Question1.

An athlete completes one round of circular track of diameter 200 m in 40 sec.

What will be the distance covered and the displacement at the end of 2 minutes 20 sec?

Answer:

Time taken = 2 min 20sec = 140sec.

Radius r = 100m.

In 40 sec the athlete complete one round.

So, in 140sec the athlete in 140sec = (2 π r) x(3.5) = 2x (22/7) x 100x 3.5=2200m.

At the end of his motion, the athlete will be in the diametrically opposite position.

• Displacement = diameter =200m

Question2.

Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec and

then turns around and jogs 100 m back to point C in another 1 minute.

What are Joseph’s average speeds and velocities in jogging (a) from A to B (b) from A to C?

Answer:

(a) For motion from A to B:

Distance covered = 300 m

Displacement = 300 m.

Time taken =2min 30sec = 170 sec.

We know that, Average speed = (Total distance / Total time taken)

= (300 m/ 170 sec) = 1.7625 ms^-1

Average velocity = (Net displacement / time taken)

= (300 m/ 170 sec) = 1.7625 ms^-1

 

(b) For motion from A to C:

Distance covered = (AB+ BC) = 300 + 100 = 400 m.

Displacement = AB - CB = 300 - 100 = 200 m.

Time taken = 170 sec + 60sec = 230 sec.

Therefore, Average speed = (Total distance / Total time taken)

= (400 / 230) = 1.739 ms^-1

Average velocity = (Total displacement / time taken) = (300-100)/ (230sec)

= (200 m / 230 sec) = 0.87ms^-1

 

Question 3.

Abdul, while driving to school, computes the average speed for his trip to be 20 kmh^-1.

On his return trip along the same route, there is less traffic and the average speed is 40 kmh^-1. What is the average speed of Abdul’s trip?

Answer:

From Home to school:-

Distance = d

Time taken =t₁

Average Speed = 20 kmh^-1 = (d/t₁)

• t₁ = (d/20) ------(1)

From School to Home:-

Distance = d

Time taken =t₂

 Average Speed = 40 kmh^-1 = (d/ t₂)

• t₂ = (d/40) ------(2)

Therefore net avg. speed = (2d)/ (t₁ + t₂) = (2d) / ((d/20)   + (d/40))

= (2d)/ (3/40) = (80/3) = 26.67m/s.

  Question 4.

A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^-2 for 8.0 s. How far does the boat travel during this time?

Answer:

Initial velocity (u) = 0m/s

Acceleration= 3m/s^-2

Time (t) = 8 sec

We know that

Therefore, Distance covered (s) = ut+ (1/2) at²

• s=0 x 8 + (1/2) 3m/s² x (8s)²
• s = (1/2) x 3 x64m
• s = 3 x 32m
• s = 96m.

Therefore, boat travel a distance of 96 m in the given time.

  Question 5.

A driver of a car travelling at 52 km h^–1 applies the brakes and accelerates uniformly in the opposite direction.

The car stops in 5 s. Another driver going at 3 km h^–1 in another car applies his brakes slowly and stops in 10 s.

On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer:

For First Driver,

Initial velocity (u) = 52 km h^–1

= (52x 1000m)/ (60 x 60s) = 14.4 ms^-1

Time, t = 5s.

Final velocity v = 0 (As car stops).

For Second Driver,

Initial velocity u = 3 km h^–1

Distance s = (3000m)/ (60x 60 s) = 9.4 ms^-1

Time taken t = 10s

Final velocity v =0.

Distance is calculated by the area under the slope of the graph.

• Distance s = (1/2) x OA X OD
• = (1/2) x 14.4 x5
• s = 7.2m/s x 5s = 36m.

Distance covered by 2^nd car = area of triangle (OBC)

Distance, s = (1/2) x OC x OB

• s = (1/2) x 9.4m/s x 10s
• s= 4.7m/s x 10s = 47m

Therefore, second car travelled farther.

 

 

In the graph, dark brown slope shows the velocity of the first car and light brown slope shows the velocity of the second car.

Question 6.

Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answers the following questions:

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

 

Answer:

• It is clear from graph that B is covering more distance in less time. Therefore, B is the fastest.
• All of three are never at the same point at the same time on the road.
• According to graph; each small division shows about 0.57 km.

A is passing  B at point S which is in line with point P and shows about 9.14km.

Thus at the point C travels about = 9.14 – 2.1375km=7km.

Thus, when A passes B, C travels about 7km.

(d)  B passes C at point Q at the distance axis which is

= 4km + (0.57x2.25)   =5.28km.

Question 7.

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms^-2,

with what velocity will it strike the ground? After what time will it strike the ground?

Answer:

Given, Initial velocity u =0

Distance s = 20m

Acceleration a = 10 ms^-2

• We know that v² = u² + 2as

=> v² = 0 + 2x 10m/s² x 20m

=> v² = 400 m² s^-2

=> v = √400 m² s^-2

=> v = 20ms^-1

(b)  We know that, v = u + at

=> 20ms^-1 = 0 + 10ms^-2x t

=> t = (20ms^-1 / 10ms^-2) = 2s

Therefore, Ball strikes the ground at the velocity of 20ms^-1.

Time taken to reach the ground = 2s.

                                                                                                                                                                                   

Question 8.

The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Answer:

• Distance covered = area under slope of the speed – time graph.

In the given graph 56 full squares and 12 half squares come under the area slope for the time of 4s.

Total number of squares = 56 + (12/2) = 62 squares.

The total area of the squares will give us the distance travelled by the car in 4s.

On X-axis (time) = 5 squares = 2s

Therefore 1 square = (2/5) s

On Y-axis (speed) = 3 squares = 2m/s

Therefore 1 square = (2/3) m/s

Therefore area of 1 square = (2/5) s + (2/3) m/s = (4/15) m

Therefore area of 62 squares = (4/15) m x 62

= (248/15) m = 16.53 m

Therefore car travels 16.53m in first 4 sec.

• MN of the slope of the graph is a straight line parallel to x-axis, thus it represents the uniform motion of the car.

Question 9.

State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

• A body can have constant acceleration even when its velocity is 0. When a body is thrown up, at the
• highest point its velocity is 0 but it has acceleration equal to acceleration due to gravity.
• Yes, acceleration which is moving in horizontal direction is acted upon by the acceleration due to gravity that acts
• vertically downwards. In case of circular motion, when an object moves on a circular path, its direction is
• along the tangent of the circle but the acceleration is towards the radius of the circle.
• And tangent always makes a right angle with the radius, so in circular motion acceleration and velocity are mutually perpendicular to each other.

  Question 10.

An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer:

Radius = 42250km = 42250000m

Time t =24h = 24 x 60 x 60 s

Using   Speed v = (2 π r)/ (t)

= (2 x 3.14 x 42250000) / (24 x 60 x 60) m/s

=3070.9 m/s = 3.07 km/s.