**NCERT Solutions for Class 7 Science Chapter 13 Motion and Time**

## Solutions of NCERT for class 7 chapter 13 motion and time exercise:

- 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) Oscillatory motion

While running, our hands move in back and forth direction and repeat its movement after some time. Hence it is an oscillatory motion.

(ii) Motion along the straight line-

A horse is pulling a cart on a straight road. Therefore it follows the motion along a straight line.

(iii) Circular motion-

Since the merry-go-round moves in a circular motion

(iv) Oscillatory motion-

The see-saw moves up and down continuously. It oscillates up-down.

(v) Oscillatory motion

(vi) Motion along a straight line-

The train is moving on a straight bridge. Therefore it follows the motion along the straight line.

- 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 constant.

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

**Answer:**

The option (ii), (iv) and (v) are not correct.

(ii) The object can move with variable or constant speed.

(iv) The time period of a pendulum depends on the length of the thread. So, it is constant for a particular length.

(v) Speed of train is measured either in km/h or m/s not in m/h

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

**Answer:**

Given that,

A simple pendulum takes 32 s to complete 20 oscillations. It means,

No. of oscillation = 20, and

Total time is taken = 32 s

According to question,

Time period = (T)

**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:**

Given that,

Distance between the two stations = 240 km

and, time taken = 4 hour

Therefore, Speed = Distance / Time

= 240/4

= 60 km/h

- 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:**

We have,

The initial reading of odometer = 57321 km

The final reading of odometer = 57336 km

So, distance covered by the Car = Final reading - Initial reading = 15 km

The total time taken by the car to travel 15 km = 8:50 AM - 8:30 AM = 20 min

According to question,

Speed of the car = Distance / time taken = 15/20 = 0.75 km/min

Again, 60 minute = 1 hr

so, 1 min = 1/60 hr

Speed in terms of 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:**

Given that,

time taken by Salma to reach her school from house = 15 min =

and, speed of the bicycle = 2 m/s

It is known that,

Distance covered = Speed time taken

= 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.

**Answer:**

(i) A car is moving with a constant speed covers equal distance in equal interval of time. Therefore,

(ii) Distance-time graph of a car parked on a roadside is such that an increase in time, there is no change in distance.

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

**Answer:**

The option (ii) is the correct answer.

Speed = distance /time , this relation holds correctly

- The basic unit of speed is:

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

**Answer:**

The option (iv) is the correct answer.

The SI unit of speed is m/s (meter per second)

- 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

**Answer:**

The correct option is (ii)

Answer = 25 km

1st case,

Speed = 40 km/hr

and time taken = 15 min

So, distance =

Case 2nd,

Speed = 60 km/h

Time taken = 15 min = 0.25 h

So, Distance travelled =

So, total distance = 15 + 10 = 25 km

- Suppose the two photographs, shown in Fig. 13.1 and Fig. 13.2, 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 fastest car.

Answer:

Given, Scale:

The blue car is the fastest.

Distance travelled by the blue car in the picture=

Therefore, the actual distance travelled by blue car =

Time taken =

Speed of the blue car =

Therefore, Speed of the blue car is

**12 **. Fig. 13.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

**Answer:**

It is known that,**(speed = distance/time)**

it means greater the speed, more distance is covered by the car in a given interval of time.

From the graph, we can see that The distance covered by the vehicle A is more than the vehicle B. Hence A is faster than B.

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

**Answer:**

The correct option is (iii)

- In graphs, the constant speed is shown by a straight line.
- A straight line parallels to the time axis, it indicates that the body is at rest position.
- A curved line on graphs means the speed is not constant.

## Study Resources Quick Links

#### Study Notes

#### Reference Books Solutions

- HC Verma Part 1
- HC Verma Part 2
- RD Sharma Class 9
- RD Sharma Class 10
- RS Aggarwal Class 10
- Lakhmir Singh Class 9
- Lakhmir Singh Class 10