# TEXTBOOK ANSWERS AND SOLUTIONS OF CBSE CLASS VII SCIENCE Chapter 13 Motion And Time

1) 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.
(i) Oscillatory
(ii) Straight Line
(iii) Circular
(iv) Oscillatory
(v) Oscillatory
(vi) Straight Line
2)Which of the following are not correct?
(i) The basic unit of time is second.
(ii) Every object moves with a constant speed.
(iii) Distances between two cities are measured in kilometres.
(iv) The time period of a given pendulum is not constant.
(v) The speed of a train is expressed in m/h.
(i) True
(ii) False. Different objects have different
speeds.
(iii) True
(iv) False. The time period of a given pendulum is fixed.
(v) False. The speed of train is usually expressed in km/hr or miles/hr.
3) A simple pendulum takes 32 s to complete 20 oscillations.
What is the time period of the pendulum?
ANSWER: – Time Period = Total Time Taken ÷ No. of Oscillations
No. of Oscillations = 20
Total Duration = 32s
Time Period = 32 ÷ 20 = 1.6s

4) The distance between two stations is 240 km. A train takes 4
hours to cover this distance. Calculate the speed of the train.
ANSWER: – Distance between two stations = 240 km
Time taken by train to cover the distance = 4 hours
Speed = Distance ÷ Time
Speed of Train = 240 / 4 = 60km/hour.
5) 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.
ANSWER:- Odometer reading at 8.30 AM (O1) =57321.0 km
Odometer reading at 8.50 AM (O2) = 57336.0 km
Distance covered by car = O 2 – O1 = 57336.0 –
57321.0 = 15 Km
Time interval between 8.30 AM to 8.50 AM = 20 min.
Speed of car = Distance ÷ Time = 15km ÷ 20 min
= 0.75 km/min
1 Hr = 60 min.
Speed of Car in km/hr = 0.75 × 60 = 45 km/hr.
6) 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.
ANSWER: – Distance = Speed ×Time
Speed of the bicycle = 2 m/s
Time taken to reach school = 15 min = 15 × 60seconds = 900s
Distance = 2 × 900 = 1800 m = 1800/1000Km = 1.8 Km
7) 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.

ii-

8) Which of the following relations is correct?
(i) Speed = Distance × Time
(ii) Speed=Distance/time
iii) Speed=time/distance
iv) Speed=1/distance*time
9) The basic unit of speed is:
(a) km/min
(b) m/min
(c) km/h
(d) m/s
10) A car moves with a speed of 40 km/h for 15minutes 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
ANSWER: – Case I
Speed of the car = 40 km/h

Time taken = 15 min =

Distance covered, d1 = Speed × Time taken = 40 × 0.25 = 10 km
Case II
Speed of the car = 60 km/h
Time taken = 15 min =

Distance covered, d2 = Speed × Time taken = 60× 0.25 = 15 km
Total distance covered by the car, d = d1 + d2 =10 + 15 = 25 km
Therefore, the total distance covered by the car is 25 km.
11) Suppose the two photographs, shown in Figure 1 and Figure 2,
had been taken at an interval of 10 seconds. If a distance of 100 metres is
shown by 1 cm in these photographs, calculate the speed of the blue car.
ANSWER: – 1 cm = 100 m
Distance covered by blue car = 2.0 cm = 2.0 ×100 = 200 m
Time taken to cover 200m = 10s
Speed of Car = Distance ÷ Time = 200 / 10 =20m/s
Speed of Car (in km/hr) = 20 × 3600s/1000m =72 km/hr

13)Which of the following distance- time graphs shows a truck
moving with speed which is not constant?

ANSWER: – Graph (iii)