Kinematics Theory

KINEMATICS

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1. KINEMATIC VARIABLES Kinematics is study of geometry of motion without considering its cause. It deals with position, distance travelled by a body and its displacement, speed and velocity and acceleration of a body. Before going in details about different kinds of motion separately, let us know about these kinematics variable first.

1.1

DISTANCE AND DISPLACEMENT

When a particle is moving its successive position in general may lie on a curve, say, ABC shown in figure. The curve is then called the path of the particle. The total length of the path followed by the body is called the distance traversed by the body. Its unit in S.I. system is ‘metre’. It is a scalar quantity. Sometimes we may not be interested in the actual path of the particle but only in its final position C relative to the initial position A. The directional distance between initial and final positions of the particle AC in the figure is called displacement. It is a vector quantity.

C

A

B

Fig. (1)

1.2

SPEED AND VELOCITY

Speed is the rate at which a moving body describes its paths. The path may be a curve or a straight line and its shape need not be considered to decide the speed. If a particle traverses a distance s during a time t, Average speed,

v

s t

…(1)

If the interval of time t is infinitesimally small approaching zero, this ratio is called of instantaneous speed or sometimes referred as speed of particle. i.e.,

instantaneous speed v  Lt t  0

s ds  t dt

…(2)

Speed is a scalar quantity and in S.I. system it is measured in metre/second (m/s). Velocity is defined as the rate of change of position. If during a time internal t, a particle changes its position by r . Average velocity,

 r v av  t

Also, instantaneous velocity,

…(3)

v  Lt t  0

r d r  t dt

…(4)

Velocity is a vector quantity having direction same as that of displacement and is measured in metre per second (ms1).

1.3

ACCELERATION

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The rate of change of velocity is called as acceleration. The change in velocity can be either in the magnitude of the velocity or in the direction of velocity or in both of them simultaneously. If v is change in velocity which takes place in time interval t, then during this interval  v Average acceleration, aav  t

Also, instantaneous acceleration, a  Lt t  0

…(5) v d v  t dt

…(6)

Acceleration is a vector quantity and is measured in metre/second2 (ms2).

Illustration 1 Question:

A particle moves along a semicircular path of radius R =  m in time t = 1s with constant speed. For the particle calculate (take 2 = 10)

(i)

distance travelled

(ii)

average speed

(iii)

average acceleration

A

R

B

Solution: (i)

Distance = length of path of particle = AB = R = 10 m

(ii)

Average speed, v =

(iii)

Average acceleration =

distance R = = 10 m/s time t Change in velocity 2v 2R 2 =  2 = 20 m/s time taken t t

Can a body have (a) zero instantaneous velocity and yet be accelerating; (b) zero average speed but nonzero average velocity; (c) negative acceleration and yet be speeding up?

2. MOTION IN ONE DIMENSION As discussed earlier for the analysis of motion, we need a reference frame made of origin and a set of co-ordinate axis depending on the type of motion. If a motion is taking place on a straight line, we need one axis for the analysis of motion. So we call this motion as motion in one dimension. Similarly if motion is taking place in a plane, we need two co-ordinate axes and we call motion as motion in two dimension. Three co-ordinate axes are required if motion is random motion in space and we call such motion as motion in three dimension. Here we are going to discuss motion on a straight line i.e., motion in one dimension. Consider a particle moving on a straight line AB. For the analysis of motion we take origin, O at any point on the line and x-axis along the line. Generally we take origin at the point from where particle starts its motion and rightward direction as positive x-direction. At any moment if particle is at P then its position is given by OP = x.

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X A

O

P

B

x

Fig. (2) Velocity is defined as,

v

Acceleration is defined as, a 

2.1

dx dt

dv d 2 x v dv  2  dt dx dt

MOTION IN A STRAIGHT LINE WITH UNIFORM VELOCITY If motion takes place with a uniform velocity v on a straight line, then displacement in time t, s = v.t

…(7)

acceleration of particle is zero.

2.2

MOTION IN A STRAIGHT LINE WITH UNIFORM ACCELERATION-EQUATIONS OF MOTION

Let a particle move in a straight line with initial velocity u (velocity at time t = 0) and with uniform acceleration a. Let its velocity be v at the end of the interval of time t (final velocity at time t). Let S be its displacement at the instant t. Now, acceleration a = or,

v u t

v = u + at

… (8)

If ‘u’ and ‘a’ are in the same direction, ‘a’ is positive and hence the velocity increases with time. If ‘a’ is opposite to the direction of ‘u’, ‘a’ is negative and the velocity decreases with time. Displacement during the time interval t = average velocity x t S

u v t 2

... (9)

Eliminating v from equations (8) and (9), we get S

or

u  u  at t 2

S  ut 

1 2 at 2

... (10)

Another equation is obtained by eliminating t from equations (8) and (9) v = u + at or,

or,

a=

v u t

S =

v u t 2

aS =

v u v u  t t 2

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= or,

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v 2 u 2 2

v2 – u2 = 2aS v2 = u2 + 2aS

... (11)

Distance traversed by the particle in the nth second of its motion The velocity at the beginning of the nth second = u + a (n – 1) The velocity at the end of the nth second = u + an Average velocity during the nth second =

u  a ( n  1)  u  an 2

= u

1 a ( 2n  1) 2

Distance traversed during this one second 1   Sn = average velocity  time = u  a ( 2n  1)  1 2   i.e., S n  u 

1 a (2n  1) 2

... (12)

The five equations derived above are very important and are to be memorized. They are very useful in solving problems in straight-line motion. Calculus method of deriving equations of motion The acceleration of a body is defined as a

i.e.,

dv dt

dv = a dt

Integrating, we get, v = at + A Where A is constant of integration. By the initial condition when t = 0, v = u (initial velocity), we get A = u v = u + at



we know that the instantaneous velocity v 

dS . dt

Displacement of body for duration dt from time t to t + dt is given by dS = v dt = (u + at) dt On integration we get, S  ut 

At,

1 2 at  B , where B is integration constant. 2

t = 0, S = 0 yields B = 0.

KINEMATICS 

Acceleration = 

1 2 at 2

S = ut +

a

dv dv dS dv  .  v. dt dS dt dS dv dS

=v

a.dS = v.dv Integrating we get, aS =

v2  C , where C is integration constant. 2

Applying initial condition, when S = 0, v = u we get 0 

or

u2 C 2

C=–

u2 2

v 2 u2  2 2



aS 



v2 – u2 = 2aS



v2 = u2 + 2aS

If S1 and S2 are the displacements during n seconds and (n – 1) seconds S1 = u n +

1 an2 2

S2 = u (n – 1) + 

1 a (n – 1)2 2

Displacement in nth second Sn = S1 – S2 = u n + Sn = u +

1 1 an2 – u (n – 1) – a (n – 1)2 2 2

1 a (2n – 1) 2

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Illustration 2 Question:

A certain automobile manufacturer claims that its super-delux sport’s car will accelerate from rest to a speed of 32.0 ms–1 in 8.0 s. Under the important assumption that the acceleration is constant.

(a)

determine the acceleration of car in ms–2.

(b)

find the distance the car travels in 8.0 s.

(c)

find the distance the car travels in 8th second.

Solution: (a)

We are given that u = 0 and velocity after 8 s is 32 m/s, so we can use v = u + at to find acceleration

a (b)

v u 32.0  0  = 4.0 ms–2 t 8.0

distance travelled in 8.0 s, We can use, S = ut +

(c)

1 1 2 at = 0 +  4.0  8 2  128 m 2 2

distance travelled in 8th second, We have, Sn = u + (2n –1) = (2 × 8 – 1) ×

2.3

a 2 4.0 = 30 m 2

GRAPHICAL REPRESENTATION OF MOTION

(i) Displacement-time graph: If displacement of a body is plotted on Y-axis and time on X-axis, the curve obtained is called displacement-time graph. The instantaneous velocity at any given instant can be obtained from the graph by finding the slope of the tangent at the point corresponding to the time. y Displacement (s=y-y0)

If the graph obtained is parallel to time axis, the velocity is zero (ab in figure). If the graph is an oblique straight line, the velocity is constant (oc and ef in figure). If the graph obtained is a curve (od in figure) whose slope increases with time, the velocity goes on increasing i.e., the motion is accelerated. If the graph obtained is a curve of type og in figure whose slope decreases with time, the velocity goes on decelerated.

a

v=0

decreasing velocity

b

g

e

c d

v = constant

v = constant increasing velocity

y0 O

t0 Time t

Fig. (3)

f x

KINEMATICS y

A Velocity constant (a = 0) B g

Velocity (v)

Velocity-time graph: Similarly a graph can be drawn between velocity and time of a moving body. The curve obtained will be similar to the one shown in figure. If the graph is a straight line parallel to time axis, the velocity is constant and acceleration is zero (AB in figure). If the graph is an oblique straight line, the motion is uniformly accelerated (positive slope) or uniformly decelerated (negative slope). The velocity-time curve may be used to determine displacement, velocity and acceleration i.e., it is used to specify the entire motion.

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c

a decreasing

d Retardation (a) constant

f

a = constant a = increasing e

Time (t)

x

Fig. (4)

y q Velocity (v)

To obtain the velocity at any time t we draw a perpendicular from given instant on the curve and noting the corresponding point and dropping a perpendicular from this point on the velocity axis. The slope of the tangent at the point corresponding to the particular time on the curve gives instantaneous acceleration. The area enclosed by velocity-time graph and time axis for a time interval gives the displacement during this time interval. In figure the shaded area pqt2t1 represents the net displacement during the time interval between t1 and t2.

P

t1

Work t t+ t

t2

x

Time (t) Fig. (5)

Acceleration-time graph: It is a graph plotted between time and acceleration. If the graph is a line parallel to time axis, the acceleration is constant. If it is a straight line with positive slope, the acceleration is uniformly increasing. The co-relation of the graph explained above follows directly dS dv from the differential expressions v  and a  . dt dt The area enclosed between acceleration-time graph and time axis gives during this time interval change in velocity.

Illustration 3 –2

Question:

At t = 0 a particle is at rest at origin. Its acceleration is 2ms for the first 3s and –2 ms2 for the next 3s. Plot the acceleration versus time, velocity versus time and position versus time graph.

Solution:

We are given that for first 3s acceleration –2 is 2 ms and for next 3s acceleration is – –2 2ms . Hence acceleration time graph is as shown in the figure.

Y a(ms-2)

O

X 3

6

t (s)

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The area enclosed between a-t curve and t-axis gives change in velocity for the corresponding interval. Also at t = 0, v = 0, hence final velocity at t = 3s will increase to 6 ms1. In next 3s the velocity will decrease to zero. Hence the velocity time graph is as shown in figure.

v (ms-1) 6

O

Note that v – t curves are taken as straight line as acceleration is constant. Now for displacement time curve, we will use the fact that area enclosed between vt curve and time axis gives displacement for the corresponding interval. Hence displacement in first three second is 9 m and in next three second is 9m. Also the x-t curve will be of parabolic nature as motion is with constant acceleration. Therefore x-t curve is as shown in figure.

Y

6

3

t (s)

X

Y x (m) 18

9

X O

6

3

t(s)

If acceleration-time graph is a straight line inclined with time axis, then what will be the nature of velocity-time graph.

2.4

VERTICAL MOTION UNDER GRAVITY

When a body is thrown vertically upward or dropped from a height, it moves in a vertical straight line. If the air resistance offered by air to the motion of the body is neglected, all bodies moving freely under gravity will be acted upon by its weight only. This causes a constant vertical acceleration g having value 9.8 m/s2, so the equation for motion in a straight line with constant acceleration can be used. In some problems it is convenient to take the downward direction as positive, in such case all the measurement in downward direction are considered as positive i.e., acceleration will be +g. But sometimes we may need to take upward as positive and in such case acceleration will be g. Projection of a body vertically upwards Suppose a body is projected vertically upward from a point A with velocity u. If we take upward direction as positive

B

(i) At time t, its velocity v = u  gt (ii) At time t, its displacement from A is given by S = ut 

1 2 gt 2

C

(iii) Its velocity when it has a displacement S is given by u

v2 = u2  2gS A

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(iv) When it reaches the maximum height from A, its velocity v = 0. This u happens when t  . The body is instantaneously at rest at the highest point B. g (v) The maximum height reached u2 H 2g 2u (vi) Total time to go up and return to the point of projection = g Since, S = 0 at the point of projection, 1 S = ut  gt 2 2 1 2 2u 0 = ut  gt or t = 2 g u 2u u u Since the time of ascent = , the time of descent =   g g g g (vii) At any point C between A and B, where AC = S, the velocity v is given by

v   u 2 2gS The velocity of body while crossing C upwards   u 2 2gS and while crossing C downwards is  u 2 2gS . The magnitudes of the velocities are the same.

A body is fired upward with speed v 0. It takes time T to reach its maximum height H. True or false. H T v H (a) It reaches in (b) It has speed 0 at (c) it has speed v 0 at 2T. 2 2 2 2

Illustration 4 Question:

A body is projected upwards with a velocity 100 m/s. Find (a) the maximum height reached, (b) the time taken to reach the maximum height, (c) its velocity at a height of 320 m from the point of projection, (d) speed with which it will cross down the point of projection and (e) the time taken to reach back the point of projection. Take g = 10 m/s2

Solution: (a)

The maximum height reached Initial upward velocity u = 100 m/s Acceleration a = ( g) = – 10 m/s2 Maximum height reached H is given by v2 = u2 + 2aS 0 = 1002 + 2(  10)H H=

(b)

100 2 = 500 m 2  10

The time taken to reach the maximum height

KINEMATICS t

(c)

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u 100  = 10 s g 10

Velocity at a height of 320 m from the point of projection v2 = u2 + 2aS v2 = 10000 + 2( 10)320 v =  60 m/s + 60 m/s while crossing the height upward and 60 m/s while crossing it downward.

(d)

Velocity with which it will cross down the point of projection v2 = u2 + 2gS At the point of projection S = 0 

v=u

While crossing the point of projection downwards, speed = u = 100 m/s The velocity has the same magnitude as the initial velocity but reversed in direction. (e)

The time taken to reach back the point of projection t

2.5

2u 2 100  = 20 s g 10

GENERAL MOTION IN A STRAIGHT LINE

Saying general motion means motion it is other then motion with uniform velocity or motion with uniform acceleration. In such motion we will be given the relation between two variables among position, velocity, acceleration and time and other are to be calculated. For this we need to use, Velocity,

v=

Acceleration, a 

dx dt dv vdv  dt dx

and then using calculus we find the unknown variables.

Illustration 5 Question:

–2

A particle moving in a straight line has an acceleration of (3t – 4) ms at time t seconds. The particle is initially 1 m from O, a fixed point on the line, with a velocity of 2 ms1. Find the time (> 4/3s) when the velocity is zero. Find also the displacement of the particle from O when t = 3.

KINEMATICS Solution:

Using

a

dv gives dt

dv  3t  4 dt v



t

 dv   (3t  4) dt 2

0



v–2=



v=

3t 2  4t 2

3t 2  4t  2 2

The velocity will be zero when

3t 2  4t  2  0 2

i.e., when (3t – 2) (t – 2) = 0 2 or 2 3



t=



t = 2s

Using

ds ds 3t 2  v we have   4t  2, dt dt 2 s

3  3t 2  ds    4t  2  dt  2   1 0









t 3  s – 1 =   2t 2  2t   1.5  2  0



s = 2.5

3

Therefore the particle is 2.5 m from O when t = 3 s.

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PROFICIENCY TESTI The following questions deal with the basic concepts of this section. Answer the following briefly. Go to the next section only if your score is at least 80%. Do not consult the Study Material while attempting these questions. 1.

2.

3.

Describe a physical situation, for example, with a ball or a car, for each of the following cases. (a) a = 0, v  0 ;

(b) v = 0, a  0;

(c) v < 0, a > 0;

(d) v = 0, a < 0.

True or false (a)

Positive slope of x versus t graph implies motion away from origin

(b)

Negative slope on the v versus t graph means the velocity of body is decreasing

(c)

Area enclosed between v versus t graph and t-axis gives displacement.

(d)

Area enclosed between a versus t graph and t-axis gives velocity.

The position time graph in figure depicts the journey of three bodies A, B and C. (a)

At 1s, which has the greatest velocity?

(b)

At 2s, which has travelled the farthest?

(c)

When A meets C, is B moving faster or slower than A?

(d)

Is there any time at which the velocity of A is equal to that of B?

A

x

B C

1s

4.

5.

2s

3s

4s

t

A ball is thrown up from the surface of earth and falls back on it. Taking origin at ground and upward as positive direction, draw (a)

x – t graph,

(b)

v – t graph and

(c)

a – t graph

A particle starts with a velocity of 6 m/s and moves with uniform acceleration of 3 m/s2. Find its velocity after 10 seconds.

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6.

A body starts from rest and moves with uniform acceleration of 4 m/s2 for one minute. What is the velocity acquired by the body?

7.

An automobile moving with a velocity of 72 km/hr is brought to rest in a distance of 10 m by the application of brakes. Find the retardation assuming it to be uniform.

8.

A train starting from rest accelerates uniformly for 100 s, runs at a constant speed for 5 minutes and then comes to a stop with uniform retardation in the next 150 seconds. During this motion it covers a distance of 4.25 km. Find its constant speed.

9.

A stone is dropped from a height of 100 m from a balloon which rises up from the ground. Find the velocity of the balloon at that moment if the stone reaches the ground 5 seconds after it was dropped.

10.

A particle moves in straight line with acceleration 2t ms–2 at time t. If it starts from rest of O on the line, find its velocity and displacement from O at time t = 3 s.

KINEMATICS

ANSWERS TO PROFICIENCY TESTI 2.

(a) True (b) True (c) True (d) False

3.

(a) B (b) C (c) Slower (d) Yes, in the interval 2 to 3 s

5.

The velocity at the end of 10 seconds is 36 m/s.

6.

240 m/s

7.

20 m/s2

8.

36 km/hr

9.

Velocity of the balloon = 5 m/s directed upward

10.

9 m/s, 9 m

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