##### NCERT-Class VII-Science

01-Nutrition in Plants

02-Nutrition in Animals

03-Fibre to Fabric

04-Heat

05-Acids Bases and Salts

06-Physical and Chemical Changes

07-Weather Climate and Adaptations of Animals to Climate

08-Winds Storms and Cyclones

09-Soil

10-Respiration in Organisms

11-Transportation in Animals and Plants

12-Reproduction in Plants

13-Motion and Time

14-Electric Current and its Effects

15-Light

16-Water A Precious Resource

17-Forests Our Lifeline

18-Wastewater Story

### Chapter 13-Motion and Time-Exercise Solutions

NCERT Book Page Number- 156

Q.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.

Answer:

(i) Motion of your hands while running – Oscillatory motion.

Explanation: –

During running the hands move to and fro and as this motion gets repeated after certain time interval.

Therefore it is oscillatory motion.

(ii) Motion of a horse pulling a cart on a straight road – Straight line motion.

Explanation: –

As the horse cart is moving on a straight road. Therefore motion is along the straight line.

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

Explanation: –

Motion of the merry-go-round is circular. Therefore kids sitting inside it experiences the circular motion.

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

Explanation: –

As the see-saw goes up and down continuously .Therefore it is oscillatory motion.

(v) Motion of the hammer of an electric bell – Oscillatory motion.

Explanation: –

When the hammer vibrates the bell it starts vibrating. It is an example of oscillatory motion.

(vi) Motion of a train on a Straight Bridge – Straight line motion.

Explanation: –

The train is moving on the straight bridge. It exhibits motion of a straight line.

Q.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.

Answer:

(i) The basic unit of time is second – Correct

Explanation: –

SI unit of time is second.

(ii) Every object moves with a constant speed – Not correct

Explanation: –

Speed of object is constant or variable.

(iii) Distances between two cities are measured in kilometres – Correct

Explanation: –

The distance between two cities is very large. And as a kilometre is a bigger unit therefore it is used

to measure the distance between two cities.

(iv) The time period of a given pendulum is not constant – Not correct

Explanation: –

The time period depends upon the length of the thread. Therefore it will be constant for a particular pendulum.

(v) The speed of a train is expressed in m/h – Not correct

Explanation: –

The speed of train is measured either in km/h or m/s.

Q.3. A simple pendulum takes 32s to complete 20 oscillations. What is the time period of the pendulum?

Answer: The time taken to complete one oscillation is known as time period of the pendulum.

$Time Period = \frac{Total time taken}{Number of Oscillations}$

Given: – 20 oscillations taking 32s to complete.

Therefore 1 oscillation will take = $\frac{32}{20}$ sec = 1.6 second

Therefore 1.6s is the time period of the pendulum.

Q.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: $\Speed=frac{Distance travelled}{ Time}$

$ = \frac{240 km}{4h}

$=60 Kmh^{-1}$

Therefore the speed of the train is $=60 Kmh^{-1}$.

Q.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: Initial reading of the odometer of the car =57321.0 km

Final reading of the odometer of the car = 57336.0 km

The car starts at 8:30 AM and stops at 8:50 AM.

Distance covered by car = (57336 – 57321) km = 15 km

Time taken between 08:30 AM to 08:50 AM = 20 minutes

= $\frac{20}{60}$ hour

= $\frac{1}{3} hour$

So Speed in km/min

$ Speed = \frac{Distance travelled}{Time}$

$ = \frac{15km}{20min}$

$ =0.75km min^{-1}$

Speed in km/h

$Speed =\frac{Distance travelled}{Time}$

$= \frac{15km}{(1/3h}$

$= \frac{(15 × 3) km}{1h}$

=45km/h

Q.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:

Speed = 2m/s

Time taken to reach school = 15 minutes = 15 × 60 seconds = 900 seconds

$Speed = \frac{Distance travelled}{Time}$

Distance = (Speed x Time)

=2 × 900 =1800 m

Also 1km= 1000m

Therefore 1800 × $\frac{1}{1000}$ = 1.8km.

The distance between her house and the school is 1.8 km.

Q.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.

Answer: (i) A car moving with a constant speed covers equal distance in equal intervals of time. It will be a uniform motion.

Distance-time graph will be as below:

(ii) A car parked on a road there is no change in the distance with the time. No motion.

Therefore the graph so obtained will be parallel to x-axis.

Distance- time graph will be as below:-

Q.8. Which of the following relations is correct?

(i) Speed = (Distance × Time)

(ii) $ Speed = \frac{Distance}{Time}$

(iii) $Speed = \frac{Time}{Distance}$

(iv) $ Speed = \frac{1}{ (Distance ×Time}$

Answer: Speed of an object is given by the relation:-

(ii) $Speed =\frac{Distance}{Time} $

NCERT Book Page Number-157

Q.9. The basic unit of speed is:

(i) km/min (ii) m/min (iii) km/h (iv) m/s

Answer:

(iv) m/s.

The unit of distance is metre (m) and of time is second(s).

$Speed = \frac{Distance}{Time}$

Therefore the basic unit of speed is m/s.

Q.10. 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:

• 100 km

• 25 km

• 15 km

• 10km

Answer:

(ii) 25 km

Case I:

Speed = 40 km/h

Time = 15 min = $\frac{15}{60}$ hour

Distance (d1) = Speed × Time = 40 × $\{\frac{15}{60}$ = 10 km

Case II:

Speed = 60 km/h

$Time = 15 min = \frac{15}{60}$ hour

Distance $ (d_{2} = (Speed × Time) = 60 × \frac{15}{60}$ = 15 km

Total distance (d) = $ (d_{1} + d_{2}$) = 10 km + 15 km = 25 km

Therefore the total distance covered by the car = 25km.

Q.11. Suppose the two photographs, shown in Fig (1) and Fig (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.

With the help of the scale we will first measure the distance.

Suppose the distance measured is 2cm.

So, the distance covered d = 2 x 100= 2m. (Because 1m=100cm).

Time taken = 10seconds.

Speed = (Distance/Time) = (200m/10s) = 20m/s.

Therefore the speed of the blue car = 20m/s.

Q.12. Fig (3) shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Fig (3):- Distance– time graph for the motion of two cars.

Answer: In distance – time graph speed is measured by its slope.

Vehicle A is moving faster as the slope of the graph of A is more than the slope of the graph B.

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

Answer:

This shows that the truck is moving with variable speed.