AP State Board new syllabus AP Board Solutions Class 9 Physics 7th Lesson Motion Questions and Answers.

## AP 9th Class Physical Science 7th Lesson Questions and Answers Motion

### 9th Class Physics 7th Lesson Motion Questions and Answers (Exercise)

Question 1.

An athlete completes one round of a circular track of diameter 200 m in 40s. What will be the distance covered and the displacement at the end of 2 minutes 20s ?

Answer:

Diameter of circular track =200 m

Time taken by athlete to complete one round = 40

Total time athlete run around the track = 2 min 20s

= 2 × 60+20

= 120+20=140s

No. of rounds he completed =

So, after completing 3 rounds starting from ‘A’ he will end up at ‘B’ in 3 \(\frac{1}{2}\) rounds in 2 min 20 sec.

i) Displacement = AB = 200 m

ii) Distance = No. of rounds ×Circumference

Question 2.

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C ?

Answer:

a) From A to B

Distance s = 300m

Time t =2 min 30sec

= 2 × 60+30

= 120 + 30 = 150sec

[Note: If direction doesn’t change, speed velocity does not change] Average velocity

(υ)=2 ms^{-1}

b) From A to C

Distance (s) =AB+BC

= 300+100=400m

Time (t) = 2min+30 sec + 1min

= 120+30+60 = 210 sec

Question 3.

Abdul, while driving to school, computes the average speed for his trip to be 20 km h^{-1}. On his return trip along the same route, there is less traffic and the average speed is 30 km h^{-1}. What is the average speed for Abdul’s trip?

Answer:

Abdul driving : Home to school (H→S)

Speed (s) =20 km h^{-1}

distance = d;

time = t_{1
}distance = speed x time [d = s x t]

d = 20 x t_{1} = 20 t_{1}⇒ t_{1}= \(\frac{d}{20}\)

School to Home [S→H]

speed (s) = 30 kmh^{-1}; distance = d

[Note : distance from H → S & S → H remains same]

time = t_{2}; d = s x t_{2}

Question 4.

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^{-2 }for 8.0 s. How far does the boat travel during this time ?

Answer:

Motor Boat : Initial velocity u =0; Acceleration a =3 ms^{-2 } ; Time t = 8sec^{
} Displacement s =

∴ Boat travels 96 m in 8 sec

Question 5.

A driver of a car travelling at 52 km h^{-1} applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h^{-1} in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

‘

Question 6.

Below figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions :

a) Which of the three is travelling the fastest ?

b) Are all three ever at the same point on the road?

c) How tar nas C travelled when B passes A ?

d) How far has B travelled by the time it passes C ?

Answer:

a) From d-t graph, d =12 ; t = 1.8

i) Speed of car-A :

ii) d = 12 ; t =1.4

Speed of car – B :

iii) d =10 ; t=1.7

Speed of car -C :

Car – B travels faster.

b) No, all three never met.

c) 7 km from graph.

d) 5 km from graph.

Question 7.

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^{-2}, with what velocity will it strike the ground? After what time will it strike the ground?

Answer:

Ball : Initial velocity u =0

Acceleration a =10 ms^{-2}

Height (displacement) s =20m

v^{2}-u^{2}=2 as

v^{2 }– 0^{2}=2 × 10 × 20

v^{2}=(20)^{2}

v = 20 ms^{-1}

It will strike the ground at 20 ms^{-1}

v = u+a t

20=0+10 t

20 = 10 t

t = 2 sec

It will strike the ground after 2 sec.

Question 8.

The speed – time graph for a car is shown is adjacent figure.

a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

A.

From s -t graph, Area under s – t graph denotes distance travelled

b) Which part of the graph represents uniform motion of the car?

Answer: Car travelling in uniform motion between time interval 6 to 9 sec.

Question 9.

State which of the following situations are possible and give an example for each of these :

a) an object with a constant acceleration but with zero velocity.

Answer: Yes. Example : If an object is in free fall, then it will accelerate with zero velocity.

b) an object moving with an acceleration but with uniform speed.

Answer: Yes. Example : When you turn a vehicle it may have acceleration with uniform speed.

c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: Yes. Example : When an object is in circular motion its acceleration is in perpendicular direction.

Question 10.

An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer:

### 9th Class PS 7th Lesson Questions and Answers (InText)

Question 1.

An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer:

Yes.

Reason : If an object travels from a point ‘ A ‘ to ‘ B ‘ and returns from ‘ B ‘ to ‘ A ‘, then its net displacement is zero.

Ex :

Distance =AB+BA=10+10=20m

Displacement s = 0

Question 2.

A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

Let the farmer moves along a field A B C D,

AB = BC = BC=AD=10m

Time of travel =40sec

Total time farmer walked around square field

= 2 min 20 sec

= 2 × 60+20 [∵ 1min = 60 sec]

= 120+20=140 sec

Number of rounds he walked =

So, after completing 3 rounds he reaches from A to C.

∴ AC is the displacement.

In ΔABC, By Proportional Theorem,

H^{2}=S^{2}+S^{2 }[∵Triangulation law of vectors ]

AC^{2 }= AB^{2}+BC^{2}

AC^{2}=10^{2}+10^{2}=100+100=200

AC = \(\sqrt{2 \times 100}= \sqrt{2 \times 10^2}=10 \sqrt{2} \mathrm{~m} = 10 ×1.414

[∵ \sqrt{2} = 1.414]\)

AC=14.14m is the displacement of the farmer.

Question 3.

Which of the following is true for displacement?

a) It cannot be zero.

b) Its magnitude is greater than the distance travelled by the object.

Answer:

a) False.

Reason : Displacement can be positive, negative and can be zero.

b) False.

Reason : Displacement is the shortest distance between initial point to final point, so it can be equal or less than distance travelled but can never be greater than distance travelled.

Question 4.

Distinguish between speed and velocity.

Speed | Velocity |

1) Distance travelled by the object in unit time. | 1) Displacement of an object in unit time. |

2) It is scalar. | 2) It is vector. |

3) It’s magnitude is always positive. | 3) It’s magnitude can be positive or negative. |

4) Speed = \(\frac{\text { distance }}{\text { time }} ; s=\frac{\mathrm{d}}{\mathrm{t}}\) | 4) Velocity = \(\frac{\text { displacement }}{\text { time }} ; v=\frac{s}{t}\) |

5) Units: ms^{-1} |
5) Units: ms^{-1} |

Question 5.

Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer:

If an object travels in straight line without changing its direction, then its average velocity is equal to average speed.

Question 6.

What does the odometer of an automobile measure ?

Answer:

- The odometer of an automobile measures the total distance that the vehicle has travelled since it was first put into use.
- It is typically located on the dashboard and displays the number of miles or kilometres that the vehicle has travelled.

Question 7.

What does the path of an object look like when it is in uniform motion ?

Answer:

The path of an object in uniform motion is a straight line.

Question 8.

During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station ? The signal travels at the speed of light, that is, 3 x 10^{8} ms^{-1}.

Answer:

Time of reaching signal = 5 min = 5 x 60 = 300 sec [ ∵ 1 min = 60 sec]

Speed of the signal s = 3 x 10^{8} ms^{-1 }Distance of the spaceship from ground d = s x t

= 3 x 10^{8} x 300

= 900 x 10^{8
}d = 9 x 10^{10} m

[Note : While doing application sums in physics make sure that all measuring units must be in same units.]

Question 9.

When will you say a body is in

i) uniform acceleration?

ii) non-uniform acceleration?

Answer:

i) If a body travels in a straight line and its velocity increases (or) decreases by equal time intervals of time, then the body said to be travelling with uniform acceleration.

ii) If a body’s velocity is non-uniform, then its acceleration is non-uniform.

Question 10.

A bus decreases its speed from 80 km h^{-1} to 60 km h^{-1} in 5 s. Find the acceleration of the bus.

Answer:

Bus:

Question 11.

A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^{-1 }in 10 minutes. Find its acceleration.

Answer:

Train starting from station refers that it is at rest.

So, its initial velocity is zero.

Initial velocity u=0

Final velocity v=40 km h^{-1}

Acceleration of train is 0.0185 ms^{-2}

Acceleration of train in km h^{-1 }=240 km h^{-1}

Question 12.

What is the nature of the distance-time graphs for uniform and non-uniform motion of an object ?

Answer:

Uniform Motion :

1. Distance-time graph for uniform motion is a straight line, which means the object covers an equal distance in equal time intervals.

2. The slope of the distance-time graph represents the speed of the object.

Non-Uniform Motion :

1. The distance-time graph for nomuniform motion is a curved line, indicating that the object covers a different distance in equal time intervals.

2. The slope of the tangent at any point on the distance-time graph represents the instantaneous speed of the object at that point.

Question 13.

What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis ?

Answer:

If an object’s distance-time graph is a straight line parallel to the time axis, this means that the object is not moving. In other words, its position is not changing with respect to time.

Question 14.

What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer:

If the speed-time graph is a straight line parallel to the time axis, it means that the object is moving at a constant speed and there is no change in its velocity over time. This type of motion is called uniform motion and it occurs when an object moves at a constant speed in a straight line.

Question 15.

What is the quantity which is measured by the area occupied below the velocity time graph?

Answer:

The area under the velocity-time graph denotes displacement.

Question 16.

A bus starting from rest moves with a uniform acceleration of 0.1 ms^{-2} for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer:

Bus starting from rest means its initial velocity is zero.

Initial velocity u=0

Acceleration a = 0.1 ms^{-2}

Time t =2min = 2 × 60=120 sec

a) Speed acquired refers to final velocity v=u+at

v=0+0.1 × 120

v = 12 ms^{-1}

b) Distance travelled refers to displacement v^{2}^{ }– u^{2}= 2 as

(12)^{2}-(0)^{2}=2 × 0.1 × s ⇒ 144=2× \(\frac{1}{10}\)× s = 720 m

The bus travelled 720 m in 2 min

Question 17.

A train is travelling at a speed of 90 km h^{-1}. Brakes are applied so as to produce a uniform acceleration of -0.5 km^{2}. Find how far the train will go before it is brought to rest.

Answer:

Train initial velocity u=90 km h^{-1 }=

Acceleration (or) deceleration a =-0.5 ms^{-2}

As the train came to rest its final velocity v=0

v^{2}-u^{2}=2 as

0^{2}-25^{2}=2 ×(-0.5) × s ⇒-625=-1 × s=625m

The train travelled 625 m

Question 18.

A trolley, while going down an inclined plane, has an acceleration of 2 cm s^{-2}. What will be its velocity 3s after the start ?

Answer:

Trolley : Acceleration a=2cm s^{-2}; Time t = 3 sec

Initial velocity u =0.

v=u+at=0+2 ×s 3=6cm s^{-1}

Question 19.

A racing car has a uniform acceleration of 4 ms^{-2}. What distance will it cover in 10s after start?

Answer:

Racing Car: Acceleration a = 4 ms^{-2 }; Time t = 10 sec

As the racing car starts from rest its initial velocity is zero.

Initial velocity u=0

Racing car covers 200 m in 10 sec

Question 20.

A stone is thrown in a vertically upward direction with a velocity of 5 ms^{-1}. If the acceleration of the stone during its motion is 10 ms^{-2} in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer:

Stone thrown upwards :

Initial velocity u=5 ms^{-1}

Acceleration a=-10 ms^{-2} [as it is thrown upwards a is – ve]

Final velocity v =0

v^{2}-u^{2}=2 as

(0)^{2}-(5)^{2}=2 × (-10) × s

– 25=20 × s [At point B, v becomes zero before the stone comes down]

25=20 ×s

Time taken reach height v=u+at

0=5-10t

So, stone will reach a height of 1.25 m in 0.5 sec.

Examples

Question 1.

An object travels 16 min 4 s and then another 16 m in 2 s. What is the average speed of the object?

Answer:

Total distance travelled by the object =16m+16 m=32 m

Total time taken =4 s + 2s = 6s

\(\text { Average speed } \stackrel{i}{=} \frac{\text { Total distance travelled }}{\text { Total time } \cdot \text { taken }}=\frac{32 \mathrm{~m}}{6 \mathrm{~s}}=5.33 \mathrm{~ms}^{-1}\)

Therefore, the average speed of the object is 5.33 ms^{-1}.

Question 2.

The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h^{-1} and ms^{-1}

Answer:

Distance covered by the car, s=2400km-2000 km =400 km

Time elapsed, t=8h

Average speed of the car is,

The average speed of the car is 50 km h^{-1} or 13.9 ms^{-1}

Question 3.

Usha swims in a 90m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.

Answer:

Total distance covered by Usha in 1 min is 180 m

Displacement of Usha in 1min = 0m

The average speed of Usha is 3 ms^{-1} and her average velocity is 0 ms^{-1}.

Question 4.

Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 ms^{-1} in 30s. Then he applies brakes such that the velocity of the bicycle comes down to 4 ms^{-1} in the next 5s. Calculate the acceleration of the bicycle in both the cases.

Answer:

In the first case :

initial velocity, u =0; final velocity, v=6 ms^{-1} ; time, t=30 s.

From Eq. (7.3), we have a =\(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\)

Substituting the given values of u, v and t in the above equation, we get

a = \(\frac{\left(6 \mathrm{~ms}^{-1}-0 \mathrm{~ms}^{-1}\right)}{30 \mathrm{~s}}\) = 0.2ms^{-2}

In the second case :

initial velocity, u=6 ms^{-1}; final velocity, v=4 s^{-1} ; time, t=5s

Then, a = \(\frac{\left(4 \mathrm{~ms}^{-1}-6 \mathrm{~ms}^{-1}\right)}{5 \mathrm{~s}} \)=-0.4 ms^{-2}

The acceleration of the bicycle in the first case is 0.2 ms^{-2} and in the second case, it is -0.4 ms^{-2}.

Question 5.

A train starting from rest attains a velocity of 72 km h^{-1} in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

Answer:

We have been given u =0 ; v=72km h^{-1 }= 20 m s^{-1} and t=5 minutes =300 s.

The acceleration of the train is \(\frac{1}{15}\) ms^{-2} and the distance travelled is 3 km

Question 6.

A car accelerates uniformly from 18 km h^{-1} to 36 km h^{-1} in 5s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Answer:

We are given that u=18 km h^{-1 }=5 ms^{-1} ; v=36 km h^{-1}=10 ms^{-1} and t=5s

The acceleration of the car is 1 ms^{-2} and the distance covered is 37.5 m.

Question 7.

The brakes applied to a car produce an acceleration of 6 ms^{-2} in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.

Answer:

We have been given a=-6 ms^{-2} ; t=2 s and v=0 ms^{-1
}From Eq. (7.5) we know that v =u+at

0 = u+(-6 ms^{-2}) × 2 s (or) u=12 ms^{-1}

From Eq. (7.6) we get s=ut+ \(\frac{1}{2}\) at^{2}

= (12 ms^{-1}) × (2 ms)+\(\frac{1}{2}\)(-6 ms^{-2})

(2s)^{2}=24m-12m=12 m

Thus, the car will move 12 m before it stops after the application of brakes.

### 9th Class Physical Science Chapter 7 Questions and Answers (Lab Activities)

(Page No. 98) Activity 7.1

Question 1.

Discuss whether the walls of your classroom are at rest or in motion.

Answer:

The walls of our classroom are at rest.

(Page No. 98) Activity 7.2

Question 2.

1) Have you ever experienced that the train in which you are sitting appears to move while it is at rest?

2) Discuss and share your experience.

Answer:

This appears when the objects outside the train starts moving.

(Page No. 100) Activity 7.3

Question 3.

1) Take a metre scale and a long rope.

2) Walk from one corner of a basket – ball court to its opposite corner along its sides.

3) Measure the distance covered by you and magnitude of the displacement.

4) What difference would you notice between the two in this case ?

Answer:

Distance & magnitude of displacement is not same.

(Page No. 100) Activity 7.4

Question 4.

1) Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km.

2) Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India.

Answer:

By map the magnitude of displacement between Bhubaneshwar and New Delhi is 1630 km (approx).

(Page No. 102) Activity 7.5

Question 5.

1) The data regarding the motion of two different objects A and B are given in below Table.

Time | Distance travelled by object A in m | Distance travelled by object B in m |

9:30 am | 10 | 12 |

9:45 am | 20 | 19 |

10:00 am | 30 | 23 |

10:15 am | 40 | 35 |

10:30 am | 50 | 37 |

10:45 am | 60 | 41 |

11:00 am | 70 | 44 |

2) Examine them carefully and state whether the motion of the objects is uniform or non-uniform.

Answer:

Car – A covering equal distances in equal time intervals.

Car – B covering unequal distances in equal time intervals.

∴ Car – A is with uniform speed.

Car – B is with non-uniform speed.

(Page No. 104) Activity 7.6

Question 6.

Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is 4km h^{-1}, estimate the distance of the bus stop or school from your house.

Answer:

Let us assume you have taken 10 min to walk from your house to bus stop.

Speed of walking (s)=4 km h^{-1
}

(Page No. 106) Activity 7.7

Question 7.

1) At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.

2) Can you answer why this happens?

3) Measure this time interval using a digital wrist watch or a stop watch.

4) Calculate the distance of the nearest point of lightning. (Speed of sound in air =346 ms^{-1}.

Answer:

1) Light travels faster than sound.

2) If thunder is heard 5 sec after flash, then the distance of the hearest point of lightning can be d=s × t ⇒ d=346 × 5 ⇒ d=1730 m

(Page No. 108) Activity 7.8

Question 8.

1) In your everyday life you come across a range of motions in which

a) acceleration is in the direction of motion,

b) acceleration is against the direction of motion,

c) acceleration is uniform,

d) acceleration is non-uniform.

2) Can you identify one example each for the above type of motion?

Answer:

1) Object falling vertically downward.

2) Object is projected vertically upward.

3) Rotation of earth.

4) Vehicle moving on a road.

(Page No. 114) Activity 7.9

Question 9.

1) The times of arrival and departure of a train at three stations A,B and C and the distance of stations B and C from station A are given in below table.

Distances of stations B and C from A and times of arrival and departure of the train

Station | Distance from A | Time of arrival (hours) | Time of departure (hours) |

A | 0 | 08.00 | 08:15 |

B | 120 | 11:15 | 11:30 |

C | 180 | 13:00 | 13:15 |

2) Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.

Answer:

The d-t graph denotes the train is travelling with non-uniform speed.

(Page No. 116) Activity. 7.10

Question 10.

1) Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route.

Answer:

Table shows the distance travelled by them in different times.

Distance covered by Feroz and Sania at different times on their bicycles.

Time | Distance travelled by Feroz (Ion) | Distance travelled by Sania (Km) |

8:00 am | 0 | 0 |

8:05 am | 1.0 | 0.8 |

8:10 am | 1.9 | 1.6 |

8:15 am | 2.8 | 2.3 |

8:20 am | 3.6 | 3.0 |

8:25 am | — | 3.6 |

2) Plot the distance-time graph for their motions on the same scale and interpret.

Answer:

Interpretation : From the graph, we can observe that Feroz and Sania started at the same point and moved in the same direction.

Feroz travelled faster and covered more distance in the same time intervals.

(Page No. 122) Activity 7.11

Question 11.

1) Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end, as shown in below figure.

A stone describing a circular path with a velocity of constant magnitude

2) Now, let the stone go by releasing the thread.

3) Can you tell the direction in which the stone moves after it is released?

4) By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not.

Answer:

1) The stone moves tangentially.

2) The direction changes every time.