# Frank Solutions for Chapter 2 Motion in One Dimension Class 9 Physics ICSE

**Exercise 2.1**

**1. Define the state of rest.**

**Answer**

A body is said to be in state of rest if it does not change its position with respect to its surrounding objects with time.

**2. What do you mean by a vector quantity?**

**Answer**

A vector quantity is that physical quantity which is represented by both magnitude and direction.

**3. Is mass a vector quantity?**

**Answer**

No mass is not a vector quantity.

**4. How is a vector represented?**

**Answer**

A vector is represented by an arrow. The length of the arrow represents the magnitude of vector quantity and arrow head gives the direction of vector quantity.

**5. A book is lying in an almirah. Is it at rest or in motion?**

**Answer**

If a book is lying in almirah then it is at rest.

**6. What do you mean by motion?**

**Answer**

A body is said to be in motion when it change its position with respect to its surrounding objects with time.

**7. Are rest and motion relative?**

**Answer**

Yes rest and motion are relative to each other.

**8. Out of force and energy, which one is vector?**

**Answer**

Out of Force and Energy, Force is a vector quantity.

**9. Give two examples of scalars.**

**Answer**

Examples of scalars are distance and mass.

**10. Different positions of a body are given below. Which of the two positions lie on the same straight line?**

**(i) 8 m, 60° NE**

**(ii) 3 m, 60°
NE**

**(iii) 5 m,
60° EN**

**(iv) 5 m, 60°
NW**

**Answer**

Out of these positions, (i) and (ii) positions of body
lie on same straight line as direction of these two are same.

**11. A quantity can be represented only when its
magnitude and direction are specified. Is it scalar or vector?**

**Answer**

A vector quantity is represented only when its magnitude
and direction are specified so this quantity is a vector quantity.

**12. Are the
passengers sitting in a train in motion or at rest with respect to each other ?
**

**Answer**

Passengers sitting in a train are at rest with respect to
each other.

**13. You are
at rest as well as in motion. Justify the statement giving your point of view.**

**Answer**

Yes we are at rest as well as motion because we are at
rest with respect to a observer which is itself at rest and we are in motion with
respect to a observer which is in motion.

**14. A train is moving out of a railway station. Is the platform at rest or in motion with respect to the moving train? Justify your answer.**

**Answer**

The platform is in motion with respect to train. As train is moving with respect to platform so platform would also look in motion with respect to train.

**15. Represent the location of an object described as at 15 m, 45° west of north, on a graph paper.**

**Answer**

**16. Represent the position of a body described at 15 m, 30° north of west, on a graph paper. **

**Answer**

**17. Name the physical quantity representing the magnitude and its direction.**

**Answer**

The physical quantity representing the magnitude and its direction is a vector quantity.

**18. Answer the following questions:**

**(i) Can we add two scalars?**

**(ii) Can we add two vectors?**

**(iii) Can we multiply two scalars?**

**(iv) Can we add a scalar quantity to a vector quantity?**

**(v) Can we subtract two scalars?**

**(vi) Can we subtract a scalar quantity from a vector quantity? Is the reverse true?**

**(vii) Can we multiply vectors?**

**Answer**

(i) Yes we can add two scalars.

(ii) Yes we can add two vectors.

(iii) Yes we can multiply two scalars.

(iv) No we cannot add a scalar quantity to a vector quantity.

(v) Yes we can subtract two scalars.

(vi) No we cannot subtract a scalar quantity from a vector quantity. Reverse is also not true.

(vii) Yes we can multiply vectors.

**19. What do you mean by the distance travelled by a body?**

**Answer**

The actual length of the path covered by a moving object irrespective of its direction of motion is called the distance travelled by the object.

**20. Can distance covered by a body be less than the magnitude of its displacement?**

**Answer**

No the distance covered by a body cannot be less than the magnitude of its displacement.

**21. Define displacement.**

**Answer**

The displacement of a moving body is defined as the change in its position along a particular direction.

**22. Write the SI units for measurement of distance and displacement.**

**Answer**

SI unit for measurement of distance and displacement is metre denoted by m.

**23. Show the difference between distance and displacement with the help of a diagram. **

**Answer**

so distance travelled is = Ï€r.

Displacement = diameter of circle

So displacement is 2r.

**24. Can a body have negative displacement?**

**Answer**

Yes a body can have negative displacement.

**25. Under what condition is the displacement of a body equal to the distance travelled by it?**

**Answer**

If a body is moving in a straight line then the displacement of a body is equal to the distance travelled by it.

**26. Write two differences between distance and displacement.**

**Answer**

(i) Distance is a scalar quantity whereas displacement is a vector quantity.

(ii) Distance is always positive but displacement can be negative, zero or positive.

**27. Calculate the distance and displacement in the following cases:**

**(i)**

**(ii)**

**(iii)**

**Answer**

(i) Distance is 4 + 4 = 8 m while displacement is 3m.

(ii) Distance is 2 + 4 + 2 + 4 = 12 m while displacement is 0 m.

(iii) Distance is 2 + 2 + 2 + 2 + 2 + 2 = 12 m while displacement is 2 + 2 + 2 = 6m.

**28. A body is moving along a circular path of radius r. What will be the distance and displacement of the body when it completes**

**(i) Half a revolution**

**(ii) 3/4 ^{th} of a revolution**

**(iii) Full revolution**

**(iv) Two full revolutions?**

**Answer**

(i)

So distance travelled is = Ï€r.

Displacement = diameter of circle

So displacement is 2r.

(ii)

Distance travelled = (3/4) of perimeter of circle.

So distance travelled is = (3Ï€r/2).

Displacement can be calculated as s = √r^{2 }+ r^{2 }= √2r

(iii)

So distance travelled is = 2Ï€r

Displacement = 0 as starting and ending point are coinciding.

(iv)

Distance travelled is two times the perimeter of circle.

So distance travelled is = 4Ï€r.

Again displacement is = 0 as starting and ending point are coinciding.

**29. Ashok starts from his house and travels 3 km to reach his school. On finding the school gates closed, he returns to his house. What is?**

**(i) The distance travelled by Ashok and**

**(ii) The displacement of Ashok?**

**Answer**

between school to house = 3 + 3 = 6 km.

(ii) Displacement is = 0 as starting and ending point are coinciding.

**30. A satellite makes a complete round along its circular orbit. What is its displacement?**

**Answer**

**31. An object is thrown vertically upwards. It rises upto a height H and then returns to its initial position. Calculate **

**(i) The total distance travelled by the object, **

**(ii) The displacement of the object.**

**Answer**

So total distance = H + H = 2H.

(ii) Total displacement = 0 as starting and ending points are coinciding.

**Exercise 2.2**

**1. Define speed of a body.**

**Answer**

Speed of a body can be defined as distance covered by the body in unit time.

**2. What do you understand by the average speed of a body?**

**Answer**

Average speed of a body can be defined as ratio of total distance covered by a body In total time.

**3. Write the SI units of speed and average speed.**

**Answer**

Both speed and average speed have same unit and that is m/s

**4. Do speed and average speed of a body have the same meaning?**

**Answer**

No speed and average speed of a body have different meaning.

**5. Which is greater?**

**(i) 60 km/hr**

**(ii) 15 m/s**

**Answer**

60 km/hr can be converted into m/s to compare with 15m/s.

60 km/hr = (60 × 1000)/3600 = 16.66 m/s. so speed 60 km/hr is greater.

**6. Convert 20 m/s into km/hr.**

**Answer**

20m/s can be converted into km/hr as:

20 m/s = (20 × 3600)/1000 = 72 km/hr.

**7. Which is greater?**

**(i) 36 km/hr**

**(ii) 8.5 m/s**

**Answer**

36 km/hr can be converted into m/s te compare with 8.5m/s.

36 km/hr = (36 × 1000) /3600 = 10 m/s which is greater than 8.5 m/s

So 36 km/hr is greater than 8.5 m/s.

**8. What are the SI units of velocity?**

**Answer**

Sl unit of velocity is ms^{-1}.

**9. Two cars are moving with speeds of 40 km/hr due North and 40 km/hr due South respectively. Do they have the same velocity?**

**Answer**

No as their direction is different they don't have same velocity.

**10. Arrange the following speeds in the increasing order:**

**(i) A bicycle moving with a speed of 36 km/h.**

**(ii) A car moving with a speed of 2 km/min.**

**(iii) An athlete running with a speed of 7 m/s.**

**Answer**

we convert all the speeds in m/s to compare them.

36 km/hr = (36 × 1000)/3600 = 10m/s.

2 km/min = (2 × 1000)/60 = 33.3 m/s.

7 m/s = 7 m/s.

So increasing order of speed is 7m/s < 10m/s < 33m/s.

**11. A boy covers half of his journey with a uniform speed of u and the other half with a uniform speed of v. What is the average speed for the whole journey?**

**Answer**

let total distance be S.

Boy covers distance S/2 with speed u then time taken by him to cover this distance would be T_{1} = S/2u.

Again boy covers rest of the distance S/2 with speed v then time taken by him to cover this distance would be T_{2 }= S/2v.

So total time taken by boy to cover the distance S is T = T_{1 }+ T_{2}.

Total time T= S/2 (1/u +1/v) = s (u + v)/2uv.

And average speed = S/T = 2uv/(u + v).

**12. Does uniform speed mean constant speed?**

**Answer**

Yes uniform speed and constant speed have same meaning.

**13. A body goes from P to Q with a uniform speed of u and immediately returns back to P at a uniform speed of v. What is the average speed for the whole journey?**

**Answer**

Let S be the distance between P and Q.

Body covers forward journey distance (P to Q) with speed u then time taken by him to cover this distance would be T_{1 }= S/u.

Again body covers backward journey distance S (Q to P) with speed v then time taken by him to cover this distance would be T_{2} = S/v.

So total time taken by body to cover the distance S is T = T_{1} + T_{2}.

Total time T= S (1/u +1/v) = s(u + v)/uv.

And average speed = 2S/T = 2uv/(u + v).

**14. In Q.13 above, what is the average velocity for the whole journey?**

**Answer**

As body goes from P to Q and then return back to P so the displacement of the body would be zero and hence average velocity would also be zero.

**15. A car travels from P to Q at a uniform speed of 20 ms ^{-1} and immediately turns back and travels towards Q with a uniform speed of 30 ms^{-1}. What is the average speed for the whole journey?**

**Answer**

let distance between P and Q is S.

Speed of car while travelling from P to Q is 20 m/s.

Let car take time T, to travel from P to Q then T_{1} = 8/20.

Speed of car while travelling from Q to P is 30 m/s.

Let car take time T_{2} to travel from Q to P then T2 = S/30.

Total time = T_{1} + T_{2} = S/20 + S/30 = S/12.

So average speed of journey = total distance/ total time = 2S/(S/12) = 24 m/s.

Average speed of journey is 24 m/s.

**16. How does speed differ from velocity?**

**Answer**

Speed is a scalar quantity whereas velocity is a vector quantity. So speed doesn't have its direction and velocity has a particular direction.

**17. When do the speed and velocity of a moving body become equal?**

**Answer**

Speed and velocity of a moving body become equal when the body moves in a straight line.

**18. What is meant by a constant velocity? Is it same as uniform velocity?**

**Answer**

When the velocity of a moving body doesn't change with time then the velocity of the body is said to be constant or uniform. Yes uniform velocity and constant velocity are one and the same thing.

**19. Define acceleration.**

**Answer**

Acceleration of a body is rate of change of its velocity with respect to the time.

**20. What is the SI unit of acceleration?**

**Answer**

Sl unit of acceleration is ms^{-2}.

**21. The acceleration of a moving body is constant in magnitude and direction. Must the path of the body be a straight line? If not, given an example.**

**Answer**

If the acceleration of a moving body is constant in magnitude and direction then the path of the body must not be a straight line because in circular motion also acceleration of a body is constant in magnitude and always directed towards the centre.

So the path of the body may be a straight line and may be a circular one.

**22. The relation S = v x t can easily be used to find the total distance covered by a body moving with non-uniform speed. (Yes/no)**

**Answer**

No the relation S = v × t cannot used to find the total distance covered by a body moving with non-uniform speed.

**23. The area under a speed-time graph in a given intervals gives the total distance covered by a body irrespective of its motion being uniform or non-uniform.**

**Answer**

Yes area under a speed time graph in a given interval gives the total distance covered by a body.

**24. The area of the right triangle under a speed-time graph is S00 m, in a time interval of 20 s. What is the speed of the body? Is the motion uniform or non-uniform?**

**Answer**

Yes the motion is uniform and the uniform speed is given by area under speed time graph divided by time interval.

So speed = 500/20 = 25 m/s.

**26. What does a positive acceleration mean?**

**Answer**

Positive acceleration corresponds to situation when velocity is continuously increasing with respect to the time.

**27. A body falls towards the earth. Does it have positive or negative acceleration?**

**Answer**

If a body falls towards earth then it would have a positive acceleration.

**28. The acceleration of a body is 8.5 ms ^{-2}. What does this statement mean?**

**Answer**

If a body has acceleration of 8.5 ms^{-2} then it means its velocity is increasing at a rate of 8.5 ms^{-2} per second.

**29. Write the SI unit of retardation.**

**Answer**

SI unit of retardation is ms^{-2}.

**30. A car traveling at 60 km/h, stops on applying brakes in 10 seconds. What is its acceleration?**

**Answer**

first convert 60 km/h in m/s.

60 km/hr = (60 × 1000)/3600 = 16.7 m/s.

This is initial velocity of car i.e. u = 16.7 m/s.

As car stops in 10 seconds so final velocity is = 0 m/s.

So acceleration = (v - u)/t = (0-16.7)/10 = -1.67 ms^{-2}

Acceleration of car is = -1.67 ms^{-2}.

**31. A quantity is measured to be -30 m/s. Is it speed or velocity?**

**Answer**

-30 m/s is speed.

**32. Give the name of the physical quantity that corresponds to the rate of change of displacement.**

**Answer**

Velocity corresponds to the rate of change of displacement.

**33. Can the speed of a body be negative?**

**Answer**

No the speed of a body cannot be negative.

**34. What kind of a velocity does a flying bird most likely to have?**

**(a) Uniform velocity, or**

**(b) Non- uniform velocity.**

**Answer**

A flying bird most likely to have a non uniform velocity.

**35. Find the initial velocity of a car which is stopped in 10 seconds by applying brakes. Retardation due to brakes is 2.5 ms ^{-2}.**

**Answer**

Let initial velocity be u.

Final velocity is v= 0 m/s.

Time taken by body to come to rest = 10 sec

Retardation =2.5 ms^{-2}

We know v = u + at.

Then u = v - at.

u = 0 - (- 2.5 × 10) = 25 m/s.

So initial velocity of the body is 25 m/s.

**36. What do you understand by the equations of motion?**

**Answer**

Equation of motion gives us the picture of motion of moving body.

**37. Write down the three equations of motion.**

**Answer**

First equation of motion is v = u + at.

Second equation of motion is s = ut + 1/2a t^{2}

Third equation of motion is v^{2} - u^{2} =2as.

**38. How many variables are present in each equation of motion?**

**Answer**

Four variables are present in each equation of motion.

**39. What are the uses of equations of motion?**

**Answer**

Four variables are present in each equation of motion and if any of three is known to us then fourth can be easily find with the help of these equation of motion.

**40. Write the SI unit of acceleration and retardation.**

**Answer**

SI unit of acceleration and retardation is ms^{-2}

**41. Name the physical quantity which is equal to the area under speed-time graph.**

**Answer**

Distance is the physical quantity which is equal to the area under speed-time graph.

**42. Distinguish between uniformly and non-uniformly accelerated motions.**

**Answer**

A uniformly accelerated motion is one in which speed is constantly increasing or decreasing with time while a non uniform motion is one in which speed is not constantly changing with time.

**43. Can you use the relation [B _{1}]?**

**Answer**

No we cannot use this relation for a body moving with uniform acceleration.

**44. How does the slope of a speed-time graph give the acceleration of a body moving along a straight line?**

**Answer**

Slope of a graph is given as rate of change of y coordinates to the x coordinate. In speed time graph speed is on the y axis and time is on the x axis. And we define acceleration as rate of change of speed with respect to time. So slope of a speed time graph gives acceleration.

**45. Give two examples of uniform circular motion from your daily life.**

**Answer**

(i) Motion of blades of an electric fan.

(ii) Motion of moon around earth.

**46. How will you use a speed-time graph to find whether the acceleration of the body is uniform or not?**

**Answer**

A straight line curve on speed time graph indicates that acceleration of the body is uniform and a zigzag or curved line indicates that acceleration of a body is not uniform.

**47. The graph of one quantity versus another result in a straight line. How are the quantities related to each other?**

**Answer**

Two quantities are directly proportional to each other.

**48. A car travels at a uniform speed of 42 km/h. In 10 minutes, how much distance would it cover?**

As we distance = speed × time.

Speed = 42 km/hr.

Time = 10 m = 1/6 hr.

Distance = 42 × 01/6 =7 km.

**49. The initial velocity of a car is 10 ****ms ^{-1}**

**. It moves with an acceleration of 2 ms**

^{-2}. What will be its speed after 10 seconds?**Answer**

Initial velocity u = 10 ms^{-1}.

Acceleration a = 2ms^{-2}.

Time t = 10 s

By using first equation of motion

V = u + at.

⇒ V = 10 + 2 ×10.

⇒ V (final velocity) = 30 ms^{-1}

**50. The speed of a car increases from 10 km/h to 64 km/h in 10 seconds. What will be its acceleration?**

**Answer:**

Initial velocity u = 10 km/hr. = (10 × 1000)/3600 = 8.33 ms^{-1}

Final velocity = 64 km/hr = (64 × 1000)/3600 = 17.77 ms^{-1}

Time = 10 s.

Acceleration = (v - u)/t = (17.77 - 8.33)/10 = 9.44/10 = 0.94 ms^{-2}

**51. Given on the side are a few speed-time graphs for various objects moving along a straight line. Refer to fig. 32 (a), (b), (c) and (d).**

**Which of these graphs represent (a) Uniform motion (b) Motion with speed increasing (c) Motion with speed decreasing and (d) Motion with speed oscillating?**

**Answer**

(ii) In speed time graph motion with increasing speed is shown by straight line with positive slope so figure (c) denotes the motion with speed increasing.

(ii) In speed time graph motion with decreasing speed is shown by straight line with negative slope so figure (b) denotes the motion with speed decreasing.

**52. Can a body have negative speed?**

**Answer**

No a body cannot have a speed negative.

**53. Is distance covered during nth second more than the distance covered in n seconds?**

**Answer**

NO_{2} distance covered by body during n,, second is not more than the distance covered in n seconds.

**54. Draw the speed-time graph of a body when its initial speed is not zero and the speed increases uniformly with time.**

**Answer**

**55. Draw the speed-time graph of a body starting from some point P, gradually picking up speed, then running at a uniform speed and finally slowing down to stop at some point Q.**

**Answer**

**56. What can you conclude if the speed-time graph of a body is a curve moving upwards starting from the origin?**

**Answer**

If speed time graph is moving upward then the body is accelerating and if it is starting from origin then it means the body has initial velocity = 0.

**57. What can you conclude if the speed-time graph of a body is a straight line sloping upwards and not passing through the origin?**

**Answer**

Speed time graph is moving upward then the body is accelerating and if it is not starting from origin then it means the body has some initial velocity.

**58. Draw the following graphs:**

**(i) Speed versus time for a stepped motion.**

**(ii) Speed versus time for a non-uniform acceleration.**

**(iii) Distance versus time for a body at rest.**

**(iv) Speed versus time for a fluctuating speed.**

**(v) Speed versus time for a uniformly retarded motion.**

**Answer**

(a)

Speed time graph for a stepped motion(b)

Speed time graph for a non uniform acceleration.

(c)

Distance time graph for a body at rest.

(d)

Speed time graph for a body showing fluctuating speed.

(e)

**59. Interpret the following : **

**(a) **

**(b)**

**Answer**

(a)

(b)

(i) If distance time graph of a body is straight line parallel to x axis then the body is said to be at rest.

(ii) If displacement time graph of a body is straight line moving upwards and starting from origin then it means body has started from and moving with a uniform velocity.

**60. Given below are the speed-time graphs. Match them with their corresponding motions:**

**(a) Uniformly retarded motion **

**(b) Non – uniform acceleration**

**(c) non-uniform motion**

**(d) uniform motion**

**Answer**

(a)

Uniformly retarded motion

(b)

(c)

Non uniform motion

Uniform motion

A – (d)

B – (a)

C – (b)

D – (c)

**61. Derive the equations (i) v = u + at and (ii) v ^{2} – u^{2} = 2as, where the symbols have their usual meanings. **

**Answer**

The initial velocity of the body is u at t=0. The velocity of the body increases at a uniform rate and this increase in velocity up to time t is depicted by a straight line

PQ, The slope of line PQ gives acceleration a.

A = QR/PR.

PR = OS = t

SR = OP = u

QR = a × PR.

⇒ a × t

The point Q corresponds to the final velocity v after time t.

V = QR + RS

and generally we write v = u + at.

This is first equation of motion.

(ii) The area enclosed under a velocity time curve gives the distance covered by a moving body. So total distance S covered by a uniformly accelerating body is given by area of trapezium OSQP.

S = area of trapezium OSQP.

AREA of rectangle OSRP + area of triangle PRQ.

S = OP x OS +1/2 PR × QR.

⇒ S = u × t + ½ × t × at.

⇒ S = ut + 1/2at^{2}

This is known as second equation of motion.

**62. Derive the equation of motion,**

**S = ut + 1/2 at ^{2}**

**Where the symbols have their usual meanings.**

**Answer**

S = area of trapezium OSQP

Area of trapezium OSQP = 1/2 (sum of parallel sides) × perpendicular distance between them.

S = 1/2 (OP + SQ) × PR.

PR = QR/a = (QS - RS)/a

PR = (v - u)/a = t

So PR = t.

Substituting these values in expression of area of trapezium we get

s = 1/2 (u + v) × t

⇒ s = ½ × (u + v) × (u - v)/a

⇒ 2as = v^{2} – u^{2},

⇒ v^{2} – u^{2} = 2 as.

This is known as third equation of motion.

**Exercise 2.3 **

**1. Write two examples of vectors.**

**Answer**

Displacement and velocity are two examples of vectors.

**2. Write the SI unit of retardation.**

**Answer**

SI unit of retardation is ms^{-2}.

**3. Name the physical quantity associated with the rate of change of displacement with time.**

**Answer**

Velocity is the physical quantity associated with the rate of change of displacement with time.

**4. Define angular velocity.**

**Answer**

The angular velocity is defined as rate of change of angular displacement (Î¸) with time (t). It is denoted by (Ï‰).

**5. How many types of rectilinear motion are there?**

**Answer**

There are three types of rectilinear motion Translational, vibrational and rotational.

**6. When a body said to have a uniform velocity?**

**Answer**

A body is said to have a uniform velocity if it covers equal displacement in equal interval of time.

**7. Out of energy and acceleration, which one is a vector?**

**Answer**

Acceleration is a vector quantity.

**8. What does the slope of speed-time graph indicate?**

**Answer**

Slope of speed time graph represents acceleration.

**9. A stone is dropped from a certain height. Is its motion uniform or non-uniform?**

**Answer**

If a stone is dropped from a certain height then it undergoes non uniform velocity motion.

**10. The distance covered by a body is directly proportional to the square of the time elapsed. What can you say about its acceleration?**

**Answer**

This means the body has a positive acceleration.

**11. Which of the following graphs represents a motion with negative acceleration?**

**Answer**Graph (c) represents a motion with negative acceleration.

**12. State which of the following situations are possible:**

**(a) A body moving with constant acceleration but with Zero velocity.**

** (b) A body moving horizontally with an acceleration in vertical direction.**

**(c) A body moving with a constant velocity in an accelerated motion.**

**Answer**

(a) No a body with constant acceleration cannot have a zero velocity.

(b) No a body with an acceleration in vertical direction cannot move horizontally.

(c) No in an accelerated motion a body cannot have a constant velocity.

**13. Fig. 2 shows displacement-time graph of two objects A and B moving in a straight line. Which object is moving faster?**

**Answer**

**14. What can you say about the nature of motion of a body of its displacement-time graph is**

**(a) A straight line parallel to line axis?**

**(b) A straight line inclined to the time axis with an acute angle?**

**Answer**

(i) In displacement-time graph a straight line parallel to time axis shows that body is at rest position.

(ii) In displacement-time graph a straight line inclined to the time axis with an acute angle means body is moving with a positive velocity.

**15. Is it possible for an accelerating body to have constant speed?**

**Answer**

No a accelerating body cannot have constant speed.

**16. What can you say about the motion of a body if**

**(a) Its time-displacement graph is a straight line.**

**(b) Its time-velocity graph is a straight line?**

**Answer**

(i) In displacement-time graph a straight line shows body is at rest if it is parallel to time axis and shows a body is moving with uniform velocity if it is inclined to x axis.

(ii) In velocity-time graph a straight line shows body is moving with uniform constant velocity if it is parallel to x axis and shows body is moving constant acceleration of it is inclined to x axis.

**17. Is the average speed during different time intervals for a uniform motion the same or different?**

**Answer**

Average speed during different time intervals for a uniform motion is same.

**18. What is the velocity of a stone thrown vertically upwards at its maximum height?**

**Answer**

Velocity of a stone thrown vertically upward at its maximum height is Zero.

**19. Why does the velocity of a stone thrown vertically upwards decreases?**

**Answer**

Velocity of a stone thrown vertically upwards decrease because acceleration due to gravity is acting on downward direction.

**20. What is the relation between linear velocity and linear speed?**

**Answer**

Linear velocity would be equal to linear speed if body is moving in a straight line.

**21. Draw distance-time graphs to show:**

**(a) Increasing velocity**

**(b) Uniform velocity and**

**(c) Decreasing velocity**

**Answer**

(a)

As slope of a distance time graph indicates velocity so a increasing velocity means a straight line having a positive slope with time axis.

(b)

(c)

As slope of this line is negative so this represent decreasing velocity.

**22. Draw velocity time graph to show**

**(a) Acceleration**

**(b) Deceleration**

**(c) Zero acceleration**

**Write a sentence to explain the shape of each graph. **

**Answer**

(a)

(b)

Deceleration is represented by a straight line having a negative slope with time axis.

(c)

As slope of velocity time graph gives acceleration. So motion with zero acceleration is represented by line having zero zero slope with time axis or a line parallel to the time axis.

**23. Interpret the following graphs: **

**(a) **

**(b)**

**Answer**

(b) In this graph portion O to A represents motion with acceleration, A to E represents motion with uniform velocity, B to C represents motion with acceleration.

**24. During circular motion, which physical quantity of the body,**

**(a) Remains constant**

**(b) Changes continuously?**

**Answer**

During circular motion

(a) Speed remains constant.

(b) Velocity changes continuously.

**25. Is the following statement correct? If not, modify it suitably to make it correct. “The earth moves round the sun with uniform velocity”.**

**Answer**

The statement is not correct , the correct statement is "the earth is moves round the sun with constant speed”.

**26. A particle moves along a circular path. How many times does it change its direction in two complete rounds?**

**Answer**

As in circular motion direction changes continuously with motion so after two complete revolutions we can say that direction has changed infinite times.

**27. A body moves along a circular path of radius r. When it completes three complete rounds, what is the ratio of distance covered to its displacement?**

**Answer**

As after completing 3 revolution in circular motion the displacement is = 0. so the ratio of distance covered to the displacement is infinite.

**28. What happens to the stepwise increasing speed-time graph when the size of each step is made very small?**

**Answer**

The graph becomes straight line with positive slope with time axis and represents almost a constant acceleration.

**29. A body has an acceleration of -3.5 ms ^{-2}. What is its retardation?**

**Answer**

Retardation is negative of acceleration so retardation the body is +3.4 ms^{-2}.

**30. The average time taken by a normal person to react to an emergency is one fifteenth of a second and is called the ‘reaction time’. If a bus is moving with a velocity of 60 km/h and its driver sees a child running across the road, how much distance would the bus had moved before he could press the brakes? The reaction time of the people increases when they are intoxicated. How much distance had the bus moved if the reaction time of the driver were 1/2 s under the influence of alcohol?**

**Answer**

Bus is moving with initial velocity of u = 60 km/hr.

60 km/hr = (60 × 1000)/3600 = u = 16.66 ms^{-1}.

Reaction time = t = 1/15 sec.

Distance would the bus had moved before pressing the bus would be = u × t.

S = 16.66 × 1/15 = 1.1m.

Now if the driver is intoxicated then reaction time would be t = 1/2 seconds.

So S becomes S = u × t = 16.66 × 1/2 = 8.33m.

**31. A girl had designed a clap switch for a science exhibition that enabled her to switch on or off an alarm just with clapping of hands. While testing her device in a hall, she noticed that once the alarm has sounded it followed with another one due to echo of the clap, that is, the sound reflected by the walls. She recorded the two soundings of alarm with her tape recorder and found out that time difference in between them is 0.1 s. If the distance of the walls be 15 m, calculate the speed of sound.**

**Answer**

Time difference of 0.1 s denotes the time taken by sound to go from device to wall and back to wall. As the distance between wall and device is 15 m so total distance covered by sound is 2 × 15 m = 30 m.

So speed of sound is = total distance covered/time taken = 30/0.1 =300 ms^{-1}

So speed of sound is 300 ms^{-1}

**32. An artificial satellite is moving in a circular orbit of radius nearly 42,250 km. Calculate its linear velocity, if it takes 24 hours to revolve round the earth.**

**Answer**

Radius of orbit is 42250 km.

Distance covered by satellite to complete 1 revolution is 2Ï€r.

So distance is = 2 × 3.14 × 42250 = 265330 km.

time taken by satellite to complete one revolution is 24 hr = 24× 60×60 = 86400 sec.

So linear velocity is = distance/time = 265330/86400 ms^{-1} = 3.07 kms^{-1}.

**33. A circular cycle track has a circumstance of 314 m with AB as one of its diameter. A cyclist travels from A to B along the circular path with a velocity of constant magnitude 15.7 m/s. Find:**

**(a) The distance moved by the cyclist.**

**(b) The displacement of the cyclist if AB represents north-south direction.**

**(c) The average velocity of the cyclist.**

**Answer**Circumference of track = 314 m.

We know that circumference = 2Ï€r

So, 2Ï€r = 314

⇒ R = 314/2Ï€ = 50 m

Diameter of track would be = 2 x 50 = 100 m.

Length of path AB = 100 m.

(i) Distance moved by cyclist= half the circumference of track i.e. 314/2 = 157 m.

(ii) The displacement of cyclist is equal to the diameter of circle i.e. AB = 100 m.

(ii) Average velocity would be equal to = 15.7 ms^{-1}.

**34. What does the slope of velocity-time graph represent?**

**Answer**

Slope of velocity time graph represents acceleration of the body.

**35. Fig. 5 shows the velocity-time graphs for two objects A and B moving in same direction. Which object has the greater acceleration?**

**Answer**As acceleration is the slope of line in velocity time graph and as B has greater slope than line A. so B has greater acceleration than A.

**36. Sketch the shape of the velocity-time graph for a body moving with (a) Uniformly velocity, (b) Uniform acceleration.**

**Answer**

**(a) **

**(b)**

**37. Derive the equation: **

**S = ut + 1/2at ^{2 }**

**Using a speed - time graph. **

**Answer**

S = area of trapezium OSQP.

AREA of rectangle OSRP + area of triangle PRQ.

S = OP×OS + 1/2 PR×OR.

⇒ S = u×t + 1/2 × t ×at.

⇒ S = ut + 1/2at^{2}.

This is known as second equation of motion.

**38. Derive the equation**

^{v2} - u^{2} = 2aS

**Answer**

In figure we know,

S = area of trapezium OSQP

Area of trapezium OSQP = 1/2 (sum of parallel sides) × perpendicular distance between them.

S = 1/2 (OP + SQ) ×PR.

⇒ PR = OR/a = (QS - RS) /a

⇒ PR = (v – u)/a = t

So PR = t.

Substituting these values in expression of area of trapezium we get

S = 1/2 (u + v) × t

⇒ S = 1/2 (u + v) x (u - v)/a.

⇒ 2aS = v^{2} – u^{2}

⇒ v^{2} - u^{2 }= 2aS.

**39. For a body moving along a circular path, show that:**

**Linear speed = Angular velocity **×**radius of the circle**

**Answer**

Angular displacement Î¸ = length of arc/ radius of circle

Î¸ = I/r

⇒ l = Î¸×r

divide above equation with t

l/t = Î¸× r/t

Now,

l/t = v **(linear speed)**

⇒ Î¸/r = Ï‰

So, v = Ï‰ r.

**40. Draw the distance time graph of the bodies P and Q starting from rest, moving with uniform speeds with P moving faster than Q. **

**Answer**

**41. A body covers half the distance with the speed A and the other half with speed B. What will be the average speed for the whole journey?**

**Answer**

let total distance be S.

Boy covers distance S/2 with speed A then time taken by him to cover this distance would be T_{1} = S/2A.

Again boy covers rest of the distance S/2 with speed B then time taken by him to cover this distance would be T_{2} = S/2B.

So total time taken by boy to cover the distance S is T = T_{1} + T_{2}.

Total time T = S/2 (1/A +1/B) = s(A + B)/2AB.

And average speed = S/T = 2AB/(A + B).

**42. A car travels 30 km at a uniform speed of 60 km/h and the next 30 km at a uniform speed of 20 km/h. What is its average speed?**

**Answer**

Car travels 30 km distance with speed 60 km/hr

Time taken by car to travel this distance = 30/60 = 0.5 hr.

Car travels another distance of 30 km with speed of 20 km/hr.

Time taken by car to travel this distance = 30/20 = 1.5 hr.

Total time taken = 1.5 + 0.5 = 2 hr.

Total distance = 30+ 30 = 60 km.

Average speed of car = 60/2 = 30 km/hr.

**43. On a 120 km track, a train travels the first 40 km at a uniform speed of 30 km/h. How fast must the train travel the next 80 km, so as to average 60 km/h for the entire trip.**

**Answer**

Train travels first 40 km at speed of 30 km/hr.

Time taken by train to cover this distance is = distance/speed = 40/30 = 4/3 hr.

Let speed of train to cover next 80 km is v.

Then time taken by train to cover these 80 km is 80/v.

Total time becomes T = 4/3 + 80/v = (4v + 240)/3v.

Total distance= 120 km.

Average speed = 60 km/h (given)

However average is given by = total distance /total time.

So,

(120 × 3 × v)/(4v + 240) = 60

⇒ 360 v = 240v + 14400

⇒ 120v = 14400

⇒ V = 14400/120 = 120 km/hr.

so train has to cover those 80 km at a speed of 120 km/hr.

**44. Fig. 6 shows the distance-time graph of three students A, B and C. On the basis of the graph, answer the following:**

**(a) Which of the three is traveling the fastest?**

**(b) Will the three ever meet at any point on the road?**

**(c) When B meets A, where is C?**

**(d) How far did B travel between the time he passed C and A?**

**Answer**

(a) B is travelling fastest among all the 3 students.

(b) C is at 8 km when B meets C.

(c) B travel 4 km between the time he passed C and A.

**45. In fig. 7, the position of a body times is shown. Calculate the speed of the body as it moves from O to A, A to B, B to C and its average speed.**

**Answer**

(a) Speed as it moves from 0 to a = (6 - 0)/(3 - 0) = 2 ms-^{1}

(b) Speed as it moves from A to B = (6 - 6)/ (5 - 3) = 0 ms^{-1}

(c) Speed as it moves from B to C = (14 - 6)/ (8 - 5) = 8/3 = 2.66 ms^{-1}

**46. Two friends leave Delhi for Chandigarh in their cars. A starts at 5 am and moves with a constant speed of 30 km/h, whereas B starts at 6 am and moves with a constant speed of 40 kmh- ^{1}. Plot the distance-time graph for their motion and find at what time the two friends will meet and at what distance from Delhi.**

**Answer**

So we can see that the two friends will meet at 9 a.m. ant till then they cover a distance of u’ x t’ = u’ x t’ = 30× 4 = 40× 3 =120 km

So they are 120 km away from Delhi when they meet at 9 a.m.

**47. A car accelerates uniformly from a velocity of 18 km/h to 36 km/h in 15 min. What is its acceleration?**

**Answer**

Initial velocity of car u = 18 km/hr.

Final velocity of car v = 36 km/hr.

Time taken by body = 15 min. = 1/4 hr

Acceleration of car a = (v - u)/t = (36 - 18) × 4 = 72 kmh^{-2}

**48. A car is moving with a speed of 50 km/h. One second later, its speed is 55 km/h. What is its acceleration?**

**Answer**

Initial speed of car u = 50 km/h.

Final speed of car v = 55 km/h.

Time taken by car to attain this speed is = 1 sec. = 1/3600 hr.

Acceleration of the car is = (55 – 50) × 3600 = 18000 kmh^{-2}.

**49. Convert the following accelerations:**

**(a) 7200 km/h ^{2} into m/s^{2}**

**(b) 1/36 m/s ^{2} into km/h^{2}**

**Answer**

(a) 7200 km/h^{2} = (7200 × 1000)/(3600 × 3600) = 5/9 ms^{-2}.

(b) 1/36 m/s^{2} = (1 × 3600 × 3600)/(36 × 1000) = 3600 kmh^{-2}.

**50. A body moves along a straight road with a speed of 20 m/s and has a uniform acceleration of 5 m/s ^{2}. What will be its speed after 2 s?**

**Answer**

Initial velocity u = 20 m/s.

Acceleration = 5 m/s^{2}.

T = 2s.

We know v = u + at.

v = 20 + 5×2 = 30 m/s.

**51. A car moving with a uniform acceleration of 10 m/s ^{2} on a straight road changes its speed from 10 m/s to 30 m/s. Find the time lapsed for the change of speed.**

**Answer**

acceleration of the car = 10 ms^{-2}

Initial velocity u = 10 m/s.

Final velocity v = 30 m/s.

We knwo,

v = u + at.

⇒ t = (v – u)/a

⇒ t = (30 – 10)/10 = 2 sec.

Time taken by car is 2 sec.

**52. The graph in fig. 8 shows how the velocity of a scooter varies with time in 50 s.**

**(a) Interpret the graph**

**(b) Work out:**

**(i) Acceleration,**

**(ii) Deceleration, and**

**(iii) The distance travelled in 10s, 20s and 50s.**

**Answer**(a) From 0 to 10 sec. i.e. O to A motion is accelerating one. From 10 s to 30 sec. scooter is moving with uniform velocity, from 30 s to 50 sec scooter is retarding from B to C.

(b) Scooter is accelerating in the region O to A and acceleration is given by

a = (v - u)/t = (20 - 0)/10 = 2 ms-^{2}.

(c) Scooter is decelerating in the region B to C.

a = (v - u)/t = (0 - 20)/20 = -1 ms^{-2}

(d) The distance travelled in 10 sec is S_{1} = 1/2 ×20×10 = 100m

Distance travelled in 10s to 30s is S_{2} = 20×(30 - 10) = 20 x 20 = 400 m

Distance travelled in 30 s to 50 s is S_{3 }= 1/2 ×20× (50 - 30) = 200m.

Total distance covered by scooter is = 100m + 400m + 200m = 700m.

**53. A car accelerates to a velocity of 30 m/s in 10 s and then decelerates for 20 s so that it stops. Draw a velocity-time graph to represent the motion and find:**

**(a) the acceleration,**

**(b) the deceleration,**

**(c) the distance travelled.**

**Answer**

^{-2}.

(b) Deceleration a = (v - u)/t = (0 - 30 )/20 = -1.5 ms^{-2}

(c) Distance travelled = area under speed time graph = 1/2 ×30×30 = 450 m.

**54. A body moves along a straight track. Half of the distance it covers with a velocity of 40 ms ^{-1} and the remaining half with a velocity of 60 ms^{-1}. What is the average velocity for the whole journey?**

**Answer**

Let total distance be S.

Body covers distance S/2 with speed 40 ms^{-1} then time taken by him to cover this distance would be T_{1} = S/2 ×40.

Again body covers rest of the distance S/2 with speed 60 ms^{-1} then time taken by him to cover this distance would be T_{2} = S/2 ×60.

So total time taken by body to cover the distance S is T = T_{1} + T_{2}.

Total time T = S/2 (1/40 + 1/60) = s(40 + 60)/2 ×40 ×60 = s/48.

And average speed = S/T = 48 ms^{-1}

So average speed is 48 ms^{-1}

**55. A body goes from A to B with a velocity of 30 ms ^{-1} and comes back from B to A with a velocity of 20 ms^{-1}. What is the average velocity of the body for the whole journey?**

**Answer**

As displacement for the motion from A to B and B to A is zero so the average velocity of the body would be zero.

**56. A body starts to slide over a horizontal surface with an initial velocity of 0.5 ms ^{-1}. Due to friction, its velocity decreases at the rate of 0.05 ms^{-2}. How much time will it take for the body to stop?**

**Answer**

Initial velocity of body u = 0.5 ms^{-1}

Final velocity of the body v = 0 ms^{-1} as body comes to rest finally.

Retardation of body = 0.05 ms^{-2}.

We know that v = u + at.

0 = 0.5 - 0.05t

⇒ T = 0.5/0.05 = 10 sec.

**57. A train is moving at a speed of 90 km/h. On applying brakes, a retardation of 2.5 ms’2 is created. At what distance before, should the driver apply the brakes to stop the train at the station?**

**Answer**

Initial speed of train = 90 km/hr

Speed of train in m/s = (90 x 1000)/3600 = 25 m/s.

Retardation of the train = 2.5 ms^{-2}.

Final speed of train at platform = 0 m/s.

We know that v^{2} - u^{2} = 2as.

0 - 25×25 = 2× (-2.5)× s

⇒ S = 625/5 = 125m.

So driver should apply the brakes 125 m before the platform.

**59. A 50 m long train passes over a bridge at a speed of 30 km/h. If it takes 36 seconds to cross the bridge, calculate the length of the bridge.**

**Answer**

speed of train = 30 km/hr.

Speed in m/s = (30× 1000)/3600 = 50/6 m/s.

Length of train = 50 m.

Let length of bridge be s metre.

Train has to cover total distance of 50 +s to cross that bridge.

Time taken by train to cover this distance = 36 sec.

So as time taken = total distance/total time taken.

36 = (50 + s)× 6/50.

⇒ 1800 = 300 + 6s

⇒ 6s = 1500.

⇒ s = 1500/6 = 250m

Length of bridge is 250 m.

**60.**

**(a)**

**(b)**

**(i) In speed time graph uniform motion is given by a straight line parallel to x axis so figure (a) denotes the uniform motion.****(ii) In speed time graph motion with increasing speed is shown by straight line with positive slope so figure (c) denotes the motion with speed increasing.**

**(iii) In speed time graph motion with decreasing speed is shown by straight line with negative slope so figure (b) denotes the motion with speed decreasing.**

**(iv) In speed time graph motion with oscillating speed is shown by zigzag line so figure (d) denotes the motion with speed oscillating.**

**Answer**

Acceleration during O to P = (10 - 0)/ (10 - 0) = 1 ms^{-2}

Acceleration during P to Q = (10 - 10)/(20 - 10) = 0 ms^{-2}

Acceleration during Q to R = (0 - 10) /(25 - 20) = - 2 ms-^{2}

**61. Fig. 10 represents graphically the velocity of a car moving along a straight road over a period of 100 hours.**

**(i) How long is the body travelling with a uniform velocity?**

**(ii) Calculate the acceleration along AB and the retardation along BC.**

**(iii) Calculate the distance travelled in the last 40 h.**

**Answer**

(ii) Acceleration along AB = (100 - 100)/(40 - 20) = 0 ms^{-2}

Retardation along CO = (100 - 50)/ (100 – 60) = 50/40 = 1.25 ms^{-2}

(iii) Distance travelled in last 40 hour would be equal to area under graph during that time

= 50 ×(100 - 60) + 1/2 ×(100 - 60) ×(100 - 50).

⇒ S = 2000 + 1000 = 3000 km.

**62. Write down the type of motion of a body in each of the following distance time-graph (Fig. 11). **

**(a) **

**(b)**

**(c)**

**Answer**

**(a)**

**(b)**

**(c)**(i) Body is showing decreasing velocity that is retardation.

(ii) Initially body is showing retarded motion and then a accelerating one.

(iii) The body is in state of rest.

**63. The table below shows the distance travelled by two vehicles A and B during each second:**

**Draw a velocity-time graph for the above data and answer the following questions with reference to the graph:**

**(i) Which vehicle is moving with a uniform velocity?**

**(ii) Which vehicle is moving with a constant acceleration?**

**(iii) At what time do A and B meet?**

**(iv) Give the value of the velocity at which they are meeting.**

**(v) Which vehicle is ahead at the end of 7 ^{th} second and by how much?**

**Answer**

(i) No vehicle is moving with uniform velocity.

(ii) Vehicle B is moving with constant acceleration.

(iii) At 6 seconds both vehicles would meet.

(iv) Velocity of both the vehicles is 60 m/s when they meet.

(v) Vehicle B is ahead at the end of 7^{th} sec and by 70m.

**64. An object covers a distance of S meters in t seconds as follows:**

**Plot a graph, taking t on X-axis and S on Y-axis. Determine the velocity of the object at time**

**(i) 6s, and **

**(ii) 14s**

**Answer**

The given question is wrong as distance can never decrease with progress of time.