# Lakhmir Singh Manjit Kaur Class 9 Physics 1st Chapter “Motion” solution

## Lakhmir Singh Manjit Kaur Class 9 Physics 1st Chapter “Motion” solution

Lakhmir Singh and Manjit Kaur Physics solution: “Motion” Chapter 1. Here you get easy solutions of Lakhmir Singh and Manjit Kaur Physics solution Chapter 1 . Here we have given Chapter 1 all solution of Class 9. Its help you to complete your homework.

• Board – CBSE
• Text Book – Physics
• Class – 9
• Chapter – 01

### Motion

We are already discussed about motion. When the object changes it’s position with respect to surrounding and time then we say that the object is in motion. Such change of position will occur with time. The quantities dealing with study of motion are distance, displacement, speed, velocity, acceleration. This is the quantitative study of motion only. The branch of physics dealing with the study of motion is called kinematics. Let’s solve the exercise based on kinematics.

1> Is displacement a scalar quantity ?

Answer:  No, displacement has magnitude as well as direction and such quantities are called vector quantity therefore displacement is vector quantity.

2> State whether distance is a scalar or a vector quantity.

Answer: Distance has only magnitude so it is scalar quantity.

3> Change the speed of 6 m/s into km/h.

Solution: To solve such type of conversion let’s try to make standard form.

So we conclude two rules

• If the quantity is in km/ hr and we have to convert that quantity in m/s then simply multiply it by 5/18.
• If the quantity is in m/s and we have to convey it in km / hr then simply multiply it by 18/5.

4> What name is given to the speed in a specified direction ?

Answer:  Speed is a scalar quantity so if we give specific direction to this scalar quantity then it become vector. So speed with specific direction will becomes velocity.

5> Give two examples of bodies having non-uniform motion.

Answer:  Examples of bodies having non-uniform motion

• Accelerating lift
• Zig-zag motion of butterflies
• Motion of aero plane before take-off

6> Name the physical quantity obtained by dividing ‘Distance travelled’ by ‘Time taken’ to travel that distance.

Answer: Distance travelled by object/ Time , this is the definition of speed so this quantity must be speed.

7> What do the following measure in a car ?

• (a) Speedometer
• (b) Odometer

Answer: (a) Speedometer measures speed of that vehicle and  (b) Odometer measures the path length of that vehicle.

8> Name the physical quantity which gives us an idea of how slow or fast a body is moving.

Answer:  Speed gives the information about how slow or how fast the motion of body.

9> Under what conditions can a body travel a certain distance and yet its resultant displacement be zero ?

Answer: If a body moves certain path length and retain at its initial position then the resultant displacement of body will zero.

Example- Motion of a body along a closed loop is zero.

10> In addition to speed, what else should we know to predict the position of a moving body ?

Answer: Direction of speed predict the position of moving body. The quantity having magnitude of speed and direction is called velocity. Velocity predicts direction of position of moving body.

11> When is a body said to have uniform velocity ?

Answer:  When a body covers equal displacement in equal interval of time then we say that the body is in uniform velocity.

Example- A train perform rectilinear motion and covers equal distance in equal interval of time.

12> Under which condition is the magnitude of average velocity equal to average speed ?

Answer:  When a body perform rectilinear motion then the magnitude or average velocity equal to average velocity.

13> Which of the two can be zero under certain conditions : average speed Of a moving body or average velocity of a moving body ?

Answer: Average velocity can be zero.

Example: If a body moves and retain in its original position then average velocity can be zero. Average velocity of earth in one year is zero.

14> Give one example of a situation in which a body has a certain average speed but its average velocity is zero.

Answer: Average velocity can be zero.

Example: If a body moves and retain in its original position then average velocity can be zero. Average velocity of earth in one year is zero.

15> Give one example of a situation in which a body has a certain average speed but its average velocity is zero.

If a body movies from point A to B and return at point A in 4 seconds then

A> Average speed = Total path length/ total time

Average speed = 8/4 = 2 metre / second

In this journey average speed is 2 metre / second

B> Average velocity = displacement / Time

Average velocity = 0/ 4 = 0

In this journey average velocity is zero

16> What is the acceleration of a body moving with uniform velocity ?

Answer: Acceleration is feeling with change in velocity. So the acceleration of uniform moving body is zero.

17> What is the other name of negative acceleration ?

Answer: Negative acceleration is also called as retardation.

18> Name the physical quantity whose SI unit is :

(a) m/s

Answer meter is a unit of length and second is SI unit of time

So m/s is unit of speed or velocity.

(b) m/s2

Answer: m/s2 is a unit of acceleration.

19> What type of motion is exhibited by a freely falling body ?

Answer: Freely falling body moves under constant force so it’s motion is under constant acceleration.

20> What is the SI unit of retardation ?

Answer: Retardation is nothing but acceleration so it’s SI unit is m/s2

21> Fill in the following blanks with suitable words :

(a) Displacement is a Vector quantity whereas distance is a   quantity.

(b) The physical quantity which gives both, the speed and direction of motion of a body is called its…… Velocity……

(c) A motorcycle has a steady acceleration of 3 m/s2. This means that every…… Second……….its…..velocity………..increases  3 m/ s

(d) Velocity is the rate of change of . .. … Displacement….lt is measured in m/s

(e) Acceleration is the rate of change of…….velocity…. . It is measured in .m/s2

21> What type of motion, uniform or non-uniform, is exhibited by a freely falling body ? Give reason for your answer.

Answer: Freely falling body exhibit non uniform motion because, It moved down by gravitational force of earth so according to second law of motion when some force exert on the body , it get accelerate and the value of that acceleration is 9.8 m/s2.

22> State whether speed is a scalar or a vector quantity. Give reason for your choice.

Answer: Speed has only magnitude and we know that the physical quantity having only magnitude are considered as scalar quantity therefore speed is also a scalar quantity.

23> Bus X travels a distance of 360 km in 5 hours whereas bus Y travels a distance of 476 km in 7 hours. Which bus travels faster ?

Answer:  Firstly we calculate speed of bus X

From equation 1 and 2 we conclude that bus X travel faster than bus Y.

24> Arrange the following speeds in increasing order (keeping the least speed first) :

(i) An athlete running with a speed of 10 m/s.

(ii) A bicycle moving with a speed of 200 m/min.

(iii) A scooter moving with a speed of 30 km/h

Answer: Unit of speed in above three equations are different so firstly we convert unit of speed in SI system.

• An athlete running with a speed of 10 m/s.
• A bicycle moving with a speed of

• Scooter moving with a speed of

Speed of Scooter is 8.33 m/s

[Trick To convert Km/ hr into m/s then simply multiply that quantify by 5/18]

Now we arrange the sentences in increasing order.

• A bicycle moving with a speed of 3.33 m/s
• Scooter moving with a speed of 8.33 m/s
• An athlete running with a speed of 10 m/s.

25> (a) Write the formula for acceleration. Give the meaning of each symbol which occurs in it.

(b)   A train starting from Railway Station attains a speed of 21 m/s in one minute. Find its acceleration.

Answer: (a) Acceleration is defined as rate of change of velocity with respect to time. It gives information about change in speed and calculated by following formula,

Where,

a – Acceleration

v – Final velocity

u – Initial velocity

t – time required to change in velocity.

(b) Given ,

Train start from railway station so initial speed= 0 m/s

Final velocity = 21 m/s.

Time to change the velocity = 1 min= 60 seconds

As we know,

a = 0.35m/s2

26> (a) What term is used to denote the change of velocity with time ?

(b)   Give one word which means the same as ‘moving with a negative acceleration’.

(c)    The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero ? Give reason for your answer.

Answer: (a) Acceleration is used to denote the change of velocity with time.

(b) A body moving with a negative acceleration is called as retardation.

(c) The displacement of a moving object in a given interval of time is zero. In this statement, distance may be zero or not. When a body travels some distance and retain at its initial position then displacement will not zero but distance will.

27> A snail covers a distance of 100 metres in 50 hours. Calculate the average speed of snail in km/h.

Distance = 100 metres = 0.1 km

Time = 50 hours

28> A tortoise moves a distance of 100 metres in 15 minutes. What is the average speed of tortoise in km/h ?

Distance = 100 metres = 0.1 km

Time = 15 minutes = 0.25 Hr

29> If a sprinter runs a distance of 100 metres in 9.83 seconds, calculate his average speed in km/h,

Distance = 100 metres = 0.1 km

Time = 15 minutes = 0.00273 Hr

30> A motorcyclist drives from place A to B with a uniform speed of 30 km/hr and returns from place B to A with a uniform speed of 20 km/h. Find his average speed.

Speed in forward journey = 30 Km/Hr

Speed in backward journey = 20 Km/hr

Suppose distance between A and B is d km,

Time required for forward journey = distance/ speed

Time required for forward journey = d/30 hr.

Similarly, time required for backword journey = d/20 hr.

As we know, average speed = total distance / total time

Average speed of journey is 24 km/hr

[Trick: To solve such question, average speed = 2v1v2/v1+v2,

where,

v1– velocity for forward journey

v2 – velocity for backward journey]

31> A motorcyclist starts from rest and reaches a speed of 6 m/s after travelling with uniform acceleration for 3 s. What is his acceleration ?

Motorcyclist start from rest so initial speed= 0 m/s

Final velocity = 6 m/s.

Time to change the velocity = 3 seconds

As we know,

A = 2m/s2

Acceleration of that motorcyclist is 2 m/s2

32> An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft’s altitude, how long will it take to reach the ‘sound barrier’ ?

Answer:  Initial speed of aircraft =600 km/hr.

It get accelerate at the rate of 10 km / hr per seconds = 10 km× 60 / hr2 = 600 km/hr2

Final velocity of the aircraft = 1100 km/hr.

We have to calculate time required for acceleration as

t = 50 seconds.

50 seconds required  to reach the ‘sound barriers.

33> If a bus travelling at 20 m/s subjected to a steady deceleration of 5 m/s2, how long will it take to come to rest ?

Given,  initial speed of bus = 20 m/s

Finally bus stop so its speed will becomes zero.

Acceleration = -5 m/s2 ( minus sign indicates that the bus is decelerating) As

Time required to stop the bus is 4 seconds.

34> (a) What is the difference between ‘distance travelled’ by a body and its ‘displacement’ ? Explain with the help of a diagram.

(b)   An ant travels a distance of 8 cm from P to Q and then moves a distance of 6 cm at right angles to PQ. Find its resultant displacement.

• Difference between distance and displacement
• Distance is scalar quantity and displacement is vector quantity.
• Distance: Actual path travelled by the object is called distance
• Displacement: Difference between initial and final position is called displacement.
• Distance is path dependent but displacement is depend on initial and final position.
• Distance has only magnitude but displacement has both magnitude as well as direction.
• Distance is always greater than or equal to magnitude of displacement.

If a particle moves from A to B then distance and displacement both are 4 m.

If a particles moves from A to B and returned to A then distance = 8 m and  displacement= 0 m

• Distance is always positive but displacement may be zero, positive and negative.

B> From the given data, we draw a path of journey

The particle start from point  P and covers distance of 8 cm upto point B. Then it turn right and moves 6 cm.

As we know, displacement is a shortest distance between initial and final point. So we join point P and point R.

By Pythagoras theorem,

(PR)2= (PQ)2 + (QR)2

(PR)2 = (8)2 + (6)2

(PR)2 = 64 + 36

(PR)2 = 100

Taking square root on both sides

PR= 10

Hence the total displacement of the journey is 10 cm

35> Define motion. What do you understand by the terms ‘uniform motion’ and ‘non-uniform motion’ ? Explain with examples.

Answer: Motion: When an object changes it’s position with respect to surrounding and time continuously then we say that the object is in motion.

Uniform motion: When the object moves with constant velocity, such motion is called uniform motion.

Or

When particle covers equal displacement in equal interval of time then such motion is uniform motion.

In uniform motion the velocity of particle remains constant.

There will be no change in magnitude of speed and direction.

Non-Uniform motion: When the object moves with variable velocity or speed, such motion is called non-uniform motion.

Or

When particle covers equal displacement in unequal interval of time then such motion is non-uniform motion.

Example- The particle moves first  2 km in 2 hrs next 2 km in 2 hrs and last 2 km in 2 hrs. In such type of motion particle covers equal distance in equal interval of time.

In non-uniform motion the velocity of particle varies with time.

There will be change in either magnitude of speed or direction.

Example- The particle moves first 2 km in 1 hr next 2 km in 3 hrs and last 2 km in 6 hrs. In such type of motion particle covers equal distance in unequal interval of time.

36> (a) Define speed. What is the SI unit of speed ?

Answer: Speed- Rate of change of distance with reapect to time is called as speed. Speed gives information about how fast change the position with time.

Speed is a scalar quantity. SI unit of speed is meter/second.

(b)   What is meant by (t) average speed, and (it) uniform speed ?

Answer:  In kinematics we have to study the change in position with surrounding. We also interested rate of this change. So speed is a quantity which gives information about this rate of change.

Average speed: Average speed is defined as it is a ratio of total distance covered by an object with respect to time. It gives speed of total journey. It is not dealing with change at every instant. Average speed is calculated by

Average speed = Total distance covered by an object/ Total time required for journey.

Uniform speed: When a body moves under constant seed then we say.

37> (a) Define velocity. What is the SI unit of velocity ? (b) What is the difference between speed and velocity ? (c) Convert a speed of 54 km/h into m/s.

(a) Velocity: The term velocity gives information about how fast the change in the position of an object.

The rate of change of displacement with respect to time is called velocity. Velocity is a vector quantity. SI unit of velocity is meter/second.

(b) between speed and velocity.

• Speed is scalar quantity and velocity is vector quantity.
• Speed is the ratio of distance with time and velocity is the ratio of displacement with time.
• Speed cam be only positive but velocity may be positive, negative and zero.
• Speed is path dependent but velocity is position dependent.

(c) we have to convert 54 Km/hr into m/s

54 km/hr = 54 × 1000/3600 m/s

54 km/hr = 54 × 5/18 m/s

54 km/hr = 15 m/s.

38> (a) What is meant by the term ‘acceleration’ ? State the SI unit of acceleration.(b)Define the term ‘uniform acceleration’. Give one example of a uniformly accelerated motion

Answer: (a) Acceleration gives information about change in speed. So you observed reading in speedometer when you accelerated the motorcycle? The speedometer gives information about acceleration.

Acceleration is defined as rate of change of velocity with respect to time.

SI unit of acceleration is m/s2

(/b) Uniform acceleration:  when the object changes equal velocity in equal interval of time then we say that the object moves with uniform acceleration. Such option is exhibit under constant force.

Example of uniform acceleration- If the object is realised from certain height and ignore the air resistance, such object fall under constant acceleration.

39> The distance between Delhi and Agra is 200 km. A train travels the first 100 km at a speed of 50 km/h. How fast must the train travel the next 100 km, so as to average 70 km/h for the whole journey ?

Given, Total distance= 200 km.

Trains covers first 100.km at speed 50 km/hr. Say

V1 = 50km/hr.

Average velocity = 70km/hr

Train required time t for next 100 km. So velocity of next 100.km is V2. V2 = 100/t

If equal distance covers with different velocity then,

Average velocity =2v1v2/(v1+v2)

70 = 2×50 × v2 / (50 + v2)

70 × 50 + 70 × v2 = 100×v2

3500 = 30v2

V2 = 116.6 m/s

Train must covers 116.6 km/hr for next 100 km to gain average velocity 70 km/hr.

40> A train travels the first 15 km at a uniform speed of 30 km/h; the next 75 km at a uniform speed of 50 km/h; and the last 10 km at a uniform speed of 20 km/h. Calculate the average speed for the entire train journey.

Answer: Given, d1 = 15 Km, v1 = 30 km/hr.

d2 = 75  Km, v2 = 50 km/hr.

d3= 10 Km, v3 = 20 km/hr.

We have to calculate time for each journey.

As we know, v = d/t

t= d/v

Therefore time for first journey t1= d1/v1

Therefore time for first journey t1= 15/30 = 1/2 hr

Therefore time for second journey t2= d2/v2

Therefore time for second journey t2= 75/50 = 3/2 hrs

Therefore time for third journey t3= d3/v3

Therefore time for third journey t3= 10/20

Therefore time for third journey t3= 1/2 hrs.

As we know that average speed = (d1+d2+d3)/(t1+t2+t3)

Average speed = (15+75+10)/(1/2 + 3/2 + 1/2)

Average speed = 100/(5/2)

Average speed = 100×2 / 5

Average speed = 40 km/hr.

41> A car is moving along a straight road at a steady speed. It travels 150 m in 5 seconds :

(a)    What is its average speed ?

(b)   How far does it travel in 1 second ?

(c)    How far does it travel in 6 seconds ?

(d)   How long does it take to travel 240 m ?

Answer: car moving in straight line and covers a distance of 150 m I’m 5 seconds.

(a) Average speed = total distance / total time.

Average seed of journey = 150/5

Average seed of journey = 30 m/s

(b) The car travels distance in 1 is = velocity × time

The car travels distance in 1 is = 30×1

The car travels 30 m in one second.

(c) The car travels distance in 6 s = velocity × time

The car travels distance in 6 is = 30×6= 180 m

The car travels 180 m in six seconds.

(d) Time required to travels 240 m = distance/ velocity.

Time required to travels 240 m = 240/ 30 = 8 seconds

Time required to travels 250 m is 8 seconds.

### Part 2 Solution

1.) (a) What remains constant in uniform circular motion ?

(b) What changes continuously in uniform circular motion ?

a) Speed remains constant in uniform circular motion.

b) Velocity changes continuously in uniform circular motion

2.) State whether the following statement is true or false :

Earth moves round the sun with uniform velocity.

Answer: False, Earth does not move round the sun with uniform velocity.

3.) A body goes round the sun with constant speed in a circular orbit. Is the motion uniform or accelerated ?

Answer: A body goes round the sun with constant speed in a circular orbit. This motion is accelerated motion because in such motion, body changes direction continuously.

4.) What conclusion can you draw about the velocity of a body from the displacement-time graph shown below :

Answer: The above graph shows that the particle moves with constant velocity.

5.) Name the quantity which is measured by the area occupied under the velocity-time graph.

Answer: The quantity which measured by the area occupied under the velocity-time graph is displacement.

6.) What does the slope of a speed-time graph indicate ?

Answer: The slope of a speed-time graph is ,

Slope = (y2 -y1)/ (x2 – x1)

Slope = speed/ time = acceleration.

Slope of speed time graph is acceleration.

7.) What does the slope of a distance-time graph indicate ?

Answer: The slope of a distance-time graph is,

Slope = (y2 -y1)/ (x2 – x1)

Slope = distance/ time = speed

Slope of distance time graph is speed.

8.) Give one example of a motion where an object does not change its speed but its direction of motion changes continuously.

Answer: Examples of a motion where an object does not change its speed but its direction of motion changes continuously.

I) Motion of planets around sun.

II) Motion of artificial satellite around earth.

III) Motion of electron around nucleus.

9.) Name the type of motion in which a body has a constant speed but not constant velocity.

Answer: In uniform circular motionbody has a constant speed but not constant velocity.

10.) What can you say about the motion of a body if its speed-time graph is a straight line parallel to the time axis ?

Answer: If speed – time graph is straight line parallel to time axis then such motion is uniform motion. In this type of motion body moves with constant speed. But in case, body moves circular path then we can not say it is uniform motion.

11.) What conclusion can you draw about the speed of a body from the following distance-time graph?

Answer: The above graph represents that the particle moves with constant speed because particle covers equal distance in equal interval of time.

12.) What can you say about the motion of a body whose distance-time graph is a straight line parallel to the time axis ?

Answer: If distance– time graph is straight line parallel to time axis then In this type of motion body moves with constant speed.

13.) What conclusion can you draw about the acceleration of a body from the speed-time graph shown below?

Answer: The above speed-time graph shows that, the particle moves with increasing speed hence accelerated motion. The acceleration of particle also increases.

14.) A satellite goes round the earth in a circular orbit with constant speed. Is the motion uniform or accelerated ?

Answer: A satellite goes round the earth in a circular orbit with constant speed.In this motion, satellites changes directions continuously so it is accelerated motion.

15.) What type of motion is represented by the tip of the ‘seconds’ hand’ of a watch ? Is it uniform or accelerated ?

Answer: The motion is represented by the tip of the ‘seconds’ hand’ of a watch is accelerated motion.

16.)

(a) If a body moves with uniform velocity, its acceleration is …………….„..

Answer: If a body moves with uniform velocity, its acceleration is zero

(b) The slope of a distance-time graph indicates ……………of a moving body.

Answer: The slope of a distance-time graph indicates speedof a moving body.

(c) The slope of a speed-time graph of a moving body gives

Answer: The slope of a speed-time graph of a moving body gives acceleration

(d) In a speed-time graph, the area enclosed by the speed-time curve and the time axis gives the  by the body.

Answer: In a speed-time graph, the area enclosed by the speed-time curve and the time axis gives the  by the body path length or distance.

(e) It is possible for something to accelerate but not change its speed if it moves in a

Answer: It is possible for something to accelerate but not change its speed if it moves in a Circular motion.

17.) Is the uniform circular motion accelerated ? Give reasons for your answer.

Answer: yes, uniform circular motion is accelerated motion because at every second particles changes it’s direction.

18.) Write the formula to calculate the speed of a body moving along a circular path. Give the meaning of each symbol which occurs in it.

Answer: Speed of a body moving along circular path cm be calculated by,

Speed= n2πR/T

Where,

π= constant term having value= 22/7

n= no of revolution

T= time

19.) Explain why, the motion of a body which is moving with constant speed in a circular path is said to be accelerated.

Answer: When a body moves along circular path with constant speed such motion is called uniform circular motion. In such type of motion particle changes it’s direction at every point so velocity changes. We can understand this  by following diagram.

Such motion is called accelerate motion.

20.) What is the difference between uniform linear motion and uniform circular motion ? Explain with examples.

Answer: Difference between uniform linear motion and uniform circular motion.

a.)  In uniform linear motion, particle does not changes it’s direction but in uniform circular motion particle changes it direction continuously.

b.) Uniform circular motion is accelerated motion but uniform linear motion is not.

21.) State an important characteristic of uniform circular motion. Name the force which brings about uniform circular motion.

Answer: In uniform circular motion, speed is constant but particle changes it’s direction continuously so it becomes accelerated motion. This is important characteristic of uniform circular motion.

Forces is necessary for circular motion because it is accelerated motion. That force is centripetal force.

Examples- Satellite perform circular motion around the earth is due to gravitational force.

Electron moves along nucleus due to electrostatic force.

22.) Find the initial velocity of a car which is stopped in 10 seconds by applying brakes. The retardation due to brakes is 2.5 m/s2.

Initial velocity = u =?

Time = t = 10 seconds

Retardation = -a= -2.5 m/s2

Final velocity= v = 0

From first equation of kinematics,

v =  u + at

0 = u – 2.5×10

u = 25 m/s

Initial velocity of the car is 25 m/s.

23.) Describe the motion of a body which is accelerating at a constant rate of 10 m s-2. If the body starts from rest, how much distance will it cover in 2 s?

Answer: The body moves with constant acceleration of 10 m/s2. It means that the velocity of that body increases bye 10 at every seconds.

It start from rest so initial velocity = u =0

a = 10 m/s2.

Time = t = 2 s.

As we know second equation of kinematics

x = ut + ½  a t2

x = 0×2 + ½  10 × 22

x = 20 m

The body moves 20 m in 2 seconds.

24.) A motorcycle moving with a speed of 5 m/s is subjected to an acceleration of 0.2 m/s2. Calculate the speed of the motorcycle after 10 seconds, and the distance travelled in this time.

Initial velocity = u =5 m/s.

Acceleration = a =0.2 m/s2.

Time = t = 10.seconds.

We have to calculate final speed and distance.

As we know first equation of kinematics,

v = u +at

v = 5 + 0.2× 10

v = 5+ 2

v = 7 m/s.

Now we calculate distance from second equation of kinematics.

x = ut + ½  a t2

x = 5 × 10 + ½  ×0.2 × 102

x = 50 + 0.1×100

x = 60 m.

The speed of particles will 7 m/s and distance covered bye particle will 60 m after 10.seconds.

25.) A bus running at a speed of 18 km/h is stopped in 2.5 seconds by applying brakes. Calculate the retardation produced.

Initial velocity = u =18 km/hr = 18 × 5/18 = 5 m/s

Time = t = 2.5 seconds

Retardation = -a= ?

Final velocity= v = 0

From first equation of kinematics,

v =  u + at

0 = 5– a×2.5

2.5a =5

a = 2 m/s2

The retardation of that bus is 2 m/s2

26.) A train starting from rest moves with a uniform acceleration of 0.2 m/s2 for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

Answer: Given, acceleration = a =0.2 m/s2, time = 5 min = 300 seconds. Initial speed = u =0

Firstly we find final speed,

v = u +at

v = 0+ 0.2× 300

v = 60

v = 60 m/s.

Now we calculate distance from second equation of kinematics.

x = ut + ½  a t2

x = 0×300+ ½  ×0.2 × 3002

x = 0+ 0.1×90000

x = 9000 m.

x = 9km

Velocity after 5 minute will becomes 60 m/s and distance will be 9 km.

27.) Name the two quantities, the slope of whose graph gives :

(a) speed, and

(b) acceleration

The slope of distance time graph is speed and velocity time graph is acceleration.

28.) A cheetah starts rom rest, and accelerates 2 m/s2 for 10 seconds. Calculate

(a) the final velocity

(b) the distance travelled.

Answer: Given, acceleration = a =2 m/s2, time =10 seconds. Initial speed = u =0

Firstly we find final speed,

v = u +at

v = 0+ 2× 10

v = 20

v = 20 m/s.

Now we calculate distance from second equation of kinematics.

x = ut + ½  a t2

x = 0×10+ ½  ×2 × 102

x = 0+ 100

x = 100 m.

The speed of cheetah after 10 seconds will 20 m/s and it covers 100 m.

29.) A train travelling at 20 m/s accelerates at 0.5 m /s2 for 30 s. How far will it travel in this time?

Answer: Given, initial velocity = u = 20 m/s. Acceleration = a = 0.5 m/s2. Time= t = 30 s.

As we know the second equation of kinematics,

x = ut + ½  a t2

x = 20 × 30+ ½  ×0.5× 302

x = 600+0.5× 450

x = 600 + 225

x = 825 m.

Train travels 825 m in 30 in seconds

30.) A cyclist is travelling at 15 m s-1. She applies brakes so that she does not collide with a wall 18 m away. What deceleration must she have ?

Answer: Given, u = 15 m/s. x = 18 m, v=0 a = ?

As we know the third equation of kinematics,

v2 = u2 + 2ax

02 = 152 +2×(- a)× 18

0= 225- 36a

36a = 225

a = 225/36

a = 6.25 m/s2

The cyclists must decelerate at the rate 6.25 m/s2

31.) Draw a velocity-time graph to show the following motion

A car accelerates uniformly from rest for 5 s ; then it travels at a steady velocity for 5 s.

Answer: velocity time graph to show  uniform acceleration upto 5s and after that it moves with constant velocity.

32.) The velocity-time graph for part of a train journey is a horizontal straight line. What does this tell you about (a) the train’s velocity, and (b) about its acceleration ?

Answer: The velocity time graph shows trains journey is horizontal straight line then

(a) For velocity- The train moves with constant steady velocity. .

(b) For acceleration- Acceleration of the train must be zero.

33.) (a) Explain the meaning of the following equation of motion : v = u + at where symbols have their usual meanings.

(b) A body starting from rest travels with uniform acceleration. If it travels 100 m in 5 s, what is the value of acceleration?

v = u +at

This is the first equation of kinematics. It is derived from formula of acceleration and we can calculate final velocity of particle by using this formula.

v = u +at

Where,

v – Final velocity of the particle

u –  Initial velocity of the particle.

a – Acceleration of the particle

t – Time required to accelerate.

(b) Given, That body start from rest so u =0 m/s.

Distance = x= 100 m, time = 5 seconds.

As, we know second equation of kinematics,

x = ut + ½ at2

100 = 0×5 + ½ a52

100×2/ 25= a

a = 8 m/s2

The acceleration of the body  is 8 m/s2

34.) (a) Derive the formula : v = u + at, where the symbols have usual meanings.

(b) A bus was moving with a speed of 54 km/h. On applying brakes it stopped in 8 seconds. Calculate the acceleration.

a) As we know that the definition of acceleration,

Acceleration = (final velocity – Initial velocity )/ time

In symbolic form,

a = (v-u)/t

at = v – u

v = u + at.

Where,

v – Final velocity of the particle

u –  Initial velocity of the particle.

a – Acceleration of the particle

t – Time required to accelerate.

This is our first equation of kinematic.

b) Given, u = 54 km/hr = 54 ×5/18 m/s

u= 15 m/s,

Finally it stop after 8 seconds, v = 0 m/s

Time = t = 8 seconds.

As we know first equation of kinematics,

v = u + at

0 = 15 + a ×8

a = (-15)/8

a = – 1.875

Minus sign of acceleration indicates that speed is decreasing and it is called as retardation.

35.) (a) Derive the formula : s = ut +  at2, where the symbols have usual meanings.

(b) A train starting from stationary position and moving with uniform acceleration attains a speed of 36 km per hour in 10 minutes. Find its acceleration.

a)

As we know average velocity = ( v +u ) /2——–1

Where v- final velocity and u – initial velocity.

First equation of kinematics is,

v = u + at—————2

As we know, velocity= displacement/time

Displacement = velocity× t

s = average velocity× t

Put value of average velocity from equation 1 and final velocity in above equation.

S = ut + ½ at2.

This is second equation of kinematics.

S = ut + ½ at2.

Where,

s– displacement of the particle

u –  Initial velocity of the particle.

a – Acceleration of the particle

t – Time

b) Given, train moving from rest so u = 0 m/s.

v = 36 km/hr = 36 × 5/ 18 m/s

v = 10 m/s

Time = t =.10 min =.600 seconds.

As we know first equation,

v = u + at

10= 0 + a ×600

a = 10/600

a = 0.0167 m/s2

The acceleration of train is 0.0167 m/s2

36.) (a) Write the three equations of uniformly accelerated motion. Give the meaning of each symbol which occurs in them.

(b) A car acquires a velocity of 72 km per hour in 10 seconds starting from rest. Find (i) the acceleration, (ii) the average velocity, and (iii) the distance travelled in this time.

Answer: (a) Three equations of motions are,

1. v = u + at
2. s = ut + ½ at2.
3. v2 = u2 + 2as

Where, s– displacement of the particle

u –  Initial velocity of the particle.

v –  Final velocity of the particle.

a – Acceleration of the particle

t – Time

37.) (a) What is meant by uniform circular motion ? Give two examples of uniform circular motion.

(b) The tip of seconds’ hand of a clock takes 60 seconds to move once on the circular dial of the clock. If the radius of the dial of the clock be 10.5 cm, calculate the speed of the tip of seconds hand of the clock. ( Given π= 22/7 )

a) Circular motion: When an object moves along the circumference of circle with constant speed then such motion is called as uniform circular motion.

Examples-

I.) Motion of tip of second hand of clock.

II.) Motion of moon around earth.

b) Given, time required to complete one revolution = 60 seconds.

Radius = 10.5cm = 0.105 m

When tip completes one rotation then it covers 2 π r distance.

As we know, speed = distance / time

Speed = 2 × π r / t

Speed = 2× π × 0.105/ 60

Speed = 0.0110 m/s.

The speed of tip of second’s hand is 0.011 m/s

38.) Show by means Of graphical method that

v = u + at

Where the symbols have their usual meanings.

Answer: We have to derive first kinematics equation.

Consider a particle moving with initial velocity u, Due to force  it get accelerated with acceleration ‘a’ and gain velocity v. We plot a graph between velocity and time is as,

We can calculate final velocity from graph,

Final velocity = v = BD+ DC———1

But DC is initial velocity so , DC = u

Equation 1 becomes,

v = BD + u————–2

As we know definition of acceleration,

Acceleration= velocity/time.

Line AB represents acceleration.

BD = a ×t

So equation 2 becomes

v = at + u

v= u + at

Where,

v- final velocity

u- Initial velocity

a: A acceleration

t- time

This is a required first kinematics equation.

39.) Show by using the graphical method that :

s = ut+ ½ at2

Where the symbols have their usual meanings.

Answer: We have to derive second kinematics equation. Consider a particle moving with initial velocity u, Due to force it get accelerated with acceleration ‘a’ and gain velocity v. We plot a graph between velocity and time is as,

We know that, area under velocity time graph is displacement. So

Displacement = area of rectangle ADCO + area of triangle ABD

Displacement= base × height + ½ BASE × height.

Displacement= t × u+ ½ t × (v-u)

Multiply and divide last term by

S= ut + ½ t (v-u) (t/t)

S= ut + ½ t × a.        {Since- change in velocity with time is acceleration, (v-u)/t = a}

S = ut + ½ at×t

S = ut + ½ at2

Where,

v- final velocity

u- Initial velocity

a: A acceleration

t- t

s- displacement.

This is required equation.

40.) Derive the following equation of motion by the graphical method :

v2 = u2 + 2as

Where the symbols have their usual meanings

Answer: We have to derive second kinematics equation. Consider a particle moving with initial velocity u, Due to force  it get accelerated with acceleration ‘a’ and gain velocity v. We plot a graph between velocity and time is as,

We know that, area under velocity time graph is displacement. So

Displacement = s = area of trapezium ABCO

S = ½ (sum of parallel sides) × base.

S = ½ (AO+ BC) × OC

S = ½ ( u+ v) × t———–1

As, first equation of kinematics, v = u +at

v-u = at

(v-u )/a = t

Put this value in equation 1

S= ½ (u+v) × (v-u)/a

But we know, (a+b)×(a-b) = a2– b2

S= ½ (u2– v2 ) / a

2as = u2– v2

v2 = u2 + 2as

Where, v- final velocity

u- Initial velocity

a: A acceleration

t- t

s- displacement.

This is required equation.

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Updated: May 28, 2022 — 3:39 pm