Kinematics
IIT–JEE Syllabus:Kinematics of one and two dimensions (Cartesian Coordinates only), Projectiles, Circular motion (Uniform and non-uniform), Relative velocity.
MOTION IN ONE DIMENSION
Displacement
The displacement of a particle is defined as the difference between its final position and its initial position. We represent the displacement as ∆ x.
∆ x = xf − xi
The subscripts i and f refer to be initial and final positions. These are not necessarily the positions from which the particle starts its motion nor where its motion ceases. The i and f designate the particular initial and final positions we are considering out of the entire motion of the object. Note the order : final position minus initial position . Whenever we calculate “delta” anywhere, we always take the final value minus the initial value.
Average Velocity and Average Speed
The average velocity of an object travelling along the x−axis is defined as the ratio of its displacement to the time taken for that displacement.
vav = i f i f t t x x Δt Δx − − = (1)
The average speed of a particle is defined as the ratio of the total distance travelled to the time taken. Average speed = t ∆ travelled distance Total
Note that velocity and speed have different meanings. Example : 1
A bird flies toward east at 10 m/s for 100 m. It then turns around and flies at 20 m/s for 15 s. Find
(a) its average speed (b) its average velocity
Solution
Let us take the x axis to point east. A sketch of the path is shown in the figure. To find the required quantities, we need the total time interval. The first part of the journey took ∆ t1 = (100 m)/ (10 m/s)
= 10 s, and we are given ∆ t2 = 15 s for the second part. Hence the total time interval is
−200 −100 0 100
x(m) Fig. 1
∆ t = ∆ t1 + ∆ t2 = 25 s
The bird flies 100 m east and then (20 m/s) (15 s) = 300 m west
(a) Average speed = 16
25 m 300 m 100 Distance = + = ∆t s m/s
(b) The net displacement is
∆ x = ∆ x1 + ∆ x2 = 100 m − 300 m = −200 m So that vav = 25 s 8 m 200 − = − = ∆ ∆ t x m/s
The negative sign means that vav is directed toward the west.
CAUTION
Sometimes students try to calculate the average velocity by just adding the two given velocities and dividing by two. This procedure is wrong, and it can be clearly illustrated with the following example. A college student drives a car 1 kilometer at 30 kmph. How fast must the student drive a second kilometer in order to average 60 kmph for the 2 minute trip.
If you believe that the average velocity is the average of the velocities, then the answer will be 90 kmph because
2 90
30 +
= 60 kmph. But the correct answer is “not possible”. There is no way the
student can average 60 kmph for the trip! Sixty kmph means 1 km/min. In order to average 60 kmph
for 2 km, the trip must be driven in 2 min. But going the first kilometer at 30 kmph takes 2 min. So the driver has no time left at all to go the second kilometer.
Example: 2
A jogger runs his first 100 m at 4 m/s and the second 100 m at 2 m/s in the same direction. What is the average velocity ?
Solution
A sketch of his motion is shown in figure. His net displacement
∆ x = ∆ x1 + ∆ x2
= 100 m + 100 m = 200 m x (m)
v1 v2
0 100 200
Fig. 2
The first half took
∆ t1 = (100 m)/(4 m/s) = 25 s, while the second took
∆ t1 = (100 m)/(2 m/s) = 50 s, The total time interval is
∆ t = ∆ t1 + ∆ t2 = 75 s Therefore, his average velocity is
vav = 75s 2.67 m 200 = = ∆ ∆ t x m/s
Since 2.67 ≠ 21 (4+2), we see that the average velocity is not, in general, equal to the average of the velocities.
Average Acceleration is defined as the ratio of the change in velocity to the time taken.
aav = Δt Δv = i f i f t t v v − − (2)
Instantaneous Velocity is defined as the value approached by the average velocity when the time
interval for measurement becomes closer and closer to zero, i.e. ∆ t → 0. Mathematically
v(t) = Δt Δx lim v lim 0 Δt av 0 Δt → = →
The instantaneous velocity function is the derivative with respect to the time of the displacement function.
v(t) =
dt dx(t)
(3)
Instantaneous Acceleration is defined analogous to the method for defining instantaneous
velocity. That is, instantaneous acceleration is the value approached by the average acceleration as the time interval for the measurement becomes closer and closer to zero.
The Instantaneous acceleration function is the derivative with respect to time of the velocity function a(t) = dt dv(t) (4) Example: 3
The position of a particle is given by
x = 40 − 5t − 5t2, where x is in metre and t is in second
(a) Find the average velocity between 1 s and 2 s (b) Find its instantaneous velocity at 2 s
(c) Find its average acceleration between 1 s and 2 s (d) Find its instantaneous acceleration at 2 s
Solution (a) At t = 1 s; xi = 30 m t = 2 s; xf = 10 m vav = 2 1 20 30 10 − = − − = − − i f i f t t x x m/s (b) v = t dt dx 10 5− − = At t = 2 s; v = −5−10(2) = −25 m/s (c) At t = 1 s; vi = −5−10 (1) = −15 m/s t = 2 s; vf = −5−10 (1) = −25 m/s vav = 2 1 10 ) 15 ( 25 − = − − − − = − − i f i f t t v v m/s2
(d) a =
dt dv
= −10 m/s2
GRAPHICAL INTERPRETATION OF DISPLACEMENT,
VELOCITY AND ACCELERATION
Average Velocity
The average velocity between two points in a given time interval can be obtained from a
displacement versus time graph by computing the slope of the straight line joining the coordinates
of the two points.
Instantaneous Velocity
The instantaneous velocity at time t is the slope of the tangent line drawn to the
position−versus−time graph at that time.
O t
x
Fig.(3) The x versus t graph of a particle whose velocity is not constant. The slope of the chord joining two points on the curve is the average velocity for that interval. The instantaneous velocity is the slope of the tangent at that point.
Chord ∆x ∆t Tangent line v = dt dx vav = t x ∆ ∆ O t x v > 0 v = 0 v < 0 v = 0
Fig.(4) The sign of the instantaneous velocity depends on the slope of the tangent.
Average Acceleration
The average acceleration between two points in a time interval is equal to the slope of the chord connecting the points on a velocity versus time graph.
Instantaneous Acceleration
The instantaneous acceleration at time t is the slope of the tangent drawn to the velocity versus
time graph. ∆t aav = t v ∆ ∆ O t v
Fig.(5) On a v-versus-t graph, the slope of the line joining two points is the average acceleration for that time interval. The instantaneous acceleration at a given time is the slope of the tangent at that time.
a = dt dv ∆v O t v
Fig.(6) The sign of the instantaneous acceleration depends on the slope of the tangent.
a = 0
a < 0 a = 0
Displacement from Velocity Time Graphs Given a velocity versus time graph, the
displacement during an interval between time ti
and tf is the area bounded by the velocity curve
and the two vertical lines t = ti and t = tf, as
shown in the Fig.(7 a).
∆t v
t vo
O ti tf
Fig.(7 a) The area under the v-versus-t curve is
the displacement ∆x = vo∆t
∆t1 v
t O
Fig.(7 b) For each segment of motion, the velocity is a
different constant. The displacement ∆x1
during the tth interval is the area v1∆t1. So the
total displacement is ∆x = v1∆t1 + v2∆t2 + v3∆t3 + v4∆t4 v1 v2 v4 v3 ∆t2 ∆t3 ∆t4 v t O ti tf
Fig.(7 c) When v versus t graph is a smooth
complex curve. the area under the curve may be obtained using integration.
Velocity from Acceleration Graphs
Given an acceleration−versus−time graph, the change in velocity between t = ti and t = tf is the area bounded by the acceleration curve and the vertical lines t = ti and t = tf.
∆t a
t ao
O ti tf
Fig.(8 a) The area under the a versus t curve is
the change in velocity ∆V = ao∆t
a
t
O ti tf
Fig.(8 b) When a versus t graph is a smooth
complex curve. the area under the curve may be obtained using integration.
Table Variation of Displacement (x), velocity (v) and acceleration (a) with respect to time for
different types of motion.
Displacement Velocity Acceleration
1. At rest O t c x x = c O t v O t a
2. Motion with constant velocity O t x x = vot + xo xo O t v vo O t a
3. Motion with constant acceleration O t x x = vot +(1/2)aot2 O t v v = aot+ vo vo O t a ao
4. Motion with constant deceleration. O t x x = vot - (1/2)aot2 O t v vo v = vo − aot O t a ao Example: 4
At t = 0 a particle is at rest at the origin. Its acceleration is 2 m/s2 for the first 2 second
and −2 m/s2 for the next 2 s. Plot the x versus t and v versus t graphs.
Solution
It is given that x = 0 and v = 0 at t = 0. The acceleration versus time graph is plotted in Fig. (9 a). The velocity at t = 2 s is equal to the sum of velocity at t = 0 and the area under the acceleration−time graph between
t = 0 and t = 2 s. a +2 t (s) −2 m/s2 O 2 4 (a) Fig.(9 a) Example (4)
3 v (m/s) t (s) 4 O 1 2 4 Fig. 9(b) 3 t (s) O 1 2 4 4 8 x (m) Fig. 9(c) v2 = v0 +
[
Area of graph]
tt 20 = = −t aSince at t = 0: velocity is zero, vo = 0 ∴ v2 = 0 + (2) (2) = 4 m/s Similarly velocity at t = 4s is v4 = v0 +
[
Area of graph]
tt 40 = = −t a = 0 +2 (+2) + 2 (−2) = 0 Note that area below the x−axis is taken as negativeThe displacement at t = 2 s is x2 = x0 +
[
area of v−tgraph]
tt==20s ∴ x2 = 0 + 21 (2) (4) = 4 m and x4 = 0 + 2 1 (4) (4) = 8 mMOTION WITH CONSTANT ACCELERATION
In this section, we are going to derive equations relating to displacement, velocity, acceleration and time.
We know that, a
dt
dv =
or dv = adt
On integrating both sides,we get
∫
dv =∫
adtv = at + c1 where c1 = constant of integration
If v0 be the initial velocity of the particle then at t = 0, v = v0. This implies that c1 = v0
v = v0 + at (5)
Also we know that
at v v dt dx = = + 0 or dx = (v0 + at)dt
On integrating, we get
∫
dx =∫
v0dt +∫
atdt x = v0t + 2 1 at2 + c 2 where c2 = constantIf x0 be the initial position of the particle, then at t = 0; x = x0. This implies that c2 = x0 x = x0 + v0t +
2 1
at2 (6)
Eliminating t from equations (5) and (6), we get
v2 = 2 ( )
0 2
0 a x x
v + − (7)
The equations (5) to (8) are called the equations of kinematics in the direction along the x−axis. The equations of kinematics are summarised as
v = v0 + at (5) x = x0 + v0t + 2 2 1 at (6) v2 = 2 ( ) 0 2 0 a x x v + − (7) x = x0 + 2 1 (v0 + v)t (8)
where x0 = Initial position coordinate x = Final position coordinate vo = Initial velocity
v = Final velocity
a = Acceleration (constant) t = Elapsed time
Problem Solving Strategy
1. Make a simple sketch of the situation described
2. Set up a co−ordinate system and clearly indicate the origin. 3. (a) List the given quantities with appropriate signs.
(b) List the unknown quantities.
4. Find the equation that has the quantity you need as the only unknown. (This is not always possible).
5. It is often helpful to obtain a rough graphical solution 6. Solve the equation (s) to find the desired unknown(s). Example 5
A car accelerates with a constant acceleration from rest to 30 m/s in 10 s. It then continues at a constant velocity. Find
(a) its acceleration
(b) how far it travels while speeding up
(c) the distance it covers while its velocity changes from 10 m/s to 20 m/s
Solution
A sketch and coordinate system are shown in figure(a). Note that x0 = 0
(a) Given
Unknown: a = ? x = ?
From equation (5) we have
a = − 0 =+3 t v v m/s2 (b) Given : v0 = 0; v = 30 m/s; t = 10 s; a = 3 m/s2 Unknown: x = ?
The position coordinate x appears as the only unknown in equations (6) and (8).
x = 2 1 at2 = 2 1 (3) (10)2 = 150 m
If we had not found the acceleration in (a), we would have to use equation (8).
x = x0 + 2 1 (v0 + v) t = 0 + 2 1 (0 + 30) (10) = 150 m (c) Given: v0 = 10 m/s; v = 20 m/s; a = 3 m/s2 Unknown : x0 = ?; x = ?; t = ? xo = 0, vo = 0 a x v Fig. 10 (a) v = 20m/s xo ∆x x vo = 10m/s Fig. 10 (b)
If we maintain the origin as shown in figure, we have to find x0 to this part of the trip.
However, we need only the difference ∆ x = x − x0, which can be found from equation (7). v2 = v2 +2a∆x
0
202 = 102 + 2(3)∆ x ∆ x = 50 m
Example 6
A particle is at x = 5 m at t =2 s and has a velocity v = 10 m/s. Its acceleration is constant at −4 m/s2. Find the initial position at t = 0
Solution
Given x = 5 m; v = 10 m/s ; a = −4 m/s2; t =2 s Unknown: x0 = ?; v0 = ?
In this case none of the equations of kinematics yields x0 immediately. The quantity x0
appears in three equations, but always with the other unknown, v0. We have to find v0 first,
From equation (5).
v = v0 +at
10 = v0 + (−4) (2)
Thus v0 = 18 m/s. Any of the other equations will give x0. From equation (8). x = x0 + 2 1 (v0 + v)t 5 = x0 + 2 1 (18 +10) (2) Thus, x0 = −23 m
0 5 10 15 20 a = -4 m/s2 t = 2 s v = 10 m/s
Fig. 11
Example 7
Car A moves at a constant velocity 15 m/s. Another car B starts from rest just as the car A passes it . The car B accelerates at 2 m/s2 until it reaches its maximum velocity of 20 m/s.
Where and when does the car A is caught by other car B ?
Solution
When two particles are involved in the same problem, we use simple subscripts to distinguish the variables, as shown in figure. The motion of the car B has two phases: one at constant acceleration and other at constant velocity. In such problems it is convenient to use ∆ t instead of t in the equations. The car B may or may not catch the car A during the
acceleration phase. This has to be checked. We set the origin at the car B, which means xoA
= xoB = 0 vA = 15m/s vA 0 A A B aB B vB x t = ∆t1 t = ∆t1 + ∆t2 Fig. 12 Acceleration Phase:
Let it takes a time interval ∆ t1
Given: vA = 15 m/s; aB =2 m/s2; v0B = 0; vB = 20 m/s Unknown: xA = ?; xB = ? vB = vOB + aBt ∴ t = 2 10 20 = = B B a v s At this time, the positions are given by x = x0 + v0t +
2 1 at2 xA = (15) (10) = 150 m xB = 2 1 (2) (10)2 100 m
The car A is still ahead
Constant velocity phase:
Let it take a time interval ∆ t2
Given: xoA = 150 m; xoB =100 m; vA = 15 m/s; vB = 20 m/s. aA = aB =0 Unknown: xA = ?; ∆ t2 = ?
The cars meet when they have same position
Here xA= 150 + 15 (∆ t2) xB = 100 + 20 (∆ t2) ∴ 150 + 15 (∆ t2) = 100 + 20 (∆ t2) or ∆ t2 = 10 s Also, xA = xB = 300 m
FREE FALL
Motion that occurs solely under the influence of gravity is called free fall.
In the absence of air resistance all falling bodies have the same acceleration due to gravity, regardless of their sizes or shapes.
The value of the acceleration due to gravity depends on both latitude and altitude. It is approximately 9.8 m/s2 near the surface of the earth. For simplicity a value of 10 m/s2 is
being used in this package.
The equations of kinematics may be modified as
v = v0 − gt (9) y = y0 + 2 1 (v0 + v) t (10) y = y0 + v0t − 2 1 gt2 (11) v2 = 2 ( ) 0 2 0 g y y v − − (12)
The signs of v and v0 are determined by
their directions relative to the chosen +y axis.
Note that the sign of the acceleration does not depend on whether the body is
going up or coming down.
ay = -g
O x
Fig.(13) If the y axis is chosen to point upward; the acceleration of a particle in free-fall is ay = -g, where g = 10m/s2 is the
magnitude of the acceleration due to gravity.
y
Example 8
A ball thrown up from the ground reaches a maximum height of 20 m. Find (a) its initial velocity
(b) the time taken to reach the highest point (c) its velocity just before hitting the ground (d) its displacement between 0.5 and 2.5 s
(e) the time at which it is 15 m above the ground.
Solution
A coordinate system is shown in figure. Note that at the highest point the ball is instantaneously at rest, that is v = 0
(a) Given: y0 = 0; y = 20 m; v = 0; a = −10 m/s2 Unknown: v0 = ? t = ?
Both equations (9), (10) and (11) involve the other unknown, namely t. So from equation (12),
O y x a = -g Fig.14 0 = v02 +2(−10)(20 −0)
Thus, 2 400
0 =
v m2/s2, and v
0 = ± 20 m/s. Since “up” is positive, v0 = + 20 m/s
(b) Given: All information in part (a) and v0 = 20 m/s Unknown : t = ?
Either Equation (9) or equation (11) would lead to t. Since equation (11) involves a quadratic it is quicker to use equation (9)
0 = 20 − 10 t Thus t = 2 s
(c) Given: y0 = 0; y = 0; vo =20 m/s Unknown : v = ?; t = ?
From equation (12) we have
v2 = 2 2(10)(0 0)
0 − −
v
Thus, v = ±v0. At the single point y = 0, the velocity has two values +v0 initially and −v0
when it lands
(d) To find the displacement, ∆ y = y2− y1, we need equation (11); y1 = 20(0.5) − 5 (0.5) 2 = 8.75 m
y2 = 20(2.5) − 5 (2.5) 2 = 18.75 m
Hence ∆ y = + 10 m
(e) Given: y = 15 m; y0 = 0; v0 = 20 m/s Unknown: t = ?
Since equations (9), (10) and (12) also contain the other unknown, we use equation (11) 15 = 0 + 20t − 5t2
The solutions of the quadratic equations are
t = 10 15 5 4 ) 20 ( 20 ± 2 − × × = 1 s, 3 s Example 9
A ball is thrown upward with an initial velocity of 10 m/s from a rooftop 40 m high. Find (a) its velocity on hitting the ground.
(b) the time of flight (c) the maximum height
(d) the time to return to roof level (e) the time it is 15 m below the rooftop.
Solution
The origin is assumed at the ground level so that all positions are positive
Given: y0 = 40 m; v0 = +10 m/s a = −10 m/s2
(a) When the ball lands, its final position coordinate is y = 0
The final velocity v appears as the only unknown in equation (12)
v2 = 102 + 2(−10) (0−40) = 900 m2/s2
or v = −30 m/s
(b) Since v is now known, we can use equation (9) O 40 m x ay = -g 10 m/s y Fig. 15 −30 = 10 − 10t
which gives t = 4 s. If v were unknown we could use equation (11) 0 = 40 +10t − 5t2
(c) At the maximum height v = 0, so from equation (12) 0 = (10)2 + 2 (−10) (y − 40)
Thus y = 45 m.
(d) At the roof level, the final position is y = 40 m, From Equation (11) 40 = 40 + 10t − 5t2
Therefore, t = 0 s, 2 s. Of course we pick t =2 s. This is just double the time needed to reach the maximum height.
(e) Again from equation (11), with y = 25 m 25 = 40 + 10t − 5t2
After solving, we get t = 3 s. Example 10
Two balls are thrown towards each other, ball A at 20 m/s upward from the ground, and two second later ball B at 10 m/s downward from a roof 30 m high.
(a) Where and when do they meet (b) What are their velocities on impact?
Solution
Remember in this kind of problem we have to find when before where. We need to write the general expression for the position coordinates. The coordinate system is shown in figure.
(a) Given : y0A = 0; v0A = +20 m/s
y0B = 30 m; v0B = −10 m/s; a = −10 m/s2.
If A has been in motion for time t, then B has been in motion for time (t − 2). From equation (11),
yA = 20t − 5t2 yB = 30 − 10(t−2) −5(t−2)2 10 m/s O 30 m x y A B 20 m/s ay = -g Fig .16
They meet when yA = yB. This condition immediately leads to t = 3 s. Substituting into
either yA or yB gives y = 15 m
(b) Since t = 3 s, we have vA = 20 + (−10) (3) = −10 m/s
and vB = −10 + (−10) (3−2) =−20 m/s. Notice that A is already moving downward when it
collides with B.
Whatever we have studied in the kinematics of one dimensional motion the same is applicable for motion in two and three dimensions.
In two dimensions the position vector r of a particle whose coordinate are (x, y) is
r = xiˆ + y jˆ
If the particle moves from P1 at position r1 to
P2 at position r2, as shown in figure its
displacement is given by ∆ r = r2− r1 = ∆ xiˆ + ∆ y jˆ P2 path followed P1 ∆r r2 r1 O x y
Fig.(17) Displacement of a particle
The student must carefully note the correct direction of ∆ r.
∆ r is the vector that must be added to
the initial position r1 to give the final
position r2,
i.e. r2 = r1 + ∆ r
The average velocity is defined as the ratio of the displacement over the time interval
vav = ∆t
∆r O x
y
Fig.(18) The instantaneous velocity v is directed
along the tangent to the path, but its magnitude is not the slope of that line.
dt dr v= t + ∆t ∆r t t r vav ∆ ∆ = or v = dt dr = vxiˆ + vy jˆ (13) where vx = dt dy v dt dx y = and
The direction of v is along the tangent to the path. The instantaneous acceleration is the ratio of change of the velocity with respect to time. a = v axi ayj dt d = + (14) where ax = dt dv a dt dv y y x ; =
Note that one cannot determine
acceleration directly from the path of the particle. One needs to know how each
component of the velocity varies as function of space and time.
O x
y
Fig.(19) Possible directions of acceleration of a particle traveling along a curved path.
a
a a
path
General equations of kinematics for constant acceleration
v = v0 + at (15)
r = r0 + v0t +
2 1
r = r0 +
2 1
(v0 + v)t (17)
For two−dimensional motion in the plane, the x and y components of these equations are:
vx = v0x + axt vy = v0y + ayt x = x0 + v0x + 2 1 axt2 y = y0 + y0yt + 2 1 ayt2 x = x0 + 2 1 (v0x + vx)t y = y0 + 2 1 (v0y + vy)t ) ( 2 0 2 2 v a x x vx = ox + x − v2y =voy2 +2ay (y−yo) Example : 11
A particle moves one quarter of a circular path of radius 20 m in 10 s. Its initial position is given by ri = (20 m)i and its
final position is rf = (20m)j .
(a) Find its displacement ∆ r and average velocity vav
(b) Find the magnitude of its average velocity and its average speed.
Solution
(a) The displacement is given by
i j r r r= f − i =20 ˆ−20 ˆ ∆ or ∆r=−20 ˆi+20 ˆj(m) and the average velocity is
Fig.(20) Average speed depends on total distance traveled, but average velocity depends only on the location of the initial and final –points.
x y S ri rf ∆r j i j i r v 2ˆ 2ˆ 10 20 20 + − = + − = ∆ ∆ = t av (m/s)
(b) The magnitude of the average velocity is
83 . 2 ) 2 ( ) 2 ( | | = = − 2+ 2 = av v av v m/s Average speed = 3.14m/s 10 )] 20 ( 2 [ 4 1 Taken Time travelled distance Total = π =
PROJECTILE MOTION
A projectile motion near the surface of the earth consists of two independent motions, a
horizontal motion at constant speed and a vertical one subject to the acceleration due to gravity.
In order to deal with problems in projectile motion, one has to choose a coordinate system and clearly specify the origin. If the x axis is horizontal and the y−axis points vertically upward, then
ax = 0 and ay = −g
One can easily assume the origin such that the initial horizontal coordinate is zero. i.e. xo = 0
x = voxt (18) vy = voy −gt (19) y = yo + voyt − 2 1 gt2 (20) 2 2 oy y v v = −2g (y − yo) (21) Example 12
A ball is projected horizontally at 20 m/s from a cliff of height 45 m (a) Find its time of flight
(b) Find its horizontal range R (the horizontal displacement from the point of firing).
Solution
The origin is assumed to be at the base of the cliff
Given: xo =0; yo = 45 m; vox = 20 m/s and voy = 0
The coordinates at a latter time are given by
x = 20 t (i)
x = 45 − 5t2 (ii)
(a) When the ball lands, its vertical coordinate is zero i.e. y = 0 From (ii) , we get
0 = 45 − 5t2 or t = 3 s
Note that the time of flight does not depend on the value of the horizontal component of
the initial velocity. If the ball were dropped from the same height it would have reached
the ground in 3 s.
(b) To find the horizontal range we use the horizontal velocity and the time of flight.
R = voxt = (20) (3) = 60 m
Fig.(21) The horizontal motion of a projectile is at constant velocity while the
vertical motion occurs at constant acceleration. The vertical component of the motion of a ball projected horizontally is the same as that of a ball that is simply dropped.
vox t = 1s t = 2 s t= 3 s t = 1 s t = 2 s t = 3 s 10 m/s 20 m/s 30 m/s 30 m/s 20 m/s 20 m/s 20 m/s 20 m/s 10 m/s y Example 13
A ball is projected from the ground with an initial velocity vo at an angle θ above the
horizontal
(a) Find the time of flight
(c) Determine the maximum height obtained by the ball (d) Obtain an expression for the trajectory of projectile
Solution
The origin is assumed at the point of projection. The instantaneous x and y coordinates of the ball are given by
x = (v0 cos θ ) t (i)
y = (v0 sin θ ) t −
2 1
gt2 (ii)
(a) To find the time of flight, we put y = 0 in equation (ii) 0 = (v0 sin θ ) t −
2 1
gt2
or t = 2v0gsinθ ..(iii)
(b) To find the horizontal range we substitute (iii) in (i)
R = (vo cos θ ) θ g v sin 2 0 or R = g vo2 sin 2θ (iv)
(c) The maximum height H for the projectile is given by gH v vy2 = oy2 −2 Since vy =0, therefore, H = g v g voy 2 sin 2 2 2 0 2 θ = (v)
(d) To find the trajectory of the projectile we have to express y in terms of x. Therefore eliminating t from (i) and (ii), we get
vo y θ vox voy voy = vo sin θ vy vox vox vox vy
Fig.(22) The path of a projectile is parabolic. The path is
symmetrical about the highest point only if the particle lands at the same level from which it was fired.
voy = vo cos θ y = x tan θ −
(
)
2 2 0 cos 2 v x g θ (vi)IMPORTANT
1. The time of flight is given by
T = 2vogsin θ (22)
2. The horizontal range is given by
R = g vo2sin 2θ
(23)
Equations (22) and (23) are valid only when the projectile returns to the initial vertical
level.
For a given initial speed vo, the range is a maximum when sin 2θ =1, that is when θ = 45°.
For a given velocity vo same range occurs at two angles of projection, viz. θ = 45 ± α
Fig.(23) The horizontal ranges for angle of projection (45o - α) and
(45o + α) are equal.
α α 45o
3. The maximum height of the projectile is H = g vo 2 sin2 2 θ (24) 4. The trajectory of a projectile is a parabola
y = xtan θ −21 2 2 0 cos ) (v x g θ (25) Example: 14
A hunter with a blowgun wishes to shoot a monkey hanging from a branch. The hunter aims right at the monkey, not realizing that the dart will follow a parabolic path and thus fall below the monkey. The monkey, however seeing the dart leave the gun, lets go of the branch and drops out of the tree, expecting to avoid the dart. Show that the monkey will be hit regardless of the initial velocity of the dart so long as it is great enough to travel the horizontal distance to the tree before hitting the ground.
Solution
Let the horizontal distance to the tree be x and the original height of the monkey be H, as shown in Fig. (24). Then the dart will be projected at an angle given by tan θ = H/x
If there were no gravity, the dart would reach the height H in the time t taken for it to travel horizontal distance x:
y = yoyt = H in time t = ox v x with no gravity y = voyt − 2 1 gt2 or y = H −12 gt2
This is lower than H by
2 1
gt2, which
is just the amount the monkey falls in this time. The initial velocity of the dart is varied so that for large v0 the
target is hit very near its original height and for small v0 it is hit just
before it reaches the floor.
Dart y 1/2gt2 H = voyt voy vox Fig.(24) Example 14
CIRCULAR MOTION
In its simplest kind, a particle moves in a circular path of constant radius with constant
speed. Although the magnitude of the velocity vector remains constant but the direction
of the velocity vector changes direction continuously as shown in Fig. (25). Thus a particle is continuously accelerated. Since the acceleration produces a change only in the direction of the velocity vector, therefore, it must always be at right angles to the
direction of motion. It is because any component in the direction of the motion would
Fig.(25)(a) Velocity vector (tangent to the circle) represents the direction of a body will
move if it is not pulled toward the centre.
(b) Two velocity vectors are subtracted to find the change in velocity ∆v = vf - vi
(c) The change ∆v in velocity vectors, when placed at the average position, between initial and final points toward the centre of the circular motion.
(a) ∆θ (c) (b) vi vf −vi vf ∆v vi vf ∆v ∆θ θ = 0 g
Fig.(26) Comparison of uniform circular motion with straight line motion and projectile motion
(a) In straight line motion the acceleration vector is either parallel or antiparallel to the velocity vector. (b) In projectile motion: 90o < θ < 180o during upward motion.
0 < θ < 90o during downward motion and θ = 90o at the highest position.
(c) In uniform circular motion acceleration is always perpendicular to the velocity vector.
(a) (b) (c) v a g 180o v g θ v v g v g 90o≤θ < 90o θ v θ = 90o θ
Let us consider a particle which moves in a circular path of radius r and constant speed v as shown in Fig. (27). In a time t the line joining the particle and the centre subtends an angle θ which is given by
θ (radian) = RadiusArc lengthlength =rs On differentiating both sides, we get
dt ds r dt d 1 = θ or ω = v/r (26)
where ω is the angular speed From Fig.(25 b), ∆ θ = (∆ v/v)
|vf| = |vi| = v Since ∆ s = v∆ t = r∆ θ ⇒ ∆ t = v r∆θ Thus, acceleration, a = r v t v= 2 ∆ ∆Note that this acceleration point inward toward the centre of the circle. It is called centripetal acceleration (means ‘centre
seeking’)
Fig.(27) A particle moves with constant speed in a circular path of constant radius r.
x y O θ r r t = t′ s t = 0
Alternatively the magnitude of the centripetal acceleration can be found using calculus
and vector notation.
r = xi + yj r = r cos θ i + r sin θ j r(t) = r cos ω t i + r sinω t j v(t) = dt dr = −rω sin ω t i + rω cosω t j v = ω r a(t) = dt t dv )( = −rω 2 cosω t i − rω 2 sinω t j a(t) = −ω2 r(t) a = ac = ω 2r ac = r v2
Fig.(28) A particle moves with constant angular speed ω in a circular path of constant radius
r.
y
r rsin ωt r cos ωt θ = ωt
To summarize, an object moving in uniform circular motion at radius r and constant angular speed ω has
1. a constant linear speed v = rω tangent to the circle and
2. a centripetal acceleration ac = ω 2r = v2/r directed inward toward the centre of the circle
Example: 15
A stone tied to the end of a string 75 cm long is whirled in a horizontal circle with a constant speed. If the stone make 15 revolutions in 25s. What is the magnitude and direction of acceleration of the stone ?
Solution
Radius of the circular path, r = 75 cm = 0.75 m Number of revolutions per second n = 0.6s 1
25 15 − = Now, v = 2π nr or v = 2(3.14) (0.6) (0.75) = 2.83 m/s Magnitude of acceleration, a = r v2 = 75 . 0 ) 83 . 2 ( 2 = 10.68 m/s2
The direction of acceleration is radially inwards.
Tangential and Normal Components of Acceleration in Two Dimensions
1. The velocity vector is always tangential to the path
2. The acceleration vector may have two components: one tangential to the path and one perpendicular to the path.
(a) The component of the acceleration parallel to the path is due to a change in speed. When the speed is increasing, the tangential component at points in the same
direction as the velocity; when the speed is decreasing, the tangential component
points opposite to the velocity.
(b) When the path of an object curves, there is a component of the acceleration perpendicular to the velocity. This component of the acceleration ac points toward
the inside of the curve.
(c) The total acceleration is the vector sum of the tangential and centripetal components.
RELATIVE VELOCITY
The motion of any body has to be described relative to some frame of reference, such as the ground. Sometimes it becomes necessary to study the motion of one body relative to another body which is also moving relative to the ground. Let us study the relative motion in general.
Consider the particle P which is being observed from two frames A and B. Its positions with respect to frame A and B are rPA and rPB,
respectively.
The position vector triangle shown in the Fig. (29) explains that
rPB = rPA + rAB (27)
Position of P Position of P Position of A w.r.t. B w.r.t. A w.r.t. B Differentiating the above equation, we get
vPB = vPA + vAB (28)
Fig.(29) A point P is located with respect to
two frames A and B
P yB yA xA xB rAB rPA rPB OA OB
Note that equation (28) is a vector sum we worry about signs only when we take
components. Let us consider the case where a man walks across a train with velocity vo.
The ground forms a stationary frame (x, y) while the train forms a moving frame (x′, y′). We want to find the velocity of the man relative to the ground (v).
Using equation (28)
An observer on the train will see the man walking along the x′ axis. An observer on the ground will see him moving in a direction given by tan θ =
o rel
v v
to the x−axis.
Fig.(30)(a) The diagram shows the position of man (M) at three instants as he walks across a train
with velocity vo relative to the ground.
(b) His velocity w.r.t. ground is shown by the velocity vector diagram.
o rel v v v= + O x′ y′ vo vrel vrel vrel M M M (a) (b) v vo vrel θ x y Example 16
A boat can travel at 10 m/s relative to the water. It starts at one bank of river that is 100 m wide and flows with a velocity u = 7.5 m/s. If the boat points directly across, find
(a) Its velocity relative to the bank (b) How far downstream it travels.
Solution
In this problem the train is replaced by the river and the man is replaced by the boat (a) From the velocity vector diagram, the
magnitude of the velocity of boat w.r.t. ground is
v=
( )
m s uvrel2 + 2 = (10)2 +7.5 2 =12.5 /
The direction is indicated by the angle θ
with the y−axis.
tan θ = =710.5=43 rel v u b = 100 m x y u vrel Fig.(31 a) Example 16 or θ = tan−1(3/4) = 37°
(b) The distance travelled along the bank is
x = ut
where t is the time taken to cross the river.
t = rel v b Thus, x = m v ub rel 75 ) 100 ( 10 5 . 7 = = vrel v u θ Fig. 31 b
Example 17
In which direction the boat must point so as to cross the river in a direction perpendicular to river currents. Also, find out the time taken to cross the river. Use the information given in the previous Example 16.
Solution x u Fig.(32 ) Example 17 y vrel vrel v u θ
To get directly across the river, the boat must be pointed upstream (a)
(b)
From the velocity diagram
The magnitude of the velocity w.r.t. bank is
v = v2 −u2= ( )102−(7.5)2=6.61
rel m/s
The direction of motion of the boat w.r.t. the river current is represented by the angle θ
measured anticlockwise from the y−axis. tan θ = 1.13 61 . 6 5 . 7 = = v u or θ = tan−1 (1.13) = 48.6°
and time taken t = 15.13 61 . 6 100 = = v b s
Objective Solved Examples
1. A particle moves along the x–axis in such a way that its coordinate (x) varies with time (t) according to the expression
x = 2 – 5t + 6t2 m. The time t is in second.
The initial velocity of the particle is
(a) –5 m/s (b) –3 m/s (c) 6 m/s (d) 3 m/s Solution x = 2 – 5t + 6t2 ∴ v = dt dx = -5 + 12 t
Initial velocity means velocity at t = 0
∴ vinitial = -5 m/s
∴ (a)
2. In the arrangement shown in the figure, the ends P and Q of an unstretchable string move downward with uniform speed u. Pulleys A and
B are fixed. At the instant shown,
the mass M moves upward with a speed of (a) 2u cos θ (b) u / cos θ (c) 2u / cos θ (d) u cos θ A M θ θ B P Q Solution
In time ∆ t, ends P and Q move downward
through a distance u∆ t and the mass M
moves upward to M′. During this time, the strings AM and BM are reduced to AM′ and
BM′ and MC = MD = u∆ t. Now MM' MCθ uΔΔθ cos cos = = Speed of M = MM'Δt =cosuθ ∴ (b) A M P u∆t C D M′ θ
3. A ball is released from a height h above the ground. It takes a time T to reach the ground. Where is the ball at the time T/2
(a) at a height h/4 from the ground (b) at a height h/2 from the ground (c) at a height 3h/4 from the ground (d) none of these Solution h = 2 2 1 gT s = 4 2 2 1 T 2 h g =
∴ height from the ground = h -
4 3 4 h h = ∴ (c)
4. Two bodies, one held 30 cm directly above the other, are released simultaneously and fall freely under gravity. After 2 s their separation will be
(a) 10 cm (b) 20 cm
(c) 30 cm (d) zero
Solution
Motion parameters i.e. the initial velocity and acceleration are same for both the bodies. Therefore, they will fall by the same distance in a given time, maintaining their initial separation.
∴ (c)
5. A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the water in km/hr is
(a) 1 (b) 3
(c) 4 (d) 14
Solution
Shortest possible path is AB.
∴ the boat must move in the direction shown in the figure. For this, we must have 5 cos θ = 15 minutes1km = 4 km/hr and 5 sin θ = v ⇒ v = 3 km/hr ∴ (b) A B 1 km θ v 5
6. An object is dropped from rest. Its velocity versus displacement graph is v (a) s (b) (c) (d) v s v s v s Solution
For free fall
v2 = u2 + 2gs
Here u = 0 so v2 = 2gs
∴ graph between v and s will be a parabola symmetric about s axis.
∴ (c)
7. A particle moves in a circle of radius 25 cm at 2 revolutions per second. The acceleration of the particle in m/s2 is
(a) π 2 (b) 8 π 2
(c) 4 π 2 (d) 2 π 2
Solution
The angular velocity ω =
1 2 2× π = 4 π rad/s a = ω 2R = (4π )2× (1/4) = 4 π 2 m/s2 ∴ (c)
8. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a is correctly shown in:
(a) a (b) a (c) a (d) a
Solution
As displacement of the bob is less than maximum, the acceleration will have two components, radial and tangential. So resultant will be inclined at some angle with string.
∴ (c) a t a r a
9. A particle starts from rest. Its acceleration (α ) versus time (t) is as shown in the figure. The maximum speed of the particle will be
(a) 110 m/s (b) 55 m/s (c) 550 m/s (d) 660 m/s 10 m/s2 11 t (s) α Solution
The area under the acceleration time graph gives change in velocity.
∴ (b)
10. A small block slides without friction down an inclined plane starting from rest. Let Sn be
the distance travelled from time t = n - 1 to t = n. Then
1 + n n S S is (a) n n 2 1 2 − (b) 1 2 1 2 − + n n (c) 1 2 1 2 + − n n (d) 1 2 2 + n n Solution
