Chapter 9: Motion and Time

Q1Written answer

Classify the following as motion along a straight line, circular or oscillatory motion:

(i) Motion of your hands while running.

(ii) Motion of a horse pulling a cart on a straight road.

(iii) Motion of a child in a merry-go-round.

(iv) Motion of a child on a see-saw.

(v) Motion of the hammer of an electric bell.

(vi) Motion of a train on a straight bridge.

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Worked solution

i) Oscillatory

ii) Motion along a straight line

iii) Circular motion

iv) Oscillatory motion

v) Oscillatory motion

vi) Motion along a straight line.

Q2Multiple choice

Which of the following are not correct?

(v) The speed of a train is expressed in m/h.

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Worked solution

Incorrect statements are:

(ii) Every object moves with a constant speed.

(iv) The time period of a given pendulum is constant.

(v) The speed of a train is expressed in m/h.

Q3Written answer

A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?

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Worked solution

Number of oscillations = 20

Total time taken to complete 20 oscillations = 32 s

time period calculation

Q4Written answer

The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

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Worked solution

Distance between two stations = 240 kms

Total time take = 4 hrs/240 minutes

Speed calculation

= 60 km/h

Q5Written answer

The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.

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Worked solution

Initial reading of the odometer = 57321.0

Final reading of the odometer = 57336.0

Distance covered by the car = Final reading of the odometer – Initial reading of the odometer

= 57336.0 – 57321.0 = 15 kms

Starting time of car is 8:30 and it stops at 8:50

Hence, time taken by car = 20 mins

speed calculation

= 45 km/h

Q6Written answer

Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.

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Worked solution

Time taken by Salma to reach her school by bicycle = 15 mins= 15 x 60 = 90 s

Speed of Salma’s bicycle= 2m/s

Speed formula

Distance covered = speed x time taken

= 2 x 900 = 1800 m

1000m = 1 km

distance covered calculation

= 1.8 kms

Q7Written answer

Show the shape of the distance-time graph for the motion in the following cases:

(i) A car moving with a constant speed.

(ii) A car parked on a side road.

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Worked solution

Distance time graph

Q8Multiple choice

Which of the following relations is correct?

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Correct option: B

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Worked solution

Answer is (ii) Speed = Distance/Time

Q9Written answer

The basic unit of speed is:

(i) km/min

(ii) m/min

(iii) km/h

(iv) m/s

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Worked solution

Answer is (iv) m/s

Q10Written answer

A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:

(i) 100 km (ii) 25 km (iii) 15 km (iv) 10 km

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Worked solution

The answer is (ii) 25 km

Calculation:

When the speed of the car is 40 km/h

Time taken = 15 min = 15/60 = 0.25 h

Speed formula

Distance covered d₁ = speed x time taken

= 40 x 0.25 = 10 kms

When the speed of the car is 60 km/ h

Speed formula

Distance covered d₂ = speed x time taken

= 60 x 0.25= 15 kms

Total distance covered by the car = d₁ + d₂

= 10 + 15

= 25 kms

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