on is :

(a) 20 m (b) 22 m (c) 24 m (d) 28 m

6. A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the 3 m length of a window some distance from the top of the building. If the speed of the ball at the top and at the bottom of the window are v_T and v_B respectively, then (g = 9.8 m/sec2₎ (a) v_T + v_B = 12 ms– 1 _{(b) v} T – vB = 4.9 ms – 1 _{(c) v} BvT = 1 ms – 1 _(d) B T v v = 1 ms – 1

7. A particle moving along a straight line with constant acceleration is having initial and final velocity as 5 m/s and 15 m/s respectively in time interval of 5 s. Find the distance travelled by the particle and the acceleration of the particle. If the particle continues with same acceleration, find the distance covered by the particle in the 8th second of its motion. 8. An object is thrown vertically upward. It has a speed of 10 m/s when it has reached one half its maximum height.

(a) How high does it rise?

(b) What is its velocity and acceleration 1 sec after it is thrown? (c) What is its velocity and acceleration 3 s after it is thrown? (d) What is the average velocity during the first half second?

9. A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s. 4 s later, a boy throws a stone vertically upwards from the top of the shaft with a speed of 30 m/s. Find when and where stone hits the lift.

(Take: g = 10 m/s2₎ 10. (a) Can an object reverse the direction of its motion even though it has const. acceleration?

(b) Can body have (i) Zero instantaneous velocity and yet be accelerating (ii) Zero avg. speed but non-zero velocity

(iii) Negative acceleration and yet be speeding up

3.1.5 Graphs

For better understanding and visualization of motion of a particle we use various graphs like 1. x – t graph (position – time graph)

2. v – t graph (velocity – time graph)

3. a – t graph (acceleration – time graph)

Before proceeding for graphs let us first study various terms that will be used frequently while dealing with graphs. 1. Slope of a Line : If a line makes angle θ with the positive direction of x-axis, we define its slope = tan θ .

(a) If θ is acute, slope is positive. (b) If θ is obtuse, slope is negative.

(c) If θ is zero or line is horizontal then slope is zero.

x Line with positive slope

x

2. Slope at a point on the curve : Slope at a point on the curve is slope of the tangent on the curve at that point. x Curve (1) y ( , )x y0 0 x Curve (2) y ( , )x y0 0

Slope at (x₀, y₀) on the curve (1) is tan θ i.e., Slope at (x₀, y₀) on the curve (2) is tan θ i.e., slope of tangent at (x₀, y₀) (+ve in this case) slope of tangent at (x₀, y₀) (–ve in this case)

Parabola

Various types of parabola :

1. y = ax2 _{(a > 0) 2. y = ax}2 _{(a < 0) 3. y = ax}2 _{+ b (a > 0, b > 0)} Mouth opening upwards y x Mouth opening downwards y x Mouth opening upwards y x (0, )b 4. y = ax2 _{+ b (a < 0, b > 0) 5. y = ax}2 _{+ bx + c (a > 0)} Mouth opening downwards y x (0, )b

Mouth opening upwards

y

x

Graph for Various Cases

1. Motion with uniform velocity : Consider a particle moving along x-axis with uniform velocity u from the point x = x_i at t = 0. Then x(t) = x_i + ut ; v(t) = u ; a(t) = 0.

x – t graph : Straight line with slope = u passing through x_i at t = 0

v – t graph : Horizontal line with slope = 0 a – t graph : a = 0 (Q _{velocity is constant)}

Slope = tan = = velocity u xi u is positive t x Slope = tan = = velocity u xi u is negative t x T H E O R Y

u t v u > 0 u t v u < 0 t a a = 0

Uniformly Accelerated Motion

Case 1: (acceleration > 0) xi t x Initial velocity v i= 0 xi t x Initial velocity v >i 0 _x i t x Initial velocity v < t i 0 at = 0

Slope of x – t graph = 0 at t = 0 Slope of x – t graph is v_i > 0 at t = 0 Slope of x – t graph is v_i < 0 at t = 0

vi t Slope = a0 v vi t v Slope = a0 vi t v Slope = a0 t a0 a a0 t a a0 a t

⇒ For constant acceleration (a₀ > 0), x – t graph is a parabola with mouth opening upwards.

⇒ v – t graph is a straight line with slope equal to acceleration.

xi t x Initial velocity v i= 0 xi t x _{Initial velocity} v >i 0 xi t x Initial velocity v <i 0

Slope of x – t graph at t = 0 is v_i = 0 Slope of x – t graph at t = 0 is v_i > 0 Slope of x – t graph at t = 0 is v_i < 0

vi t v Slope = a0 vi t v Slope = a0 vi t v Slope = a0 a0 t a negative acc. a0 t a negative acc. a0 t a negative acc.

⇒ For constant acceleration (a₀ < 0), x – t graph is a parabola with mouth opening downwards.

⇒ v – t graph is a straight line with negative slope equal to acceleration (a₀).

Key points

• The slope of tangent at any point on x – t graph gives velocity at that point and the slope of tangent in v –

t graph is the acceleration.

• Area under a – t graph gives change in velocity.

• Area under v – t graph and the time axis gives the distance travelled by the particle if we take all areas as positive and displacement if areas below t-axis are taken negative.

Illustration 16: a (m/s )2 3 6 2 – 2 t(sec) 0

Acceleration - time graph of a particle moving along x-axis is shown in the figure. Draw its x – t graph and v – t graph. Initially particle is at x = 4 and moving with speed 8 m/sec towards positive x-axis. T H E O R Y

Solution :

For t : [0, 3], a = – 2 m/s2_.

∴ v – t graph is a straight line with negative slope and x – t graph is a parabola

(mouth opening downwards).

x – t graph is a parabola (mouth opening downwards). v(3) = v(0) + a × t = 8 + (– 2) × 3 = 2 m/sec x(3) = x(0) + v(0) × t + 1 2 × a × t 2 _{= 4 + 8 × 3 +} 1 2 × – 2 × 3 2 _{= 19 m.} For t : [3, 6], a = 2 m/s2_.

∴ v – t graph is a straight line with positive slope and x – t graph is a parabola (month opening upwards) v(6) = v(3) + a × t = 2 + 2 × 3 = 8 m/sec x(6) = x(3) + v(3) × t + 1 2 × a × t 2 _{= 19 + 2 × 3 +} 1 2× 2 × 3 2 _{= 34 m.} Also slope of x(t) graph at t = 0 is 8 m/sec and at t = 3 it is 2 m/sec.

4 10 19 30 34 0 3 6 x m( ) t(sec) 2 4 6 8 0 3 6 v(m/sec) t(sec) Illustration 17:

For the given velocity-time graph, select the correct alternatives.

t

0

v

t0 2t0 3t0

(a) Particle turns back exactly once during its motion. (b) Acceleration is always –ve.

(c) Average acceleration for [0, t₀] is +ve. (d) Average velocity for [0, t₀] is +ve.

(e) Average speeds for [0, t₀] and [t₀, 2t₀] are equal.

Solution :

(a) Correct : For t < t₀, v > 0, at t = t₀, v = 0 and for t > t₀, v < 0 ∴ Particle’s velocity changes sign at t = t₀.

Before t = t₀, particle is moving along the +ve x-direction and after t = t₀ it is moving along –ve x-direction.

∴ Particle turns back at t = t₀.

(b) Correct : Velocity is always decreasing, therefore, a < 0.

Alternatively : Slope of v – t graph is always –ve, therefore acceleration is always –ve. (c) Incorrect : Slope of ‘v – t’ is constant

∴ a = constant < a > = a

(d) Correct : For [0, t₀] : Displacement is +ve. ⇒ average velocity is +ve.

(e) Correct : For [0, t₀] and [t₀, 2t₀] : Magnitudes of areas (i.e. distances) are equal, therefore average speeds are equal. (lengths of intervals are also equal).

Illustration 18:

A man throws a ball upwards with speed 20 m/sec from the top of a tower of height 105 m. Taking base of the tower as origin and upward direction as positive draw v – t graph and position-time graph of the ball for next 12 sec.

(Take g = 10 m/s2 _{and assume elastic collision with ground)}

T

O

W

E

R

105 m g = 10 m/s2 20 m/sec Solution :

For upward motion :

v = u + a × t 0 = 20 – 10 × t ⇒ at t = 2 sec v = 0 y_max = y₀ + ut + 1 2at 2 _{= 105 + 20 × 2 +} 1 2(– 10) × 2 2 _{= 125 m.}

For downward motion :

125 = 1

2 × 10 × t

2 _{⇒ t = 5 sec}

∴ Ball will reach ground at t = 7 sec.

125 0 2 4 6 7 8 10 12 y t 105 v = u + at = 0 + 10 × 5 = 50 m/sec 0 – 10 10 20 30 40 50 – 20 – 30 – 40 – 50 1 2 3 4 5 6 8 9 10 11 12 v(m/sec) t(sec) 7

∴ Velocity of the ball just before hitting ground is 50 m/sec downwards. or (v = – 50 m/sec)

After hitting ground ball will rebound with same speed upwards or

v = 50 m/sec. It will reach the top most point after 5 more sec or at t

= 12 sec.

Illustration 19:

Plot x – t and a – t graphs for given v – t graph. It is given that initially particle is at x = 10.

Solution : For t : [0, 2] 0 1 2 3 4 5 t 1 2 (sec) v(m/sec)

v – t graph is straight line with slope = 2

2 = 1 m/s 2_. (constant acceleration)

x – t graph is a parabola (mouth opening upwards)

T H E O R Y

x(2) = x(0) + ut + 1 2at 2 _{= 10 + 0 × 2 +} 1 2 × 1 × 2 2 _{= 12m.} For t : [2, 5]

Velocity remains constant = 2 m/sec ∴ acceleration = 0

x – t graph is a straight line. x(5) = x(2) + ut + 1 2at 2 _{= 12 + 2 × 3 +} 1 2 × 0 × 3 2 _{= 18 m.} 0 1 2 3 4 5 a(m/s )2 t 1 2 (sec) 0 1 2 3 4 5 x t 5 10 12 15 18 20 (sec) Illustration 20:

A car starts moving rectilinearly, first with acceleration a = 5 m/s2 _{(initial velocity = 0), then uniformly, and finally,} decelerating at the same rate ‘a’ comes to a stop. The total time of motion is equal to 25 sec. The average velocity during that time is equal to <v> = 72 km/hr. How long does the car move uniformly?

Solution :

Let the particle accelerate for t₀ sec with acceleration = 5 m/s2_. Its velocity after t₀ sec is v = 0 + 5t₀ = 5t₀.

Then it will move with constant velocity for some time and finally retard for t' sec to come to rest. While decelerating, 0 = v – 5t' ⇒ 5t₀ – 5t' = 0

⇒ t' = t₀.

∴ It will retard for t₀ sec and hence will move with constant speed for (25 – 2t₀) sec. Average velocity = 72 × 5 18 = 20 m/sec v 5t0 t t0 25 – 2t0 t0 20 m/sec = Displacement Total time = Displacement 25 sec ∴ Displacement = 500 m.

Now Displacement = Area under v – t graph

∴ 500 = 2 0 0 1 5 2 t t   × ×     + 5t0 × (25 – 2t0) t₀2 _{– 25t} 0 + 100 = 0

⇒ t₀ = 5 sec (t₀ = 20 sec is not possible)

In-Chapter Exercise - 3

1. The figure below shows displacement-time graph of a particle. Find the time interval (starting from t = 0) during motion such that the average velocity of the particle during that period is zero.

0 5 10 15 20 25 x(m) t 10 20 30 (sec)

2. The displacement of a particle as a function of time is shown in figure.

The figure indicates.

20 10 0 10 20 30 40 D is p la ce m e n t

(a) the particle starts with a certain velocity, but the motion is retarded and finally the particle stops

(b) the velocity of the particle is constant throughout

(c) the particle starts with a zero velocity, the motion is accelerated. (d) the acceleration of the particle is constant throughout.

3. For the given position-time (x – t) graph find the interval in which

O t1 t2 t3 t4 t5 t6

t x

(a) velocity is zero

(b) both velocity and acceleration are negative (c) velocity is positive but acceleration is negative (d) velocity is negative but acceleration is positive.

4. The following shows the velocity-time graph for a moving object. The maximum acceleration will be :

(a) 1 m/sec2 v (m /s e c ) t in sec. 40 60 20 0 10 20 30 40 50 60 70 D C B A (b) 2 m/sec2 (c) 3 m/sec2 (d) 4 m/sec2

5. The following figure shows the linear motion velocity-time graph of a body. The body will be displaced in 5 seconds by:

(a) 2 m v( m /s ec ) B 2 1 0 – 1 – 2 1 2 3 4 5 D t in sec. C A E (b) 3 m (c) 4 m (d) 5 m T H E O R Y

6. Acceleration-time graph of a body is shown. The corresponding velocity-time graph of the same body is : a t v t v t v t v t (a) (b) (c) (d)

7. A parachutist jumps out of an aircraft, falls freely for some time and then opens his parachute. Identify the graph which correctly represents his acceleration (a) versus time (t) graph :

t a 0 t a 0 t a 0 t a 0 (a) (b) (c) (d)

3.1.6 Motion with Non-Uniform Acceleration

Use of definite integral

To find displacement when the particle is moving with a variable velocity in an interval ∆t = t₂ – t₁, we can divide this interval into n subintervals.

i.e., ∆t = t₂ – t₁ = ∆t₁ + ∆t₂ + ∆t₃ + ... + ∆t_n.

As the number of intervals is increased (i.e., n →∝), each interval becomes infinitely small (∆t_i→ 0, i = 1, 2, 3 ...) and we can treat velocity as constant during one such interval.

∴ ∆x = (v₁∆t₁ + ∆v₂∆t₂ + ... + v_n∆t_n) as n →∝ and ∆t_i→ 0 ∴ ∆x = 1 lim n i i n i v t →∞ ₌   ∆   

∑

 =

( )

2 1 t t v t dt

∫

∴ Displacement in time interval [∆t = t₂ – t₁] = ∆x =

( )

2 1 t t v t dt

∫

Similarly change in velocity ∆v = v_f – v_i =

( )

f i t t a t dt

∫

Technique for problems involving non-uniform acceleration

Case 1: Acceleration depends on time. Steps to find v(t) from a(t)

We have a = dv

dt .

If a is a function of time, then dv = a(t)dt

Integrating both sides with appropriate limits we have

( ) ( ) 0 v t v dv

∫

=

( )

0 t a t dt

∫

[v(0) = initial velocity at t = 0] On integrating we get an expression for v(t).

Steps to find x(t) from v(t)

To get x(t), we put v(t) = dx

dt ⇒ dx = v(t)dt

Integrating on both sides with appropriate limits we have

( ) ( ) 0 x t x dx

∫

=

( )

0 t v t dt

∫

(x(0) = position at t = 0)

On integrating we get an expression for x(t).

Case 2: Acceleration depends on position x. Steps to find v(x) from a(x).

We have a = dv dt = dv dt dx dx = dx dt dv dx = v dv dx ⇒ a dx = v dv

On integrating both sides with appropriate units we have 0 x x a dx

∫

= 0 v v v dv

∫

[x₀ = position when velocity is v₀] On integrating we get an expression for v(x).

We have v(x) = dx

dt ⇒

( )

dx v x = dt

Integrating on both sides with appropriate units we have

_{( )}

( ) ( ) 0 x t x dx v x

∫

= 0 t dt

∫

On integrating we get an expression for v(x).

Case 3: Acceleration depends on velocity. Steps to find v(t) from a(v)

We have a(v) = dv

dt

( )

dv a v = dt

Integrating on both sides with appropriate limits we have

( )

( ) ( ) 0 v t v dv a v

∫

= 0 t dt

∫

.

On integrating we get an expression for v(t).

Steps to find v(x) from a(v)

a(v) = dv dt = dv dt dx dx = dx dt dv dx = v dv dx or a(v) = vdv dx

( )

vdv a v = dx

Integrating both sides with appropriate limits we have

_{( )}

0 v v vdv a v

∫

= 0 x x dx

∫

[v₀ = velocity when x = x₀] On integrating we get an expression for v(x).

T H E O R Y

Illustration 21:

Acceleration of a particle moving on x-axis depends on time as a = 2t m/sec2_{. Find velocity and position of particle as a} function of time if the particle is at x = 5 m at t = 0 moving with v = 2 m/sec.

Solution :

Acceleration depends on time [case (a)]

a = 2t = dv

dt ∴ dv = 2t dt

Integrating both sides with appropriate units we have 2 v dv

∫

= 0 2 t tdt

∫

⇒

[ ]

2 v v = 2 0 t t     ⇒ v – 2 = t2 ⇒ v = (t2 _{+ 2) m/sec} Now, v = t2 _{+ 2 ⇒ v =} dx dt = t 2 _{+ 2} or dx = (t2 _{+ 2)dt On integrating we have} 5 x dx

∫

=

(

2

)

0 2 t t + dt

∫

⇒

[ ]

5 x x = 3 0 2 3 t t t   +     ⇒ x – 5 = 3 3 t + 2t ⇒ x = 3 3 t + 2t + 5 Illustration 22:

Acceleration of particle depends on its positive as a(x) = 2 1

4

x + m/s

2_{. Find its velocity as a function of position. If the} particle starts from rest at x = 2m.

Solution :

Acceleration depends on position [case (b)]

a(x) = dv dt = 2 1 4 x + ⇒ dx dt dv dx = 2 1 4 x + ⇒ v dv = 2 4 dx x +

On integrating with appropriate units we have

0 v v dv

∫

= 2 2 4 x dx x +

∫

⇒ 2 0 2 v v       = 1 2 tan 2 x x −       _{ }   ⇒ 2 2 0 2 2 v ₋ = tan 1 tan 1 2 2 2 x −  ₋ −           ⇒ 2 2 v = tan 1 2 4 x −  ₋π     ⇒ v = 2 tan1 2 2 x −  ₋π     Illustration 23:

A particle is given an initial speed v₀. There is a resisting acceleration k v , where v is instantaneous velocity and k is some positive constant. Find the time taken to stop and the maximum distance covered by the particle.

Solution :

We have a = dv

dt . As there is resisting acceleration, dv dt is negative. i.e., dv dt = – k v ⇒ dv v = – k dt On integrating with appropriate limits we have

0 0 v dv v

∫

= 0 t k dt −

_∫

⇒ 0 0 2 v v     =

[ ]

0 t k t − ⇒ 0 – 2 v0 = – kt or t = 2 v0 k (b) Stopping distance a = dv dt dx dx = dv v dx i.e., vdv dx = −k v ⇒ v dv v = – k dx ⇒ vdv = – k dx

On integrating with appropriate limits we have

0 0 v v dv

∫

= 0 x k dx −

_∫

⇒ 32 0 2 3v = kx ⇒ x = 3 2 0 2 3 v k Illustration 24:

For a particle moving in straight line with constant acceleration if t = x+ 3, find the position of the particle when its velocity is zero.

Solution :

t = x + 3 ⇒ x = (t – 3)

or x = t2 _{– 6t + 9} Now, velocity of the particle is v = dx

dt = (2t – 6) m/sec. Velocity is zero when 2t – 6 = 0 or t = 3 sec.

∴ Position of the particle when its velocity is zero is x(3) = 32 _{– 6 × 3 + 9 = 0} i.e., Particle is at origin.

Illustration 25:

Instantaneous velocity of a particle moving in +ve x direction is given as v = 2 3

2

x + . At t = 0, particle starts from origin.

Find the average velocity of the particle between the two points P(x = 2) and Q(x = 4) of its path.

Solution :

v

< > = average velocity = Displacement_{Time Interval} =

4 2 Time Interval

−

So, basically we have to find the time interval between the instants when particle is at position Q(x = 4) and position P(x = 2). Now, v = dx dt = 2 3 2 x + (x 2 _{+ 2)dx = 3dt}

On integrating with appropriate limits we have

T H E O R Y

(

)

4 2 2 2 x + dx

∫

= 2 1 3 t t dt

∫

or 4 3 2 2 3 x x   +     =

[ ]

2 1 3 t t t 3 4 2 4 3   + ×     – 3 2 2 2 3   + ×     = 3(t2 – t1) t₂ – t₁ = ∆t = 68 9 sec. ∴ < >v = average velocity = 2 t ∆ = 2 68 9       = 9 34 m/sec.

In-Chapter Exercise - 4

1. For a particle moving along x-axis, acceleration is given as a = 2x3_{. If the speed of the particle is 4 m/sec at x = 0, find speed} as a function of x.

2. For a particle moving along x-axis acceleration is given as a = 2v. Find its position as a function of time if at t = 0, it is at x = 0 moving with velocity 5 m/sec.

3. An object moves along the x-axis such that is acceleration is given as a = 3 – 2t. Find the initial speed of the object such that the particle will have the same x-coordinate at t = 5.0 s as it had at t = 0. Also find the object’s velocity at t = 5.0 s. 4. Velocity of a particle moving along x-axis is given as v = x2 _{– 5x + 4 (in m/s) where x denotes the position of the particle in}

meters. The magnitude of acceleration of the particle when the velocity of the particle is zero is :

(a) 0 m/s2 _{(b) 2 m/s}2 _{(c) 3 m/s}2 _{(d) None of these.}

5. A particle moves according to the equation t = ax2 _{+ bx, then the acceleration of particle when x =} b a is (a) 3 a b − (b) 3 2 8 a b − (c) 3 2 27 a b − (d) None of these.

Section - 3.2

Motion in Two and Three Dimension

Consider a particle which move along the curve in the plane of the paper. To locate its position we must specify the origin (the reference point). Here, we have choosen O as the origin.From point O, we draw two perpendicular lines. One of the lines is called the x-axis and the other is called the y-axis. If particle is at point P at time t, we have two ways of specifying its position.

1. Rectangular coordinate ( , )x y P O M N x y r θ

From point P we draw two perpendicular segment, PM and PN which are perpendicular to x and y axis respectively. OM and ON are called the x and y coordinates respectively.

2. Polar coordinates (r, θ)

Another way to locate the position P uniquely is to specify, the distance of P from O and angle θ which OP makes with the x-axis.

It is clear from the figure that polar and rectangular coordinates are related by the equations

x = r cos θ y = r sin θ

or, r = _x2+_y2

Position vector rr ( ) OP

r t =

uuur r

is called the position vector of the particle. From the construction in figure, it is clear that OP=OM+MP

uuur uuuur uuur

. If ˆi and ˆj are unit vector (of magnitude unity) along x & y direction then OM=xiˆ

uuuur

and MP=y jˆ

uuur

.

⇒ _{r t}r_{( )=OP}uuur_{=x t i( )}ˆ_{+y t j( )}ˆ The distance of the particle from the origin is

2 2

( ) ( )

x t +y t = rr Displacement

As the particle moves , its position vector changes in such a way that it always extends from the origin to the particle. If particle is at position r1

r

at time t and at 1 r2 r

at time t , then its displacement during the time interval is2

2 1

r r r

∆ = −r r r ———— (1)

For example, if the particle is at _rr_{1= +2i}ˆ ₃ˆ_j at time t = 0 at r₂ =4i +5j

r r

r

at t = 1 sec, then its displacement is _{∆ = − = +r}r _rr_{2 r}r_{1 2i}ˆ ₃ˆ_j. The magnitude of displacement is 22+32 = 13units.

Velocity

If a particle undergoes a displacement r∆rin time t∆, then average velocity is

ˆ ˆ r x y v i j t t t ∆ ∆ ∆ < >= = + ∆ ∆ ∆ r r ———— (2)

The instantaneous velocity vr is the value <vr> approaches in the limit as we approach ∆ →t 0.

0 0 lim lim t t r dr v v t dt ∆ → ∆ → ∆ = < > = = ∆ r r r r

It can be also written as the derivative

(

ˆ ˆ

)

ˆ ˆ d dx dy v xi y j i j dt dt dt = + = + r ———— (3) which can be rewritten as vr=v i_xˆ+v j_yˆ

in which x dx v dt = and y dy v dt

= are called the x-components and y-components of the velocity.. The magnitude of vr is called the speed of the particle.

Graphical Approach P Q Tangent at P 1 r 2 r r ∆ Figure shows the path of a particle moving in the plane of the paper. Let the particle

be at P at time t and, at Q at time t+ ∆t. The position vector at P and Q is rr₁ and 2

rr respectively. The displacement of the particle PQ

uuur

is ∆ = −r r2 r1 r r r

. The average velocity <vr> is given by eqn (2). It is in the same direction as r∆r. As we approach

t

∆ towards zero, the position vector rr₂ moves towards rr₁ so that r∆r approach towards zero. The direction of r∆r approaches the direction of tangent line. The average velocity v r t ∆ < > = ∆ r r

approaches the instantaneous velocity vr. In the limit 0

t

∆ → , we have v< > =r vr and most importantly vavg r

takes the direction of tangent line.

Acceleration

If the velocity vr of a particle changes from vr₁ to vr₂in a time interval t∆ , its average acceleration < >ar during t∆ is 2 1 v v v a t t − ∆ < > = = ∆ ∆ r r r r T H E O R Y

The instantaneous acceleration ar, is the value < >ar approaches in the limit as we shrink t∆ .It can be written as the derivative.

(

ˆ ˆ

)

xˆ y ˆ ˆ ˆ x y x y dv dv d a v i v j i j a i a j dt dt dt = + = + = + r in which x x dv a dt = and y y dv a dt

= are called the x-component and y-component of acceleration respectively..

Illustration 26:

A particle moves in the x – y plane in such a way that its x(t) and y(t) coordinates are given by x(t) = – 5t2 _{+ 20 meters and}

y(t) = 2t + 3 meters. Find

(a) Average velocity in the time interval t = 0 to t = 2 sec. (b) Average velocity in the time interval t = 0 sec to t = 3 sec. (c) Instantaneous velocity at t = 2 sec and t = 3 sec. (d) Average acceletation in the time interval t = 0 to t = 3 sec. (e) Instantaneous acceleration at time t = 2 sec and t = 3 sec.

Solution :

The position vector of the particle at time t is r tr( )=x t i( )ˆ+y t j( )ˆ = ( 5− t2+20)iˆ+(2t+3)ˆj

(a) The average velocity is given by 2 1 2 1 ( ) ( ) r t r t v t t − < > = − r r r

( )

_{0 20}ˆ ₃ˆ rr = i+ j and rr

( )

2 = − × +5 22 20iˆ+ × +

(

2 2 3

)

jˆ= +0iˆ 7ˆj=7ˆj

(

)

ˆ ˆ ˆ 7 20 3 ˆ ˆ 10 2 2 0 j i j v − + i j < > = = − + − r m/s and < > =vr 104m/s

The angle θ with the x-axis is 1 1 1

θ tan tan 5 y x v v − < > −   =  = −   < >    

(b) Here, we have t1=0s and t2 =3s.

Hence, (3) (0) 3 0 r r v − < > = − r r r

(

25 20

) (

ˆ 9 3

)

ˆ 3 0 i j − − + − = − = −15iˆ+2ˆj

(c) To find the instantaneous velocity, we differentiate ( )r tr with respect to time. On differentiating, we have

ˆ ˆ 10 2 dr v t i j dt = = − + r r Hence, v tr( =2)= − × +10 2iˆ 2ˆj= −( 20iˆ+2 )ˆj ˆ ˆ ˆ ˆ ( 3) 10 3 2 ( 30 2 ) v tr = = − × +i j= − i+ j (d) 2 1 2 1 v v a t t − < > = − r r r

(

3

) (

0

)

30ˆ 2ˆ 2ˆ _ˆ 10 3 0 3 v t v t i j j a = − = − + − i < >= = = − − r r m/s2

The average acceleration in time interval t = 0 to t = 3 is 10 m/s2 _{and its direction is in the negative x-direction.} (e) To find the instantaneous acceleration, we differentiate vr with respect to time.

ˆ ˆ ˆ ( 10 2 ) 10 dv d a ti j i dt dt = = − − = − r r m/s2 _(constant)

Illustration 27:

Figure below shows a particle, moving with constant speed v₀ m/s in a circle of radius R meters.

(a) Find the instantaneous velocity at point A, B, and C.

x y B R A C D O

(b) Find the displacement , average velocity , average acceleration in the time interval when particle moves from A to B.

(c) Find the same quantities when particle moves from point A to point C.

Solution :

(a) As we know the speed (magnitude of velocity) of the particle we only need to the direction of velocity, which is tangent to the curve along which particle moves.

Thus, vr_A =v j₀ˆm/s, vr_B= −v j₀ˆm/s and vr_C = −v j₀ˆm/s

(b) The distance travelled by the particle from A to B is 2πR πR 4 = 2 meters Thus, the time it will take from A to B =

0 πR 2v sec displacement = ∆rA→B= − =rB rA Rˆj−Riˆ 2 2 A B R R 2 r→ ∆ = + = R meters 0 0 B A 0 ˆ ˆ _{2 2} R R _{ˆ ˆ} time taken πR / 2 π π v v r r j i v j i v − − < > = = = − r r r m/s 0 0 B A 0 ˆ ˆ time taken πR / 2 v i v j v v a v − − − < >= = r r r = 2 0 2 _{ˆ ˆ} ( ) πR v i j − + m/s (c) We leave it as an exercise for you. Verify that

A C 2Rˆ r→ i ∆ = − , 2 0 _ˆ π v v − i < > =r m/s and 2 0 2 _ˆ πR v a − j < >=r m/s Illustration 28:

Position vector of a point relative to origin varies with time t as _ˆ 2_ˆ,

r=ati−bt j

r

where a and b are positive constants. Find the equation of the point’s trajectory.

Solution : It is given that _ˆ 2_ˆ, r=at i−bt j r therefore x = at and y = – bt2 Eliminating t, we get y = – b 2 x a       y = 2 2 bx a −

3.2.1 Motion in Two Dimension with Constant Acceleration

We consider the special case of motion in a plane (two dimension) with constant acceleration. As the particle moves, the acceleration ardoes not vary either in magnitude or direction. Consequently, the components of acceleration a_x and a_y remain constant. We then have a situation which can be described as the sum of two component motion occurring simultaneously with constant acceleration along each of two perpendicular directions independently.

T H E O R Y

The equations v = u + at s = ut + 1 2at 2 v2 _{– u}2 _{= 2a s}

can be applied separately for x-axis and y-axis The two sets of equations are

For motion along x-axis For motion along y-axis

v_x = u_x + a_xt v_y = u_y + a_yt x – x₀ = u_xt + 1 2axt 2 _{y – y} 0 = uyt + 1 2ayt 2 v_x2 _{– u} x 2 _{= 2a} x(x – x0) vy 2 _{– u} y 2 _{= 2a} y(y – y0) Illustration 29:

For a particle moving in the x - y plane, if _ar_=(3iˆ_{−2 )}ˆ_j m/s2 _{and ˆ ˆ} (2 4 ) m

vr = i+ j m/s, then find position of particle when its

y-coordinate is maximum. Solution : We have, a_x = 3 m/s2_{, a} y = – 2 m/s 2 _{and u} x = 2m/s, uy = 4 m/s. therefore, v_y = u_y + a_yt = (4 – 2t) and y = u_yt + 1 2ayt 2 _{= 4t – t}2 Similarly v_x = u_x + a_xt = 2 + 3t and x = u_xt + 1 2axt 2 _{= 2t +} 3 2t 2 When y is maximum, dy dt = 0 ⇒ 4 – 2t = 0 ⇒ t = 2s. At t = 2s, x = (2 × 2) + 3 22 2   ×    = 10 m and y = (4 × 2) – (2)2 = 4 m.

Hence, position vector of the particle is rr= + =xiˆ yjˆ (10iˆ+4 )ˆj m

In-Chapter Exercise - 5

1. A particle travels with speed 50 m/s from the point (3, – 7) in direction ˆ7i−24ˆj. The position vector of particle after 3 seconds will be ... .

2. The position vector rr of a moving particle at time t after the start of the motion is given by rr= (5 + 20t)ˆi + (95 + 10t – 5t2_{) ˆj . At time t = T, the particle is moving at right angles to tis initial direction of motion. The value of T is ... and} the distance of the particle from its initial position at this time is ... .

3. Read the passage carefully and answer the following questions.

“In a three dimensional co-ordinate system a particle moves such that during motion at time t the position vector of a particle of mass m = 3 kg is given by rr = 6tˆi– t3ˆj + cos t ˆk ”