# Class – 11 Physics, Chapter 4 :Uniform  and Non-Uniform Motion

Kinematics : The branch of mechanics which deals with study of motion of a particle without knowing the cause of motion, is called as kinematics.

Dynamics : The branch of mechanics which deals with the study of motion with knowing the cause of motion.

Concept of point object : An object is said to be a point object if it changes its position by distances which are much greater than its size.  A car travelling a few hundred kilometer may be regarded as the point object . Earth revolving around the sun is the point object.

Motion in one dimension : The motion of a particle in a straight line is termed as the motion in one dimension. It is also known as rectilinear or linear motion. To describe it, we require the two rectangular coordinate axes. Examples :

1. A ball thrown vertically up or a ball dropped from a certain height above the ground moves in a straight line. So, the motion of the ball is 1-D motion.
2. A mass suspended from a vertical spring can oscillate in a straight line. So, its oscillatory motion is 1-D motion.

Motion in two dimensions : If a particle moves in a curve path in plane , its motion is regarded as 2 – D motion. as for example a planet is revolving around the sun has two dimensional motion. Motion in three dimensions : when a particle moves in space then its motion is said to be three dimensional motion. To describe it, we require all the three rectangular coordinates.  Some Important Terms Regarding Motion

1)  Distance and displacement :

Distance : The separation between two points is called as distance or the path  travelled by a particle between two points is termed as its distance. It is scalar quantity as it has no direction. It has only magnitude . It can not be negative. It is always a positive quantity. It may be fractional number. It can never be zero. It never decrease with time. Its SI unit is metre and CGS unit is centimetre.  Displacement : The shortest distance between two points is referred to as displacement. It is vector quantity as it has both direction and magnitude. It may be negative, positive and fraction or even zero. It is less or equal to distance. It can not be greater than distance as it is the shortest distance between two points. It is the change in the position of a particle in a particular direction. Its SI unit is metre while CGS unit is centimetre.  2) Speed and velocity :

Speed : The rate of change of distance is called as speed. It is distance covered in unit  time. It is the scalar quantity as it has no direction. It has only magnitude.

It is always positive. It can never be negative, zero and fraction.

Speed = distance / time
SI unit : m / sec
CGS unit : cm / sec.

Instantaneous speed : The speed of an object in a particular moment or in a particular instant of time, is known as instantaneous speed. It may also be defined as the limit of average speed as the time interval becomes infinitesimally small.

Instantaneous speed = limit Δt→0  Δx/Δt = dx / dt  Average speed : It may be defined as the speed over long interval of time.

Average speed = total distance/total time = Δx / Δt

It is the harmonic mean of the individual speed. It is the arithmetic mean of the Individual speeds. It is the ratio of the path length travelled and the corresponding time interval.  Uniform sped : If  equal distance is travelled by an object in equal interval of time, (howsoever these interval of time may be ) then the speed of the object is called as Uniform sped.  Non uniform or variable speed : If an object travels equal distance in unequal interval of time or unequal distance in equal interval of time or unequal distance in unequal Interval of time (however these interval of time may be ), then the object is said to have variable or non uniform speed. Velocity : The rate of change of displacement is termed as velocity. It is the change in distance  of an object in unit time in a particular direction. It is a vector quantity as it has both magnitude and direction. It can be zero, fraction, negative and positive.  Uniform velocity : If an object travels equal displacement in equal interval of time, then the object is said to be have uniform velocity.

Non uniform velocity: If an object travels equal displacement in unequal interval of time or unequal displacement in equal interval of time or unequal displacement in unequal interval of time ( however these interval of time may be ), then the object is said to have variable or non uniform speed.

Instantaneous velocity :  The velocity of an object in a particular moment or in a Particular instant of time, is known as instantaneous velocity. It may also be defined a time limit of average velocity as the time interval becomes infinitesimally small.

Instantaneous velocity = limit Δt→0  Δx/Δt = dx / dt

Average velocity:  It may be defined as the velocity over long interval of time .

Average velocity= total displacement/total time=x/t

Acceleration :  The rate of change of velocity is called as acceleration.It is the increase in velocity in unit time.

SI unit : m / sec2

CGS unit: cm / sec2

Non uniform acceleration or variable acceleration : A body is said to have non uniform acceleration if its velocity increases by unequal amount in equal interval of time.

Examples :

Circular motion of all types or a man driving in random direction or with varying the race of his vehicle or both.

Average acceleration : Average acceleration is the change in velocity divided by an elapsed time. For instance, if the velocity of a car increases from 0 to 60 cm/s in 3 seconds, its average acceleration would be 20 cm/s2. This means that the car’s velocity will increase by 20 cm/s every second.

Average acceleration is determined over a “long” time interval. The word long in this context means finite — something with a beginning and an end. The velocity at the beginning of this interval is called the initial velocity, represented by the symbol v0(vee nought), and the velocity at the end is called the final velocity, represented by the symbol v . Average acceleration is a quantity calculated  two velocity measurements.

a = Δv / Δt = (v – v0 ) / Δt

Instantaneous acceleration : Instantaneous acceleration is measured over a “short” time interval. The word short in this context means infinitely small or infinitesimal — having no duration or extent whatsoever. It’s a mathematical ideal that can can only be realized as a limit. ( The limit of a rate as the denominator approaches zero is called a derivative ). Instantaneous acceleration is then the limit of average acceleration as the time interval approaches zero — or alternatively, acceleration is the derivative of velocity.

a = limit Δt→0 Δv / Δt = dv/dt

Linear acceleration : An object moving in a straight line is accelerated only if a force acts on it. For example, imagine a ball rolling across a smooth flat surface with a velocity of 5 meters per second. Then suppose someone hits the ball lightly with a bat. The additional force on the ball provided by the bat will cause the ball to move faster.  Circular acceleration :  Consider an object moving in a circle of radius r with constant angular velocity. The tangential speed is constant, but the direction of the tangential velocity vector changes as the object rotates . The acceleration corresponding to this situation is called as circular motion.  The acceleration of an object depends on two factors, velocity and direction. An object that moves with constant speed but that changes direction is also accelerating. A car traveling around a curve in the road is accelerating even though its speed remains constant.

Graphical representation

Position-time graph :

The graph obtained by plotting the position along y-axis and time along x-axis , is called as position-time graph. The slope of the graph represents the velocity. Compare the displacement-time equation for constant velocity with the classic slope-intercept equation

s = s0 + vΔt ————————–(1)

y = a + bx ——————————-(2)

where s is the final displacement

s0 is the initial displacement

v is the velocity

Δt is the change in time

Case-1: When the object is initially is at rest, the position – time graph is a straight line  passing through origin.

Thus velocity corresponds to slope and initial displacement to the intercept on the vertical axis (commonly thought of as the “y” axis). Since each of these graphs has its intercept at the origin, each of these objects had the same initial displacement. This graph could represent a race of some sort where the contestants were all lined up at the starting line. If it were a race, then the contestants were already moving when the race began, since each curve has a non-zero slope at the start. Note that the initial position being zero does not necessarily imply that the initial velocity is also zero.

As each graph has constant velocity, the object has zero acceleration.

Case-2 : Graph of the displacement of an object with a constant, non-zero acceleration starting from rest at the origin. The relation between displacement and time is quadratic when the acceleration is constant and therefore this curve is a parabola. (The variable of a quadratic function is raised no higher than the second power.)

s = s0 + uΔt + 1/2a( Δt )2

y = a + bx + cx2

When a displacement-time graph is curved, it is not possible to calculate the velocity from it’s slope. Slope is a property of straight lines only. Such an object doesn’t have a velocity because it doesn’t have a slope. The words “the” and “a” are bold and red coloured here to stress the idea that there is no single velocity under these circumstances. The velocity of such an object must be changing. It’s accelerating.It has variable velocity.

Case-3 :  Position time graph of stationary object When an object is at rest then its position remains unchanged with time, hence the graph will be a straight line.

Case-4 : Position time Graph of de-accelerating object

When the slope of the position time graph is negative then it will represent the motion of a de accelerating body. Note :

On a displacement-time graph…

• straight lines imply constant velocity.
• curved lines imply acceleration.
• an object undergoing constant acceleration traces a portion of a parabola.
• average velocity is the slope of the straight line connecting the endpoints of a curve.
• instantaneous velocity is the slope of the line tangent to a curve at any point.
• positive slope implies motion in the positive direction.
• negative slope implies motion in the negative direction.
• zero slope implies a state of rest.
• when two curves coincide, the two objects have the same displacement at that time

velocity or speed time graph

Note :

On a velocity-time graph…

• slope equals acceleration.
• the “y” intercept equals the initial velocity.
• when two curves coincide, the two objects have the same velocity at that time.
• straight lines imply uniform acceleration.
• curved lines imply non-uniform acceleration.
• an object undergoing constant acceleration traces a straight line.
• average acceleration is the slope of the straight line connecting the endpoints of a curve. a = Δv / Δt
• instantaneous acceleration is the slope of the line tangent to a curve at any point.   a = limit Δt→0 Δv /Δt = dv / dt
• positive slope implies an increase in velocity in the positive direction.
• negative slope implies an increase in velocity in the negative direction.
• zero slope implies motion with constant velocity.
• the area under the curve equals the change in displacement.

Case-1: The object having zero acceleration i.e constant velocity The most important thing to remember about velocity-time graphs is that they are velocity-time graphs, not displacement-time graphs. There is something about a line graph that makes people think they’re looking at the path of an object. A common beginner’s mistake is to look at the graph to the right and think that the the v = 9.0 m/s line corresponds to an object that is “higher” than the other objects. Don’t think like this. It’s wrong.

Don’t look at these graphs and think of them as a picture of a moving object. Instead, think of them as the record of an object’s velocity. In these graphs, higher means faster not farther. The v = 9.0 m/s line is higher because that object is moving faster than the others.

These particular graphs are all horizontal. The initial velocity of each object is the same as the final velocity is the same as every velocity in between. The velocity of each of these objects is constant during this ten second interval.As the velocity is constant, acceleration is zero.

Case-2 : The object having variable velocity but constant acceleration

When the curve on a velocity-time graph is straight but not horizontal, the velocity is changing. The three curves in the above figure each have a different slope. The graph with the steepest slope experiences the fastest change in velocity. That object has the greatest acceleration.

Compare the velocity-time equation for constant acceleration with the classic slope-intercept equation .

v=u+at—————————————-(1)

y=c+mx—————————————-(2)

Where v is the final velocity, u is the initial velocity and t is time

Equation (1) is  classic slope – intercept equation and equation (2) is the general form of the straight line.

We see that acceleration corresponds to slope and initial velocity to the intercept on the vertical axis. Since each the graphs has its intercept at the origin, object was initially at rest. The initial velocity being zero does not mean that the initial position must also be zero, however. This graph tells us nothing about the initial position of these objects. A straight line is a curve with constant slope. Since slope is acceleration on a velocity-time graph, each of the objects represented on this graph is moving with a constant acceleration. When the graphs curved, the acceleration would not have been constant.

Case-3 : The object having decreasing acceleration

When a velocity-time graph is curved, it is not possible to calculate the slope from it’s slope. Slope is a property of straight lines only. Such an object doesn’t have a acceleration because it doesn’t have a slope. The words “the” and “a” are bold and red coloured here to stress the idea that there is no single acceleration under these circumstances. The acceleration of such an object must be changing. It’s motion has decreasing and variable acceleration.The graph has hyperbolic nature.  Case-4 : The object having increasing acceleration

When a velocity-time graph is curved, it is not possible to calculate the slope from it’s slope. Slope is a property of straight lines only. Such an object doesn’t have a acceleration because it doesn’t have a slope. The words “the” and “a” are bold and red coloured here to stress the idea that there is no single acceleration under these circumstances. The acceleration of such an object must be changing. It’s motion has increasing and variable acceleration.The graph has parabolic nature. Case-5 : The object with constant negative acceleration

The nature of the above has straight line and have y-intercept . It has negative but constant slope . So, it is the graph of retarding object  with constant rate The nature of the graph is also a straight line but it has positive and constant slope . So , this graph represents the object having some initial velocity and constant acceleration.

Case-7 : Area under curve

The area under curve of the velocity-time graph represents the displacement of the object . In the above graph

Area of the red triangle=½*base*height

=½*(5sec)*(0.8 m/sec)

=2.0 m

` =displacement of the object during 0 sec – 5sec

Acceleration-time graph:

Note:

On an acceleration-time graph…

• slope is meaningless.
• the “y” intercept equals the initial acceleration.(figure on the right hand side )
• when two curves coincide, the two objects have the same acceleration at that time.
• an object undergoing constant acceleration traces a horizontal line.(figure B)
• zero slope implies motion with constant acceleration.(figure B )
• the area under the curve equals the change in velocity.(last figure) fig : 2   Equation of motion

In circumstances of constant acceleration, these simpler equations of motion are usually referred to as the SUVAT equations, arising from the definitions of kinematic quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).

1. s = ut + ½ at2
2. v = u + at
3. v= u+ 2as
4. snth = u + ½(2n-1)

where n is the particular instant of time and snth is the displacement at that instant of time.

We have already proved these equations of motion in class 9

Numerical problems

Question 1 :A small coin is dropped down a well. The splash is heard 2.3 sec later . How deep is the well ? (assuming that the sound travels quickly enough for any delaying effect to be ignored. )

Solution :

Given :

time of hearing ( t ) = 2.3 sec

Acceleration due to gravity ( g ) = 9.8 m /sec2

Initial velocity of the coin ( u ) = 0 m /sec

Equation of motion is given by

s = ut +1/2gt2

Where s is the total displacement which is the depth of the well

s = (0*2.3) + ½ * 9.8 * 2.3 * 2.3

s = 0 + 25.921

Hence, the required depth of the well is 25.921 m.

Question 2 : The motion of a particle along x-axis is given by the equation x = 9 + 5t2 , where x is the distance in cm and t is the time is the time in sec. Find (a) the displacement after 3 sec and 5 sec. (b) average velocity during the interval from t = 3sec to t = 5 sec.( c ) instantaneous velocity at t = 3 sec.

solution :

Given : equation of motion is

x = 9 + 5t2————-(1)

(a) displacement a fter 3 sec i.e at t = 3sec, equation (1) gives

x =9 + 5 * 32

x = 9+(5*9)

x=9+45

x=54

Hence, the required displacement is 54 cm

Similarly, we can find the value of displacement after 5 sec

At t = 5 sec equation (1) gives

x = 9 + (5*52)

x = 9 +125

x =134

So, the required displacement after 5 sec is 134 cm

(b) x=9+5t2

Differentiating both sides w.r.t ‘t’ we get

dx/dt=0+5*(2t)

v=10t

At t=3 sec

v3=10*3

v3=30

And at t=5 sec

v5=10*5

v5=50

Average velocity during 3sec to 5sec =(v3+v5)/2

=(30+50)/2

=80/2

=40 cm/sec

(c) x=9+5t2

Differentiating both sides w.r.t ‘t’ we get

dx/dt=0+5*(2t)

v=10t

At t=3 sec

v=10*30

v=30

Therefore , instantaneous velocity at t=3sec is 30cm/sec

Question 3 : An object is thrown vertically upward with a velocity of 19.6 m/sec. Calculate the distance and displacement of the object after 3 sec.

Solution :

Given :

Upward velocity = final velocity of the object (v) = 19.6m/sec

Initial velocity of the object (u) = 0 m / sec

Acceleration due to gravity (g) = 9.8 m / sec2

Time = 3 sec

Let

time to reach maximum height = T sec

From equation of motion , we have

v = u+gT

19.6 = 0+9.8*T

19.6/9.8 = T

T = 2 sec

Therefore , in first 2 sec the object covers a distance of 19.6 m

And let us assume that , in the next second the object will cover s metre

Therefore , equation of motion is given by

s = ut + 1/2gt2

s = 0+½(9.8*1*1)

s=4.9 = BC

Total distance=AB+BC =19.6+4.9

=24.5 m

displacement=AB-BC= 19.6-4.9

=14.7 m

Question 4 : The displacement of a particle moving in one dimension under the action of constant force is related to time t by the equation t=x +3, where x is in metre and t is in sec . Find the displacement of the object when its velocity is zero . [IIT 1979]

Solution :

t = + 3

Differentiating both sides w.r.t x we get

dt/dx={1/(2x)}+0

1/v=1/(2x)

v=2x

When velocity is zero i.e at v=0 m/sec

0=2x

Squaring both sides we get

x=0

When velocity is zero, then the displacement is also zero .

Question 5 :A  train was moving at the rate of 36 km/hr . When brakes were applied , it came to rest in a distance of 200m. Calculate the retardation produced in the train .

Solution :

Given :

Initial velocity (u) = 36 km/hr=36* (1000m/3600 sec)

=10 m/sec

Displacement (s) = 200 m

Final velocity (v) =0 m/sec

From equation of motion we have

v2 = u2+2as

Where a is acceleration

0=102+2a*200

400a=-100

a=-(100/400)

a=-0.25

Hence , retardation produced in the train is 0.25m/sec2.

Question 6 : A train moving with a velocity of 48 km/hr is brought to rest in 11 sec to avoid collision . What is the retardation of the train in m/s2 and km/hr2?

Solution:

Given:

Initial velocity (u) = 48 km/hr

Time (t)  = 11 sec= 11/3600 km/hr

Final velocity (v) = 0 km/hr

Let a denotes acceleration

From equation of motion , we have

v=u+at

0=48+a*(11/3600)

a=-{48*(3600/11)}

a=-1.57*104 km/hr2

Retardation in km/hr2=1.57*104 km/hr2

Retardation in m/sec2=1.57*104 *1000/(3600)2 m/sec2

=1.57/362 * 103

=1.21 m/sec2

Question 7 : A weight is dropped from the top of a building 135m height . How fast is the weight moving just before it hits the ground?

Solution:

Given :

Height of the building (h) = 135 m

Acceleration due to gravity (g) = 9.8 m/sec2

Initial velocity (u) = 0 m / sec

Final velocity (v) =  ?

The equation of motion is given by

v2 =u2 +2gh

= 02 + (2*9.8*135)

=2646

v = 2646

= 51.439

Required velocity is 51.439 m /sec

Question 8 : A boy throws a ball upwards with velocity of 9.8 m/s. How high does it go ? How long was it in the air?

Solution:

Given:

Initial velocity (u)=9.8 m/sec

Acceleration due to gravity (g) = 9.8 m/sec2

Final velocity = velocity at the top (v)=0 m/sec

Maximum height (h) =?

The equation of motion is given by

v2 =u+ 2(-g)h

0 = 9.8– (2*9.8)h

h = 96.04 / 19.6

h=4.9

The ball goes 4.9 m high

Let us assume that T is the total time period.

Time taken to cover maximum height = time taken to return its initial position=t (say)

We have

v=u+(-g)t

0=9.8-9.8t

t=9.8/9.8

t=1

T=t+t

T=1+1

T=2 sec

Therefore, the ball was 2 sec in the air

Question 9 : A juggler throws a ball into air . He throws one whenever the previous one is at its highest point.How high do the balls rise if he throws n balls eac sec. Acceleration due to gravity is g .

Solution:

Given:

Initial velocity (u)=0 m/sec

Acceleration due to gravity=g

Final velocity=v m/sec

Maximum height=h

Time taken to reach maximum height = t sec

No of balls thrown balls = n

Now

v=u+gt

v=0+gt

v=gt

And

v2=u2+2(g)s

(gt)2=0+2gs

g2t2=2gs

s=gt2 /2————————————-(1)

Each ball is thrown in each second. Therefore t=1/n

Putting the value of t in equation (1)

s = g / 2n2

Question 10 : A stone is thrown upwards from the top of a tower 85m high . It reaches the ground in 5 sec. Calculate (a) the greatest height above the ground (b) the velocity with which it reaches the ground and (c) the time taken to reach maximum height . Given: g=10m/s2.

Solution:

Given :

Height of the tower (h)=85 m

Time taken to reach the ground (t) = 5 sec

Initial velocity (u) = the velocity with which the stone is thrown =???

Acceleration due to gravity (g) = 10 m/sec2

Height from the top of the tower, which the ball has attained =s

(a) Greatest height above the ground (H)=?

The equation of motion is given by

v2=u2+2gs

The equation of motion is given by

h=ut+½ at2

85=5u+½ *10*5*5

u=-8 m/sec

And v2=u2+2gs

0= (-8)2+2*10*s

s= -64/20

s = -3.2

So the greatest height above the ground = h – s

=85-(-3.2)

=85+3.2

=88.2 m

(b) the velocity with which it reaches ground (v)=??

Equation of motion is given by

v=u+gt

v=-8+(10*5)

v=-8+50

v=42 m/sec

(c) time taken to reach the maximum height (T)=????

At the maximum height , velocity is zero

v=u+gT

0=-8+10T

T=8/10

T=0.8 sec

Question 11 :

(a) A stone is dropped from a balloon moving upwards with a velocity of 4.5m/sec . The stone reaches the ground in 5 sec . Calculate the height of the balloon when the stone was dropped .

(b) A stone falls freely under gravity , starting from rest . calculate ratio of distance

travelled by the stone during the first half of any interval of time to the distance

travelled during the second half of the same interval .

Solution:

(a)Given:

Initial velocity (u)=-4.5 m/sec    ( as the motion of the stone is in downward direction )

Time to reach the ground (t)=5 sec

Final velocity(v)=0 m/sec

The height of the balloon when the stone was dropped = maximum height attained by

the stone (h) =????

The equation of motion is given by

h=ut+½ gt2

h=-(4.5*5)+(½ *9.8*5*5)

h=100

The required answer is 100 m

(b) Initial velocity (u)=0 m/sec.

Acceleration due to gravity =g

Time to reach the maximum height =t sec

x= distance travelled  during first half

y= distance travelling  during the second half

For the first half motion , the equation of motion is given by

x=u(t/2)+½ g(t/2)2

x=0+gt2/8

For the second half motion of the same interval , the equation of motion is

y=ut+½ gt2-x

y=0+½ gt2-gt2/8

y=gt2/2-gt2/8

=⅜ gt2

The required ratio is given by

x:y=gt2/8:⅜ gt2

Question 12 : A stone is dropped from a balloon at an altitude of 300m. How long will the stone take to reach the ground if (a) the balloon is ascending with a velocity of 5 m/sec. (b) the balloon is descending with a velocity of 5 m/sec (c) the balloon is stationary ?

Solution:

Given:

Height (h) =300m

Time to reach the ground (t)=????

(a)  ascending velocity (u)=initial  velocity =-5m/sec

Acceleration due to gravity (g)=9.8 m/sec2

The equation of motion is given by

h=ut+½ gt2

300=(-5)*t+½ *9.8*t22

4.9t2-5t-300=0

It is a quadratic equation . On solving this equation we get

t=8.35sec ( rejecting negative values )

(b) descending velocity (u) =+5msec

Acceleration due to gravity (g) = +9.8 m/sec2

The equation of motion is given by

h=ut+½ gt2

300=5t+(½ *9.8)t2

4.9t2+5t-300=0

It is the quadratic equation . By rejecting the negative value we get

t=7.33 sec

(c) when the balloon is stationary , the initial velocity (u) 0 m/sec

Acceleration due to gravity (g) = +9.8 m/sec2

The equation of motion is given by

h=ut+½ gt2

300=0+½ *9.8*t2

t2=300/4.9

t=300/4.9

t=7.825 sec

Question 13 : A ball rolls down an inclined plane 2 m long in 4 second.Find (a) its acceleration (a) the time taken to cover the second metre of the track (c) the speed of the ball at the bottom of the track.

Solution:

Given :

Length of the plane (s)=2m

Time (T)=4 sec

(a) initial velocity (u) = 0 m/sec

Acceleration (a)=???

The equation of motion is given by

s= ut +½ at2

2=0+½ *a*4*4

a=2/8

a=0.25 m / sec2

(b) Time taken to cover the first metre of the track (t) =???

Distance (s) = 1m

Initial velocity (u) =0 m/sec

Acceleration (a)= 0.25 m/sec2

The equation of motion is given by

s=ut+½ at2

1=0+½ *0.25*t2

t=1/0.125

t=2.83 sec

Time taken to reach the second metre of the track = T – t

=4-2.83

=1.17 sec

(c) speed of the ball at the bottom of the track (v) =???

Initial velocity (u) =0 m/sec

Acceleration (a)= 0.25 m/sec2

Time to reach at the bottom (t) = 4 sec

The equation of motion is given by

v=u+at

v=0+0.25*4

v=1 m/sec

Question 14 : A body travels 20 cm in the first 0.2 second and 22 cm in the next 0.4 second.What will be the velocity at the end of the 0.7th sec from rest?

Solution :

Given:

Let

Initial velocity =u

Final velocity=v

Displacement =s

Time =t

Acceleration =a

The equation of motion is given by

s=ut + ½ at22

When s=20 cm and t=0.2 sec

20=0.2u+0.02a————————-(1)

When s=22 cm and t=0.4 sec

22=0.4u+0.08a————————-(2)

On solving the the above two equations , we get

u=115 cm/sec

And a=-150 cm/sec2

v=u+at

=115+(-150*0.7)

=10 cm /sec

Question 15 : A stone falls from a cliff and travels 34.3 m in the last sec before it reaches the ground .calculate the height of the cliff.

Solution:

Given :

Let n is the last second

Height of the cliff (s)=????

u= initial velocity =0 m /sec

snth=displacement in the last second = 34.3 m

g=acceleration due to gravity = 9.8m/sec2

The equation of motion is given by

snth = u+g(2n-1)/2

34.3 = 0+9.8*(2n-1)/2

(34.3*2)/9.8 = 2n-1

2n-1  = 7.15

n = (7.15+1)/2

n = 4.075

n may be considered as 4 sec

Again we have

s=ut+½ gt2

=0+½ *9.8*4*4

=78.4 m

Question 16 : During the last second of its free fall , a body covers half of the total distance travelled . Calculate (a) the approximate height from which the body falls (b) the duration of the fall.

Solution:

Given :

Let

n = last second

(a)  let the approximate height from which the body falls (h)=??

Distance covered in the last second (s)=h/2

Initial velocity =u

Acceleration due to gravity (g)= 9.8 m/sec2

The equation of motion is given by

s=u+g(2n-1)/2

h/2=0+9.8(2n-1)/2

h=9.8(2n-1)—————————-(1)

Again we have

h=ut+½ *gt2

h=0+4.9n2

h=4.9n2————————(2)

Putting the value of h in the equation (1)  we get

4.9n2=19.6n-9.8

4.9n2-19.6n+9.8=0

It is the quadratic equation . By rejecting the negative value of n we get

n=3.41 sec

Now putting the value of n in equation (2) we get

h=4.9n2

=4.9*3.41*3.41

=56.98 m

(b) the duration of the fall= n

=3.41 sec

Questions 17 Between two stations a train  accelerates uniformly at first , then moves with constant speed and finally retards uniformly . If the ratio of time taken is 1:8:1 and the greatest speed is 60 km/hr , find the average speed over the whole journey .

Solution:

Given:time taken ratio is given by

x:y:z=1:8:1

x=t

y=8t

z=t

Where t is non zero constant

Total time (T)=x+y+z

=t+8t+t

=10t

Greatest speed(v)=60 km/hr

Total distance (s)=½ *(y+T)*v

=½*(8t+10t)*60

=540t km

Average speed =s/T

=540t/10t

=54 km /hr

Question 18 : A car starts from rest and accelerates uniformly for 10 sec to a velocity of 8 m/sec . It then runs at a constant velocity and is finally brought to rest in 64 m with a constant retardation . The total distance covered by the car is 584m . Find the value of acceleration , retardation and total time taken .

Solution:

Given:

Calculation of acceleration :

Initial velocity (u) = 0 m/sec

Final velocity (v) = 8m/sec

Time (t)=10 sec

Acceleration (a) =???

The equation of motion is given by

v=u+at

8=0+10a

a=8/10

a=0.8 m/sec2

Therefore, the required acceleration is 0.8 m/sec2

Calculation of retardation :

Distance (z)=64 m

Initial velocity (u)= 8m/sec

Final velocity (v)= 0 m/sec

The equation of motion is given by

v2=u2+2az

0=82+2a*64

a=-(64/128)

a=-0.5 m/sec2

The required retardation is 0.5 m/sec2

Calculation of total time :

Let

x=distance covered when the acceleration is uniform

y=distance covered when the velocity is constant

z= distance covered when the retardation is constant =64 m

p=time taken to cover x distance =10 sec

q=time taken to cover y distance

r=time taken to cover z distance

Total distance (s) = 584 m

s=x+y+z

⇒x+y+z=584————————-(1)

x=ut+½ ap2

x=0+½ *0.8*10*10

x=40 m

And y=584-(x+z)

y=584-(40+64)

y=480 m

q=y/8

=480/8

=60 sec

z=8r+½ *(-0.5)*r2

64=8r+½ *(-0.5)*r2

0.25r2-8r+64=0

It is the quadratic equation . on solving we get

0.25r2-4t3-4r+64=0

0.25r(r-16)-4(r-16)=0

(r-16)( 0.25r-4)=0

Either r=16 sec

Or r=4/0.25

=16 sec

Total time = p + q + r

=10+60+16

=86 sec

=1 minute 26 sec

Question 19 : A hundred metre sprinter increases her speed from rest uniformly at the rate of 1 m/sec2 upto three quarters of the total run and covers the last quarter with uniform speed . How much time does she take to cover the first half and second half of the run ?

Solution :

Given :

Let total distance = (25+25+25) metre

=75 m

In the first and second quarters acceleration (a) is 1 m/sec2

initial velocity (u)=0 m /sec

The equation of motion is given by

For the first half of the total run

s=ut +½ at2

50=0+½ *1*t2

100=t2

t=100

t=10 sec

Again we have

v=u+at

v=0+1*10

v=10 m/sec

For the first 75 m distance

v2=u2+2as

v2=0+2*1*75

v=12.2 m/sec

The time taken to cover the first 75 m distance = (v-u)/a

=(12.2-0)/1

=12.2 m/sec

Time taken to cover last 25 m distance = distance /constant speed

= 25/12.2

=2 sec

Total time taken to run 75 m distance = 12.2+2

=14.2 sec

Time taken to cover the second half of the run =total time -10 sec

= 14.2-10

=4.2 sec

Question 20 : A stone is dropped from the top of tower 100m high . At the same time , another stone is thrown vertically upwards with a velocity of 50 m/sec . When and where the stones will meet ?

Solution:

Given:

Height of the tower (h)= 100 m

Let the two stones will meet at a distance of x metre from the ground at t sec

For the dropping stone :

Initial velocity (u)=0 m/sec

Height (h)=(100-x) m

Acceleration due to gravity (g)=9.8 m/s

The equation of motion is given by

(100-x)=ut+½ *gt2

100-x=4.9t2———————(1)

For the motion of the stone thrown vertically upward :

Acceleration due to gravity (g)= -9.8 m/sec

Initial velocity (u)=50 m/sec2

Height (h)= x m

The equation of motion is given by

h=ut+½ *gt2

x=50t+½ *(-9.8)*t2

x=50t-4.9t2———————-(2)

Putting the value of  in the equation (1)

100-50t+4.9t2=4.9t2

100-50t=0

t=2 sec

Putting the value of t = 2 sec in the equation (2) we get

x=50*2-4.9*2*2

x=80.4 m

Therefore , the stones will meet at a distance of 80.4 m from the ground after 2 sec

Question 21 : A stone is thrown vertically upwards with a velocity of 19.6m/sec. After two sec another stone is thrown upwards with a velocity of 9.8m/sec . When and where the stones will collide ?

Solution:

Let two stones were collide at a distance of x metre from the ground at t sec after

throwing the first stone

For the first stone :

Initial velocity (u)= 19.6 m/sec

Acceleration due to gravity (g)= -9.8 m/sec2

The equation of motion is given by

x=ut+½ gt2

x=19.6t+½ *9.8 *t2

x=19.6t+4.9t2—————————-(1)

For the second stone which is thrown 2 sec later:

Initial velocity (u) = 9.8 m/sec

Acceleration due to gravity (g)=- 9.8 m/sec2

The equation of motion is given by

x=u(t-2)+½ g(t-2)2

x=9.8*(t-2)-4.9(t-2)22e two equations we get

19.6t-4.9t2=9.8(t-2)+4.9(t-2)2

19.6t-4.9t2=9.8t-19.6-4.9t2+19.6t-19.6

9.8t=19.6+19.6

9.8t=39.2

t=39.2/9.8

t=4 sec

Putting the value of t in equation (1) we get

x=19.6t-4.9t2

x=(19.6*4)-(4.9*4*4)

x=0 m

Hence, the two stones will collide at the ground at 4 sec after throwing the first stone .

Question 22 : Two balls are thrown simultaneously, A vertically upwards with a speed of 20 m/sec from the ground and B vertically downwards from a height 40m with the same speed and along the same line of motion . at what point do the two balls collide? Given g=9.8 m/.sec2

Solution:

Let the two balls collide at a distance of x metre from the ground

For ball-A

Initial velocity (u) = 20 m/sec

Acceleration due to gravity (g) = -9.8 m/sec2

The equation of motion is given by

x=ut+½ gt2

x=20t-4.9t2——————–(1)

For ball-B

Initial velocity (u)= -20 m/sec

Height of throwing = 40m

Acceleration due to gravity (g) = -9.8 m/sec2

The equation of motion is given by

-(40-x)=ut+½ gt2

-(40-x)=-20t-4.9t2

x=40-20t-4.9t2——————-(2)

Comparing the two equations we get

20t-4.9t2=40-20t-4.9t2

20t+20t=40

t=40/42

t=1 sec

Putting the value of t in equation (1) we get

x=20t-4.9t2

x=(20*1)-(4.9*1*1)

x=20-4.9

x=15.1 m

So the two balls collide at a height of 15.1 m above the ground

Question 23 : A stone is dropped from the top of tall tower and after 1 sec , another stone is dropped from balcony 20 m below the top . If both the stones reach the ground at the same instant , calculate the height of tower . Given g=10 m/sec2.

Solution:

Let the height of the tower is x metre

For first stone :

Initial velocity (u)= 0 m/sec

Acceleration due to gravity (g)= +10 m/sec2

The equation of motion is given by

x=ut+½ gt2

x=5tv—————————-(1)

For the second stone :

Initial velocity (u)=0 m/sec

Acceleration due to gravity (g)= +10 m/sec2

The equation of motion is given by

x-20=ut+½ gt2

x-20=5(t-1)2

x-20=5t2-10t+5————————–(2)

Subtracting equ.(2) from (1) we get

20=10t-5

20+5=10t

t=2.5 sec

Putting the value of t in equation (1)

x=5*2.5*2.5

x=31.25 m

Hence , the height of the tower is 31.25 m.

1 ) Is it true that a body is always at rest in a frame which is fixed to the body itself ?

2 ) Two particles A and B are moving along the same straight line B is ahead of A , velocities remains unchanged , what would be the effect on the magnitude of relative velocity if A is ahead of B ?

Answer: There will be no effect on the magnitude of relative velocity .

3 )What conclusion you draw if the average velocity is equal to instantaneous velocity ?

Answer: the particle is moving with constant velocity

4 ) Under what condition the magnitude of average velocity is equal to the average speed ?

Answer: when the particle is moving with constant velocity , the average velocity and

Speed are equal to each other .

5 ) Why the speed of an object can never be negative ?

Answer: speed is distance per unit time and distance can never be negative .

Therefore speed can also never be negative .

6 ) Name a physical phenomenon in which earth can not be regarded as a point object?

7 ) Can a particle have varying speed but constant velocity ?

Answer: no. It is not possible as both magnitude and direction are constant in uniform

Velocity .

8 ) The average velocity of a particle is equal to instantaneous velocity . What is the shape of displacement – time graph ?

Answer: when the average velocity and instantaneous velocity are equal to each

other , the velocity is uniform . Therefore , the displacement-time graph is a

Straight line .

9 ) Is it possible that the velocity of an object be in a direction other than the direction of acceleration?

10 ) Is it possible to have the rate of change of velocity constant while the velocity itself changes both in the magnitude and direction?

Answer: yes . In projectile motion , it is possible .

11 ) If the acceleration of the particle is constant in the magnitude but not in direction , what type of path does the body follow?

Answer: If the acceleration is constant in magnitude not in direction then the body

follows circular path .

12 ) Is there any quantity in physics which deals with the time rate of change of acceleration ?

Answer: No . There is no such quantity

13 ) An iron ball and wooden ball of same radius are released from a height in vacuum . which of the two balls would take more time ? [ Orissa 1991 ]

Answer: both will take same time .

14 ) What is the general relation between the direction of velocity and acceleration ?

Answer : There is no specific relation between the direction  of velocity and acceleration.

15 ) In the case of uniform acceleration , the displacement is the product of two quantities name these quantities .

16 ) Is the kinematic equation s=ut+½ at2 true if acceleration is not constant ?

Answer : No. All the kinematic equation will be true only for uniform motion .

17 ) What is the velocity-time graph for a body moving with uniform velocity ? [EAMCET 1682]

Answer: straight line parallel to time – axis .

18 ) A piece of paper and iron piece are dropped simultaneously from the same point . They will reach the ground simultaneously if they fall in …………….. Fill in the blank         [karnataka 1995]

19 ) A body thrown up vertically reaches a maximum height of 100 m . Another body with double the mass thrown up with double the initial velocity reaches a maximum height h. What is the value of h?        [EAMCET 1987]

20 ) Two stones are dropped down simultaneously from different heights . At the time of starting , the distance between the stones is 30 cm After two seconds what will be the distance between the two stones ?

21 ) Even when rain is falling vertically downwards, the front screen of a moving car gets wet . On the other hand , the back screen remains dry . ………………..Why ?

Answer: The rain strikes car in the direction of relative velocity of rain with respect to

car.

22 ) Comment on the numerical ratio of speed and velocity of an object ?

Answer: The ratio is greater than or equal to one .

23 ) What is the acceleration of a particle moving with uniform velocity ?

24 ) Mention a one consequence of the fact that instantaneous acceleration does not depend upon instantaneous velocity .

Answer: The direction and velocity are not related to each other .

25 ) What is the relative acceleration of two equally accelerated bodies?