Selina Concise Class 9 Physics Chapter 2 Motion in One dimension Solution

Selina Concise Class 9 Physics Solution Chapter No. 2- ‘Motion in One dimension’ For ICSE Board Students.

Exercise – A Solution:

1.) Solution:

Ans:

Scalars:

  • Scalar quantities are those quantities which are having magnitude only.
  • Scalar quantities has magnitude and also unit.
  • Some purely numbers are scalars which are not having unit.

For example:

  • Π, e2 are the purely numbers which are scalars but having no unit.
  • When we measure distance then we consider only magnitude not the direction.
  • When we measure mass of any quantity then we consider magnitude only.
  • Thus, mass, distance, length, temperature, work, energy, pressure etc all are the scalar quantities which are not having direction but has only magnitude.

Vectors:

  • Vectors are the physical quantities which are having both direction as well as magnitude also.
  • They are expressed in the form of a vector which gives magnitude, direction and also has specific unit.

For example:

  • When we measure the displacement of any object then we consider both direction as well as magnitude. Hence, displacement is the vector quantity.
  • When we apply a force on object then we apply pressure per unit area in a particular direction. Hence, force is also the vector quantity.
  • Thus, velocity, displacement, acceleration, momentum, force, moment of force, weight, impulse, temperature gradient, electric field etc all are the vector quantities which has magnitude, direction and also some specific unit.

2.) Solution:

Ans:

a) Pressure

Pressure has magnitude only with unit. Hence, it is the scalar quantity.

b) Force:

Force has both direction and magnitude hence it is the vector quantity.

c) Momentum:

Momentum is the product is mass and velocity hence it is the vector quantity.

d) Energy:

Energy has only magnitude with unit also, hence it is the scalar quantity.

e) Weight:

Weight is acted in particular direction with a magnitude, hence it is the vector quantity.

f) Speed:

Speed has only magnitude not the direction and hence it is the scalar quantity.

3.) Solution:

Ans:

  • If a body does not changing its position with respect to the surrounding and time then we say that the body is at rest.
  • The body remains in its state of rest until we apply an external force.

For example:

  • When we go in garden, we see that all trees are at rest because they all are at rest with respect to their surroundings.
  • But if we see the kids playing on see- saw then we observe that it is in motion.
  • Here, is the difference between state of rest and state of motion.

4.) Solution:

Ans:

When any object or body is moving with particular velocity which may uniform or non-uniform i.e. that object is in motion with respect to its surrounding. Thus we can say that the body is in its state of motion.

For example:

When we are seating in the train which is at rest then we feel also at rest. But when train leaves the platform i.e. it starts moving then also we feel at rest because we are in the train. But the people which are seating on platform observes us in a state of motion as we are moving along with train by seating in train.

Thus, the body is said to be in a state of motion.

5.) Solution:

Ans:

When any object or body moves along the straight line or along the straight path then it is said that the body is in motion along one direction. This motion of body is also called as rectilinear motion as takes a motion along the straight line path.

For example:

  • When a body like train is moving on a track then it is said that train is in motion in one direction.
  • Also, a car moving along straight road then it is also in motion in one direction. In rectilinear motion or motion in one direction the body doesn’t changes its direction, it moves along the straight line path only.
  • And most importantly, the motion in one direction is the one dimensional motion always. If a body changes direction of motion or takes any curved path then its motion is in two dimensions.

6.) Solution:

Ans:

  • The distance between initial and final position of the body with its direction of motion is considered then it is said that the displacement.
  • It is the vector quantity as it has both magnitude as well as direction.
  • The SI unit of displacement is meter.

For example:

  • If a object moves from position A to position B, and then returns back to again A. If we considered the distance between A and B is 5m then the total distance covered by the body is 10m.
  • Because in case of distance we cannot consider the direction of motion.
  • But, if we have to find the displacement of the same body then it is here zero. As in case of displacement we consider direction and magnitude also.

7.) Solution:

Ans:

Distance:

  • It is the path length of the object covered in a particular period of time.
  • As it is the path length, it is called as scalar quantity.
  • It has only magnitude.
  • Its SI unit is meter.
  • It has magnitude and hence it is always a positive quantity.
  • Its value is always greater than or may equal to the value of magnitude of displacement.
  • If the displacement is zero, then also distance is never zero.

Displacement:

  • Displacement is the shortest distance covered by the object in specific direction and in a particular time.
  • As it depends on direction of motion and has magnitude also, it is a vector quantity.
  • Its SI unit is meter.
  • It doesn’t depends on the path followed by the particle.
  • As it depends on direction, it may be positive or negative.
  • Its value is zero if distance is zero, and displacement may be zero even if distance is not zero.
  • The magnitude of displacement is less than or sometimes equal to magnitude of distance but never be greater than the magnitude of distance.

8.) Solution:

Ans:

Yes, displacement may be zero even if distance is not zero. Because, distance is a scalar quantity and displacement is a vector quantity.

When we consider distance we take only magnitude but when we consider displacement we take both direction and magnitude and hence displacement may be zero even if distance is not zero.

For example:

  • Let, a car is moving along a straight road and reach at a particular spot. Then again it comes back to the initial spot by the same way.
  • In this case, we consider the path length is 30km.
  • Hence, during its complete motion the total distance covered is 30+ 30 = 60 km.
  • And displacement becomes, 30 – 30 = 0.
  • Thus, we can say that the displacement may be zero even if distance is not zero.

9.) Solution:

Ans:

The magnitude of displacement and distance is zero only when the motion of any object or body is in a particular fixed direction only. It doesn’t have to take a curved or zigzag path.

For example:

Let us suppose a bike is moving along a road in fixed direction, then the direction of its motion doesn’t changes which means that magnitude of distance and displacement is equal.

10.) Solution:

Ans:

Velocity is the rate of change of displacement, that means when a body changes its position with respect to time then we say that it has specific velocity which is required for its motion.

The SI unit of velocity is meter per second i.e. m/s.

11.) Solution:

Ans:

The speed of any object is defined as the rate of change of distance with respect to time. That means it is the distance covered by the object in unit time. The speed is the scalar quantity.

The SI unit of speed is m/s.

12.) Solution:

Ans:

Speed:

  • Speed is defined as the rate of change of distance or the distance covered by the object in one second when it is in motion.
  • Speed is the scalar quantity as it has only magnitude.
  • Speed always taken as positive as it gives the magnitude not direction.
  • During one complete rotation of circular path, the speed is not zero.

Velocity:

  • Velocity is defined as the rate of change of displacement that means it is the distance covered by the object in one second in a particular direction.
  • Velocity is a vector quantity as it has both magnitude and direction.
  • Velocity may be takes as positive or negative based on its direction.
  • The SI unit of velocity is m/s.
  • The average velocity of complete circular motion is takes as zero.

13.) Solution:

Ans:

Speed is the scalar quantity which has magnitude only.

While the velocity of the body has magnitude as well as direction and hence velocity gives the direction of motion of a body.

14.) Solution:

Ans:

  • Instantaneous speed is the ratio of distance covered in short time interval and the time interval.
  • It gives the speed at any instant if time.
  • While average speed is the ratio of total distance travelled and the total time taken to cover that distance.
  • When a body is moving with uniform speed i.e. with constant speed then both the instantaneous speed and average speed of the body will be same and its value is equal to the value of uniform speed.

15.) Solution:

Ans:

Uniform velocity:

When a body covers equal distance in equal interval of time in particular direction then it is said that the body is moving with constant or uniform velocity.

For example:

  • The moon is revolving around the earth with uniform velocity.
  • The earth is also revolving around the sun with uniform velocity.
  • And their motion is circular motion, hence it is called as uniform circular motion also.

Variable or non-uniform velocity:

When a body is moving with velocity which is varying alternately in equal interval of time in a particular direction then it is said to be non-uniform or variable velocity.

Sometimes, body moves with equal velocity in equal interval of time but if it’s direction of motion changes continuously then it is said that the body is moving with variable velocity.

For example:

  • When a object is falling from particular height, its direction of motion remains same but the speed of motion changes continuously. Hence, it has the variable velocity.
  • Also, a body performing circular motion with uniform speed is also the example of variable velocity, because although it is moving with constant speed but it’s direction of motion changes continuously at each point on the circular path. Hence, we can say that it has variable velocity.

16.) Solution:

Ans:

Average speed:

  • Average speed is defined as it is the ratio of total distance covered by the body and the total time taken.
  • Average speed is the scalar quantity.
  • Its SI unit is m/s.
  • Average speed is always positive as it is a vector quantity.

Average velocity:

  • Average velocity is defined as it is the ratio of total displacement covered by the body and the total time taken.
  • Average velocity is the vector quantity as it has both magnitude and direction.
  • Average velocity may be positive or negative depending on the it’s direction of motion.
  • The SI unit of average velocity is m/s.

17.) Solution:

Ans:

Variable or non-uniform velocity:

When a body is moving with velocity which is varying alternately in equal interval of time in a particular direction then it is said to be non-uniform or variable velocity.

Sometimes, body moves with equal velocity in equal interval of time but if it’s direction of motion changes continuously then it is said that the body is moving with variable velocity.

For example:

  • A body moving with uniform speed along a circular path is not having a constant velocity. Because, the direction of velocity of a body at every point along the circular path changes continuously. Hence, it has variable velocity.
  • The direction of velocity at any point is in the direction of tangent at that point as shown in figure below.
  • The earth is moving around the sun with uniform speed but with variable velocity.

18.) Solution:

Ans:

Distance and speed are the scalar quantities having magnitude only and positive value always.

While the displacement and velocity are the vector quantities which are having both magnitude and direction also. And it takes positive or negative values depending on its direction of motion.

Similarly, average speed is the scalar quantity and average velocity is the vector quantity.

When a body is moving from point A to point B and again returns to its initial state A. In this motion, the distance covered is not zero but the displacement is zero because it’s direction of motion is reversed.

19.) Solution:

Ans:

Acceleration is defined as it is the rate of change of velocity with respect to time. When the motion of body is with varying velocity then we can say that it has acceleration.

Body moving with uniform velocity never has acceleration.

The SI unit of acceleration is m/s2.

20.) Solution:

Ans:

Acceleration:

  • When the body is moving with variable velocity which changes its direction or magnitude or both then it is called as accelerated motion.
  • In accelerated motion, the velocity of the body increases with time.
  • Hence it has always positive value.

Retardation:

  • When a body is moving with variable velocity which changes its direction or magnitude or both and also its velocity is decreasing with time then it is called that it has retardation.
  • And hence retardation is the negative acceleration.

21.) Solution:

Ans:

Uniform acceleration:

Uniform acceleration is the accelerated motion in which there is change in equal velocity in equal interval of time.

Uniform acceleration is also called as constant acceleration.

For example:

The motion of a body under gravity is the uniform acceleration because here the acceleration due to gravity is same when body falls.

Variable acceleration:

Variable acceleration is the accelerated motion in which there is change in not same velocity in equal interval of time.

Hence, in variable acceleration the acceleration is different at different time interval throughout the motion of body.

For example:

A bike is moving in market is having the variable acceleration. Because its acceleration changes after equal interval of time.

22.) Solution:

Ans:

Retardation:

  • When a body is moving with variable velocity which changes its direction or magnitude or both then it is said that its motion is accelerated motion.
  • But, if in accelerated motion the velocity of body decreases with time then it is said that retardation.
  • Hence, retardation is the decrease in velocity per second.
  • Thus, SI unit of retardation is m/s.

23.) Solution:

Ans:

  • Velocity is defined as the rate of change of displacement per unit time in a particular direction.
  • Velocity of the body is the displacement covered by the body with respect to time in a particular direction.
  • Hence, only velocity gives the direction of motion of body not acceleration.
  • The SI unit of velocity is m/s.

24.) Solution:

Ans:

a) Uniform velocity:

When a body is moving with constant velocity then it is said that it has uniform velocity.

When the body covers equal distance in equal interval of time in a particular direction then the motion is said to be uniform motion.

For example:

While raining, all the raindrops falls on the earth’s surface with uniform velocity.

b) Variable velocity:

When a body covers unequal distance in equal interval of time in a particular direction then its velocity is variable velocity.

Or when the body covers equal distance in equal interval of time but its direction of motion is changing continuously then we say that its velocity is variable velocity.

For example:

When a body is performing uniform circular motion, then its speed is same along the circular path but its direction of motion is changing continuously at each point on the circular path. Hence it has variable velocity.

c) Variable acceleration:

Variable acceleration is the accelerated motion in which there is change in not same velocity in equal interval of time.

Hence, in variable acceleration the acceleration is different at different time interval throughout the motion of body.

For example:

A bike is moving in market is having the variable acceleration. Because its acceleration changes after equal interval of time.

d) Uniform retardation:

When a body is moving with variable velocity which changes its direction or magnitude or both then it is said that its motion is accelerated motion.

But, if in accelerated motion the velocity of body decreases with time then it is said that retardation.

And if there is equal decrease in velocity in equal interval of time then it is the uniform retardation.

For example:

When a car is moving with some velocity and suddenly if brakes are applied then it comes to rest. This is an example of uniform retardation.

25.) Solution:

Ans:

From figure we can say that, the oil drop pattern on road is such that, firstly the drops are seen at a equal distance which means the car is moving with constant speed. Then we see that, the drops are very close to each other which indicates that the car is moving slowly.

Thus, the car is initially moving with constant speed and then moves slowly.

26.) Solution:

Ans:

  • When any object falling freely under the action of gravity, there will be acceleration is produced inside the body due earth’s gravitational force of attraction. This acceleration produced is called as acceleration due to gravity.
  • The acceleration due to gravity is denoted by g.
  • When a body falls down freely its velocity increases with time hence it has acceleration due to gravity as +g.
  • If the body is moving upward i.e. against the direction of gravity then its velocity decreases with time, hence it has acceleration due to gravity which is -g.
  • The average value of acceleration due to gravity is 9.8 m/s2.

27.) Solution:

Ans:

The acceleration due to gravity at all places on the earth’s surface is not same because the value of acceleration due to gravity is maximum at poles and then decreases towards the equator and it is minimum at the equator.

28.) Solution:

Ans:

As stone and pencil are dropped from a tower in vacuum there will be no opposing force due to air. And we know that the acceleration due to gravity is independent of mass of the body. Hence, although the mass of the pencil and stone is different the acceleration due to gravity acting on them is same. Hence, both pencil and stone falls simultaneously on the earth.

Multiple choice type:

1) A vector quantity is

Ans: d) Velocity

Because, velocity has both magnitude and direction.

2) The SI unit of velocity is

Ans: d) m/s

Because, velocity is the rate of change of displacement with respect to the time.

3) The unit of retardation is

Ans: b) m/s2

4) When a body is projected up with an initial velocity u goes to a maximum height h in time t, and then comes back at the point of projection. The correct statement is.

Ans: d) The displacement is zero.

5) 18km/h is equal to

Ans: b) 5 m/s

Numerical:

1) Solution:

Ans:

We know that, 1km = 103 m

And, 1hr= 60min = 3600 second

Thus, 72km/h = 72*103/3600

= 20m/s.

2.) Solution:

Ans:

We know that, 1km = 103 m

And 1 hrs. = 3600 seconds

Hence, 15m/s = 15*10-3*3600 = 54km/h

3.) Solution:

Ans:

a) 1km/h

1km/h = 103/3600 = 0.000278*103 m/s = 0.278m/s

b) 18km/min

18km/min = 18*103/60 = 300m/s

4.) Solution:

Ans:

All the given speed are in different units. So first we convert them in SI unit i.e. m/s only.

10m/s

1km/min = 103/60 = 16.66m/s

18km/h = 18*103/3600 = 5m/s

Thus, the value of velocity in increasing order is 5m/s, 10m/s, 16.66m/s

That is, 18km/h, 10m/s, 1km/min.

5.) Solution:

Ans:

We know that,

Speed = distance / time

Distance = speed*time = 65*3= 195 km.

Thus, the distance between the cities Agra and Delhi is 195km.

6.) Solution:

b) The average speed of the car

Ans:

Given that, a car first travels 30km with a uniform speed of 40km/h.

We know that,

Speed = distance/time

Time= distance/speed = 30/60 = 0.5 hour

 

Then, again the car travels 30km distance with the uniform speed of 40km/h.

Thus, speed = distance/time

Time = distance/ speed

Time = 30/40 = 0.75hour

Thus,

a) The total time of journey is given by,

Time = 0.5 + 0.75 hour = 1.25 hour = 1.25* 60 minutes = 75 minutes

 

b) The average speed of the car is given by,

Average speed= total distance covered/ total time taken = 60/1.25 = 48km/h.

7.) Solution:

Ans:

Given that, the distance between two station A and B is 200km.

Train takes 2 h to reach from B to A and again return to station A in 3 h.

Thus, total distance covered = 200+200= 400 km

And total time = 2+3 = 5 hour

a) Average speed is given by,

Average speed= total distance covered/total time= 400/5 = 80km/h.

 

b) Average velocity of the train is zero. As the total displacement of the train is zero. Since it starts its motion from A then reach B and again return to A.

8.) Solution:

Ans:

Given that, distance = 1km = 1000m

And time = 100 s

Thus, speed = distance/time = 1000/100= 10m/s.

 

And, velocity is given by

Velocity = displacement/ time = 1000 due in East/ 100s = 10m/s due in East.

9.) Solution:

Ans:

Acceleration = change in velocity / time = (10-0)/2 = 10/2 = 5m/s2.

10.) Solution:

Ans:

Given that, car starts from rest and acquires velocity 180m/s in 0.05 hour.

Hence, initial velocity= 0

Final velocity = 180m/s

Time = 0.05 hour = 0.05*60*60 second = 180 second

Thus, the acceleration of the body is given by,

Acceleration = final velocity – initial velocity/ time taken

Acceleration= (180-0)/180 = 1m/s2

11.) Solution:

Ans:

As the body is moving upwards that means against the acceleration due to gravity and hence its velocity decreases with time. Hence, here is the retardation which is negative.

Thus,

Retardation = final velocity- initial velocity/ time

Retardation= 20-50/3 = -30/3 = -10m/s2.

12.) Solution:

Ans:

Here, the toy car is initially moving with constant velocity of 18km/h and comes to rest in 2 second.

As here, velocity decreases the acceleration is negative called as retardation.

Hence, r

Retardation = final velocity- initial velocity/time

But, 180km/h = 18*103/3600 = 5m/s

Thus,

Retardation = 0-5/2 = -2.5m/s2

13.) Solution:

Ans:

We know that,

Acceleration = increase in its velocity / time

Increase in velocity = acceleration* time = 5*2 = 10m/s

14.) Solution:

Ans:

Given that, initial velocity= 20m/s

Retardation = -2m/s2

Time = 5 s

Thus,

Retardation= final velocity – initial velocity/ time

-2= (final velocity – 20)/ 5

-10= final velocity – 20

Final velocity = 20-10 = 10m/s.

15.) Solution:

Ans:

Given that, initial velocity= 5m/s.

Acceleration= 2m/s2

Time = 5 s

Final velocity =?

We know that,

Acceleration= final velocity – initial velocity/ time

2= (final velocity- 5)/5

10 = final velocity – 5

Final velocity = 10+5 = 15 m/s

16.) Solution:

Ans:

Given that,

a) Speed of car is 18km/h = 18*103/3600 = 5 m/s

b)

Retardation= final velocity – initial velocity/time

Retardation = 0-5/5= -1 m/s2

c)

Initial velocity = 5m/s

Acceleration = – 1m/s2

Time = 2s

Final velocity =?

We know that,

Acceleration = final velocity – initial velocity/ time

– 1 = (final velocity – 5)/2

-2 = final velocity -5

Final velocity = -2+5 = 3 m/s

Exercise – B Solution:

1.) Solution:

Ans:

When we plot a graph of displacement vs. time for uniform velocity, then the nature of the graph is a straight line through origin. Thus, we can say that the distance is directly proportional to the time.

2.) Solution:

Ans:

When we plot the graph of displacement -time, then the nature of graph helps in deciding us what is the nature of motion which is at rest or in uniform motion.

Also, the slope of displacement-time graph gives us the velocity of motion.

In this way, from velocity and time again we can plot the graph of velocity-time also.

3.) Solution:

Ans:

a) The slope of displacement-time graph gives the velocity of the particle in motion.

b) The displacement-time graph never be parallel to the displacement axis.

Because, it means that time is constant or at rest and body is displacing which never possible.

But displacement-time graph may be parallel to time axis.

4.)

5.) Solution:

Ans:

a) We know that, acceleration is the rate of change of velocity with respect to time. And the slope of velocity-time graph gives the change in velocity with respect to time. And hence we can find acceleration of a body from the slope of velocity-time graph.

b) The total area enclosed within the velocity-time graph and the X-axis i.e. time axis gives us the total distance covered by the body in a given time, in which we cannot consider the sign of motion.

c) The total area enclosed within the velocity-time graph and the X-axis i.e. time axis taken with proper sign gives us the displacement of the body in a given time.

6.) Solution:

Ans:

a) If the displacement-time graph is straight line parallel to time axis, then we can say that the body is at rest or stationary because time is increasing but the position of the body does not changing i.e. it is at rest.

b) If a straight line inclined to the time axis with an acute angle has the positive slope, which represents the motion away from the origin with uniform velocity.

c) If a straight line inclined to the time axis with obtuse angle has negative slope, which represents the motion towards the origin with uniform velocity.

d) If the displacement-time graph is a curve then it represents motion with non-uniform velocity.

7.) Solution:

We know that, the slope of displacement-time graph gives the velocity.

And slope = tanθ

Where θ is the angle made by line with the X-axis. Hence, more the angle more is the slope i.e. velocity.

From figure, line A makes greater angle with X-axis i.e. time axis than line B.

Thus, slope is greater for line A.

Hence, vehicle A is moving faster.

8)

  • Velocity-time graph a shows the straight line, that means there is a accelerated uniform motion.

For example:

A stone falling downwards from particular height.

  • The velocity-time graph b shows a curve, which is motion with variable retardation.

For example:

If a car is moving and when it reach near to its destination then it’s motion is like with variable retardation.

9) Solution:

Ans:

The following graph shows the velocity-time graph of a body moving with initial velocity u and uniform acceleration a.

We know that, the distance travelled by the body is the area enclosed within velocity-time graph and the time axis.

Thus, here the region OPQS gives the area enclosed by velocity-time graph and time axis and which is the distance covered.

Now, the region OPQS is the trapezium.

We know that, the area of trapezium is given by,

Area of trapezium = 1/2* sum of parallel sides*perpendicular distance between the parallel sides

Here, OP= u and QS= v are the parallel sides. And distance between them is t.

Thus,

Area of trapezium= 1/2*(u+v)*t = (u+v)*t/2

10.) Solution:

Ans:

We know that acceleration is the rate of change of velocity with respect to the time.

And slope of velocity-time graph gives the change in velocity with respect to the time i.e. acceleration.

11.)

Ans:

In the given figure, the angle made by line with time axis is more for line B. And we know that, slope of velocity-time graph gives the acceleration. And slope = tanθ

More is the θ more is the acceleration.

Hence, here line B has more slope than A.

Hence, the car A has the greater acceleration.

12)

13) Solution:

We know that, the slope of velocity-time graph gives the acceleration.

But, here the body is uniformly retarded and the angle made by line with time axis is obtuse which gives the negative slope.

And hence by finding the negative slope we can calculate retardation of the body.

14)

a) The displacement-time graph in figure a shows the straight line whose slope is constant i.e. velocity is constant.

Hence, in this case the acceleration of body A is zero.

b) The displacement-time graph in figure b shows the straight line whose slope is also constant i.e. velocity is constant.

Hence, in this case the body B has zero acceleration.

c) In figure c the displacement-time graph shows the curve whose slope is decreasing with time i.e. velocity is decreasing with time.

Hence, in this case the body C has negative acceleration i.e. retardation.

d) In figure d, the displacement-time graph shows the curve whose slope is increasing with time i.e. velocity is increasing with time.

Hence, in this case the body D has positive acceleration.

15.) Solution:

Ans:

The following figure shows the acceleration-time graph for a uniformly accelerated motion.

It is the straight line parallel to time axis.

The area enclosed between straight line and time interval on time axis gives the value of change in speed for that interval of time.

16.)

17.) Solution:

Ans:

For the motion under uniform acceleration such as the motion of a freely falling body, the distance and time are related by the equation as s α t2.

Hence, distance is directly proportional to square of the time.

18.)

Ans:

The following figure shows the graphical relation between the distance fallen and the square of time.

And here, the distance is directly proportional to the square of time.

The slope of this graph gives the value which is half of the acceleration due to gravity g.

Thus, by doubling the slope of s-t2 graph we can find the value of acceleration due to gravity g.

Multiple choice type:

1.) The velocity-time graph of a body in motion is a straight line inclined to the time axis. The correct statement is

Ans: b) acceleration is uniform

2.) For a uniformly retarded motion, the velocity-time graph is

Ans: d) a straight line inclined to the time axis

3.) For uniform motion

Ans: c) the speed-time graph is straight line parallel to the time axis

Numerical:

1) Solution:

Velocity at t= 1s is 2m/s.

Velocity at t= 2s is 4m/s.

Velocity at t= 3s is 6m/s

2) Solution:

a) From the above displacement-time graph, the average velocity from the part PQ is given by,

Average velocity= (displacement at Q – displacement at P)/time

Average velocity= ()/

Average velocity = 5m/s

b) From the graph the displacement of the car at t= 2.5 s is 12.5m

And the displacement of the car at t= 4.5s is 22.5 m.

3) Solution:

a) Total distance travelled in interval 1s to 5s is given by,

Total distance = (distance at 5s – distance at 1s)

Total distance= (18-6) = 12m

b) Average velocity in interval 1s to 5s is given by,

Average velocity= total distance/time= 12/4= 3m/s

4.) Solution

From figure we solve the following:

i) 0 to 5 s

At t= 0 , s= 0

And at t= 5s, s= 3

Thus, velocity= (3-0)/(5-0)= 3/5= 0.6m/s

ii) 5 to 7s

at t= 5s , s= 3

At t = 7s, s= 3

Velocity= (3-3)/(7-3) = 0m/s

iii) 7 to 9 s

At t= 7s, s= 3

At t= 9s, s= 7

Velocity= (7-3)/(9-7) = 4/2= 2 m/s

 

b) During 5 to 9 s the total distance covered is s= (7-3)= 4

And total time = (9-5)= 4s

Thus, average velocity= 4/4= 1m/s

5.) Solution:

Ans:

a) average velocity in the first 4s:

Total distance covered in first 4 s is s= 5+5=10

Thus, average velocity= 10/4= 5/2= 2.5m/s

 

b) the displacement from initial position to 10s:

The total displacement covered in 10 s is given by,

A to C + C to D + D to E + E to F = 10 + 0 + (-10) + (-10)

= -10m

Thus, the displacement from initial position and last 10s is -10m

 

c) After 7s and also after13s the cyclist reaches to starting point.

6) Solution:

Ans:

a) the distance by which initially car B is ahead of car A is the 40km.

 

b)

Initially the velocity of car A is given by,

Velocity of car A = (40-0)/(1-0)= 40 km/h

 

Initially the velocity of car B = (60-40)/(1-0)= 20 km/h

 

c) In 2 hrs the car A catches the car B.

d) the distance from start when car A will catches car B is 80 km.

7) Solution:

Ans:

The displacement-time graph is showing that the motion is non-uniform motion.

8) Solution:

Ans:

From the graph a, the velocity of the body at time t=1s is 1m/s

Displacement= velocity*time = 1*1=1m

 

Velocity of body at t= 2s is 2m/s.

Displacement=2*2= 4m

 

Velocity of body at 3s is 3m/s.

Displacement = 3*3= 9m

 

Velocity of body at 4s is 4m/s.

Displacement=4*4= 16m

 

Thus, the graph of displacement-time is as shown in fig.

9)

Ans:

a) The distance traveled in any part is determined from the area enclosed by the graph with the time axis.

 

b) The comparison of distance traveled in part BC with the distance traveled in part AB is 2:1.

 

c) The part BC shows uniform velocity.

The part AB shows uniform acceleration.

The part CD shows uniform retardation.

 

d) The magnitude of acceleration is lower than that of retardation. Because, slope of line AB is less than the slope of line CD.

 

The comparison between magnitude of acceleration and retardation is 1:2.

10) Solution:

a)

Acceleration in part AB,

AB = change in velocity/total time= (30-0)/(4-0)= 30/4= 7.5m/s2

 

Acceleration in BC,

BC= (30-30)/(8-4) = 0m/s2

 

Acceleration in CD,

CD= (0-30)/(10-8)= -30/2= -15m/s2

 

b)

Displacement is given by the area enclosed by that line with the time axis.

 

Displacement in part AB,

AB= 1/2*base*height

AB= 1/2*4*30= 60m

 

Displacement in part BC,

BC= length*breadth= 30*4= 120m

 

Displacement in part CD,

CD= 1/2*base*height

CD= 1/2*2*30= 30m

 

c) Thus, the displacement will be,

AB+BC+CD= 60+120+30= 210m

11)

Ans:

The velocity-time graph of the ball is as shown in figure below.

 

The total distance traveled by the ball is the distance traveled in first 6s and the next 6s.

Since, it is moving with uniform speed 10m/s.

Hence, Distance = velocity*time = 10*6= 60m

 

After hitting the wall ball return to its initial position in 6s with same speed 10m/s.

Hence, again distance traveled will be,

Distance = 10*6 = 60 m

Thus, the total distance traveled by the ball will be 60+60= 120m

 

And, here the ball initially moving with uniform speed and strikes the wall and again come back to its initial position within the same time with same velocity.

Hence, displacement will be zero here. As displacement is the vector quantity.

12)

a) from graph we conclude that, the nature of line from 0s to 4s is uniformly accelerated and from 4s to 6s it is uniformly retarded.

 

b) the total displacement covered in 6 s is given by the area enclosed within the graph and the time axis, which is in the form of triangle.

Thus, displacement= 1/2*base*height= 1/2*6*2= 6m

Thus, the displacement of the particle at t=6s is 6m

 

c) The particle does not changes its direction of motion.

 

d)

The comparison between distance traveled by the particle from 0 to 4s and 4s to 6s is 2:1.

 

e)

Acceleration from 0 to 4s is given by,

Acceleration= ( 2-0)/(4-0)= 2/4=1/2= 0.5m/s2

 

Retardation from 4s to 6s is given by,

Retardation = (0-2)/(6-4)= -2/2= -1m/s2

Exercise – C Solution:

1.) Solution:

Ans:

When the body is moving with uniform or constant acceleration then it’s initial velocity u, final velocity v, acceleration a, time t and displacement s are related with each other and forms the three equations, which are called as equation of motion.

First equation of motion is given by,

v = u + a*t

The second equation of motion is given by,

s = u*t + 1/2*a*t2

The third equation of motion is given by,

v2 = u2 + 2*a*s

2.) Solution:

Ans:

The relation between displacement, velocity, acceleration and time of a moving object is given by three equations which are called as equations of motion.

It was studied by Newton.

This three equations of motion are

v = u + at

s = u t + ½ a t2

v2 = u2 + 2a s

The derivation of these three equations is given by graphical method which is as follows:

The following figure shows the change in velocity with respect to time for a uniform accelerated motion.

Fig. Graphical method to derive equation of motion

Let the object starts from point A with initial velocity u, the velocity of the object is increasing with time and finally it reaches to the point D on the graph.

Initial velocity of the object = u =OA = EC

Final velocity of the object = v = OD = EB

And t is the time required to the object to reach from initial to final position.

t = OE = AC

By definition of acceleration,

Acceleration = (change in velocity)/ time

= (final velocity – initial velocity)/ time

a = (v – u)/t

a t = v – u

v = u + a t

This is the first equation of motion.

 

From graph, the area under the curve AOEB gives the distance covered by the object during time t,

S = Area (quadrilateral AOEB)

= Area (rectangle AOEC) + Area (triangle ACB)

s = CE * OE + (1/2 *CB * AC)

= u t + ½ (v – u) t

= u t + ½ (a t) t                            since v – u = a t

s = u t + ½ a t2

This is the second equation of motion.

 

We know that, the area under the curve AOEB gives the distance covered by the object during time t and it is a trapezium,

s = Area of trapezium AOEB

s = ½ * (sum of lengths of parallel sides)* (distance between parallel sides)

s = ½ (OA + EB) * OE

s = ½ (u + v) * t

s = ½ (u + v) * (v-u/a)

2as = (v + u) * (v – u)

2as = v2 – u2

v2– u2 = 2as

This is the third equation of motion.

3.) Solution:

Ans:

Given that, initially body is at rest and then moving with constant acceleration, which covers distance s in time t.

Hence, here u = 0

The relation between distance s, time t, acceleration a and initial velocity u is given by,

s = u*t + 1/2*a*t2

Thus, here

s= 0+ 1/2*a*t2

Hence, s= 1/2*a*t2

This is required relation.

Multiple choice type:

1.) The correct equation of motion is

Ans: d) v = u + a*t

2.) A car starting from rest accelerates uniformly to acquire a speed 20km/h in 30 min. The distance traveled by car in this time interval will be

Ans: b) 5 km

Numerical:

1.) Solution:

Ans:

Given that,

Initial velocity u= 0

Acceleration a = 2m/s2

Time t= 2s

Distance s=?

We know that, the second equation of motion is

s= u*t + 1/2a*t2

Hence, s = 0 + 1/2*2*4 = 4m

Thus, the distance covered by the body in 2s is 4m.

2.) Solution:

Ans:

Given that,

Initial velocity u= 10m/s

Acceleration= 5m/s2

Time t= 5s

Distance s=?

We know that, the second equation of motion is

s= u*t + 1/2*a*t2

Hence, s = 10*5 + 1/2*5*25

s = 50+ 62.5 = 112.5 m

Thus, the distance covered by the body in 5s is 112.5m

3.) Solution:

Ans:

Initially, vehicle has velocity 30km/h and after 2 s it is 33.6km/h.

Hence, during this the acceleration of the vehicle is,

Acceleration= change in velocity/time

= (33.6-30)/2= 3.6/2 = 1.8m/s2

 

Again, the vehicle moving with velocity 33.6 km/s after 2 s it’s velocity is 37.2km/h.

During this its acceleration is,

Acceleration= change in its velocity/time

Acceleration= (37.2-33.6)/2 = 3.6/2 = 1.8m/s2.

Thus, the acceleration is uniform after equal interval of time.

4.) Solution:

Ans:

Given that,

Initial velocity u = 0

Acceleration a= 2m/s2

Time t= 5s

a) Velocity of the body is given by,

v = u + a*t

v = 0 + 2*5 = 10m/s

Thus, the velocity of body after 5s is 10m/s.

 

b) To find the distance travelled is given by,

The second equation of motion is given by,

s= u*t + 1/2*a*t2

Thus, s = 0 + 1/2*2*25 = 25 m

Thus, the distance covered by the body in 5s is 25m.

5.) Solution:

Ans:

Given that, initial velocity u = 20m/s

Final velocity v= 0

Distance s= 10cm

Retardation a=?

The third equation of motion is given by,

v2 = u2 + 2*a*s

0 = 400 + 2*a*0.1

0.2*a= -400

Hence, a= -400/0.2 = -4000/2 = -2000m/s2

Thus, the retardation caused by the object is -2000m/s2.

6.) Solution:

Ans:

Given that,

Initial velocity u= 20m/s

Final velocity v = 0

Time t = 5s

Retardation a=?

The second equation of motion is given by,

v= u + a*t

0= 20 + a*5

a= -20/5 = -4m/s2

Thus, the retardation is 4m/s2.

7.) Solution:

Ans:

We know that, the average speed is the scalar quantity and it is given by,

Average speed= total distance covered/total time taken

Let s be the distance between station A and B.

Given that, the train travels from station A to B then again come back from station B to A.

Thus, total distance covered = s+ s = 2s

 

Given that, train travels from station A to B with speed 60km/h. And returns to station A from B with speed 80km/s.

 

Hence, time required to reach from station A to B is

Time = s/60

And time required to reach from station B to station A is given by,

Time = s/80

Thus, total time required by the train is given by,

Total time = s/60 + s/80 = 140s/4800 = 7s/240

 

Thus, average speed is given by,

Average speed = total distance covered/total time taken

= 2s/ (7s/240) = 480/7 = 68.57 km/h

 

We know that, average velocity is the vector quantity.

As train travels from station A to B and then again comes back to A from B.

Thus, the average velocity will be zero.

8.) Solution:

Ans:

Given that,

Initial velocity u = 90km/h= 90*103/3600= 25m/s

Final velocity v = 0

Retardation a= -0.5m/s2

Time t= 10s

Hence, velocity after 10 s is calculated by,

The first equation of motion is given by,

v = u + a*t

v = 25 – 0.5*10 = 25- 5 = 20m/s

 

Time taken by train come to rest is given by,

We know that,

v = u + a*t

0= 25 -0.5*t

0.5*t = 25

Time t = 25/0.5= 250/5 = 50 s

Thus, the time taken by train come to rest is 50s.

9.) Solution:

Ans:

Given that,

Distance s= 100m

Average velocity = 20m/s

Final velocity v = 25m/s

Initial velocity u=?

Acceleration a=?

We know that,

Average velocity= (u+v)/2

20= (u+25)/2

40= u+25

Hence, u = 40-25= 15 m/s

Thus, the initial velocity of car is 15m/s.

 

Now, to find the acceleration of the car we use third equation of motion,

v2 = u2 + 2a*s

625= 225 + 2*a*100

625-225= 200*a

400= 200a

a = 2m/s2

Thus, the acceleration of the car is 2m/s2.

10.) Solution:

Ans:

Given that,

Retardation a= -25cm/s2 = -0.25m/s2

Time t = 20s

Final velocity v = 0

Initial velocity u =?

 

We know that,

The first equation of motion is,

v = u + a*t

0 = u -0.25*20

u= 5m/s

Thus, the initial velocity of the bus is 5m/s.

 

To find the distance travelled we use third equation of motion,

v2 = u2 + 2a*s

0 = 25 -2*0.25.s

0.5*s = 25

Thus, s = 25/0.5 = 250/5= 50m

Thus, the distance travelled by the bus during this time is 50m.

11.) Solution:

Ans:

Given that,

Initial velocity u = 0

Distance s= 270m

Time t = 3s

We know the second equation of motion,

s= u*t + 1/2*a*t2

270 = 0+ 1/2*a*9

4.5*a= 270

Thus, a = 60m/s2

 

Now, to find the velocity after 10 s we use the first equation of motion,

v = u + a*t

v = 0 +60*10 = 600m/s

Thus, the velocity of the body at 10s after start is 600m/s.

12.) Solution:

Ans:

Given that,

Body is moving with constant acceleration a, and travels distances 3m and 8m in 1s and 2s respectively.

Let, s1= 3m, s2= 8m

And t1= 1s, t2= 2s

Initial velocity u =?

We have, second equation of motion is,

s= u*t + 1/2a*t2

Thus, we write it as

s1= u*t1+1/2at12

3= u+ 1/2a i.e. 6= 2u+ a

 

And, s2= u*t2+ 1/2at22

8= 2u+ 2a i.e. 4= u+a

 

We solve simultaneous equations,

6= 2u+ a

And, 4= u+a

Thus, we get

2= u

And a= 2

Thus, initial velocity of the body is 2m/s.

And acceleration is 2m/s2.

13.) Solution:

Ans:

Given that,

Initial velocity u = 25m/s

Final velocity v = 0

The car is moving before brakes are applied with velocity 25m/s for the time 5s.

Thus, the distance travelled in this time will be,

Distance = 25*5= 125m

Thus, the distance travelled by the car before applying the brakes is 125m.

 

Now, we know that acceleration is the change in velocity with respect to time.

Hence,

Acceleration = final velocity- initial velocity/total time

Acceleration= 0-25/ (5+10)

Acceleration=-25/15= -5/3= -1.6m/s2

Thus, the retardation of the car is 1.6m/s2

 

The distance travelled by the car after applying the brakes is given by,

The third equation of motion is,

v2 = u2 + 2a*s

0 = 625-2*1.6*s

625= 3.2s

s= 625/3.2 = 195.3m

Thus, the distance travelled by the car after applying the brakes is 195m.

14.) Solution:

b) the distance travelled by the spacecraft in the first 10s after the rocket motors were started, the motors having been in action only for 6s.

Ans:

Given that,

Initial velocity u = 75km/h= 75/3.6 m/s= 20.83m/s

Final velocity v = 120km/h= 120/3.6m/s = 33.33m/s

Time t = 6s

a)

Acceleration = (v-u)/t

Acceleration = (120-75)/6= 7.5km/s2

b)

The total distance covered by the spacecraft is the sum of distance covered in first 6s and the next 4s.

Thus, distance covered in first 6s is given by,

s= ut+1/2at2

s= 75*6+1/2*7.5*36

s= 450 + 135 = 585km

And the distance covered in next 4 s is,

Distance = speed*time

Distance = 120*4= 480km

Finally, the distance covered by the spacecraft in 10 s will be = 585+480= 1065km

15.) Solution:

Ans:

Uniform acceleration a = 2m/s2

Time t = 10s

Initial velocity u = 0

a) The first equation of motion is

v = u + at

v = 0+2*10 = 20m/s

Thus, the maximum velocity reached is 20m/s.

b) At the ending of 50 s comes to rest.

Initial velocity u = 20m/s

Final velocity v = 0

Time t = 50s

Hence, acceleration = (v-u)/time

Acceleration = (0-20)/50= -0.4m/s2

Thus, the retardation in the last 50 s will be 0.4m/s2.

c) Total distance travelled is the sum of distance covered in 10s, 200s and 50s.

 

Thus, the distance covered in first 10s is,

s= ut+ 1/2at2

s= 0+1/2*2*100= 100m

 

The distance travelled in next 200s is given by,

Distance= speed*time

Distance = 20*200= 4000m

 

And finally, the distance covered in last 50 s is given by,

s= ut + 1/2at2

s= 20*50 – 1/2*0.4*2500

s= 1000- 500=500m

 

Hence, total distance covered is 100+4000+500= 4600 m

d) Average velocity is given by,

Average velocity = total distance covered/total time taken

Average velocity = 4600/ (10+200+50)

Average velocity= 4600/260= 17.69 m/s

Updated: July 28, 2021 — 4:27 pm

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