Question1.

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.

Question2.

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.

Question 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 = (Total time taken)/ (Number of Oscillations)

Given: - 20 oscillations taking 32s to complete.

Therefore 1 oscillation will take = (32)/ (20) sec = 1.6 second

Therefore 1.6s is the time period of the pendulum.

Question 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 = (Distance travelled) / (Time)

= (240 km) / (4h)

=60 Km/h

Therefore the speed of the train is 60km/hr.

 Question 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 = 20/60 hour = 1/3 hour

So Speed in km/min

Speed = (Distance travelled)/ (Time)

= (15km)/ (20min)

=0.75km/min

Speed in km/h

Speed = (Distance travelled)/ (Time)

= (15km)/ (1/3h)

= (15 x 3) km/ (1h)

=45km/h

 Question 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 = (Distance travelled)/ (Time)

Distance = (Speed x Time)

=2 x 900=1800m

Also 1km= 1000m

Therefore 1800x (1/1000) = 1.8km.

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

 Question 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:-

 

Question 8.

Which of the following relations is correct?

(i) Speed = (Distance × Time)

(ii) Speed = (Distance /Time)

(iii) Speed = (Time/Distance)

(iv) Speed = (1/ (Distance x Time))

Answer:

Speed of an object is given by the relation:-

(ii) Speed = (Distance /Time)

Question 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 = (Distance/Time)

Therefore the basic unit of speed is m/s.

 Question 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 = (15/60) hour

Distance (d₁) = Speed × Time = 40 × (15/60) = 10 km

Case II:

Speed = 60 km/h

Time = 15 min = (15/60) hour

Distance (d₂) = (Speed × Time) = 60 × (15/60) = 15 km

Total distance (d) = (d₁ + d₂) = 10 km + 15 km = 25 km

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

 Question 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.

 

Answer:

 

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.

Question 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.

Question 13.

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

Answer:

Correct option is (iii) as graph is not a straight line, it keeps on changing.

This shows that the truck is moving with variable speed.