**1) An athlete completes one**

round of a circular track of diameter 200 m in 40 s. What will be the distance

covered and the displacement at the end of 2 minutes 20 s?

round of a circular track of diameter 200 m in 40 s. What will be the distance

covered and the displacement at the end of 2 minutes 20 s?

**ANSWER**:-Diameter = 200 m, therefore, radius = 200m/2 =100m

Time of one rotation = 40s

Time after 2m20s = 2 x 60s + 20s = 140s

Distance after 140 s =?

Displacement after 140s =?

(a) Distance after 140s

We know that, distance=velocity ×time

⇒ Distance=15.7m/s ×140 s =

2198 m

2198 m

(b) Displacement after 2 m 20 s i.e. in 140 s

Since, rotating in 40 s=1

Distance covered in 2 m 20 s = 2198 m

And, displacement after 2 m 20 s = 200m

**2) Joseph jogs from one end A**

to the other end B of a straight 300 m road in 2 minutes 30 seconds and then

turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’saverage

speeds and velocities in jogging (a) from A to B and (b) from A to C?

to the other end B of a straight 300 m road in 2 minutes 30 seconds and then

turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’saverage

speeds and velocities in jogging (a) from A to B and (b) from A to C?

**ANSWER**:-Total Distance covered from AB =

300 m

Total time taken = 2 x 60 + 30s =150 s

Therefore, Average Speed from

AB = Total Distance / Total

Time =300 / 150 m s -1 =2 m s -1

Therefore, Velocity from AB

=Displacement AB / Time = 300

/ 150 m s -1 =2 m s -1

/ 150 m s -1 =2 m s -1

Total Distance covered from AC

=AB + BC =300 + 200 m

=AB + BC =300 + 200 m

Total time taken from A to C = Time

taken for AB + Time taken for BC = (2 x 60+30) +60 s = 210 s

taken for AB + Time taken for BC = (2 x 60+30) +60 s = 210 s

Therefore, Average Speed from AC =

Total Distance /Total Time

Total Distance /Total Time

= 400 /210 m s -1

= 1.904 m s -1

Displacement (S) from A to C

= AB – BC

= 300-100 m

= 200 m

Time (t) taken for displacement

from AC = 210 s

from AC = 210 s

Therefore, Velocity from AC = Displacement

(s) / Time(t)

(s) / Time(t)

= 200 / 210 m s -1

= 0.952 m s -1

**3) Abdul, while driving to**

school, computes the average speed for his trip to be 20 km h

school, computes the average speed for his trip to be 20 km h

**−**

**1. On his return trip along**

the same route, there is less traffic and the average speed is 40 km h

the same route, there is less traffic and the average speed is 40 km h

**−**

**1. What is the average speed**

for Abdul’s trip?

for Abdul’s trip?

**ANSWER**:-The distance Abdul commutes while

driving from Home to School = S

Let us assume time taken by Abdul to commutes this distance = t 1

Distance Abdul commutes while driving from

School to Home = S

School to Home = S

Let us assume time taken by Abdul to commutes this distance = t 2

Average speed from home to school v 1av = 20 km

h-1

h-1

Average speed from school to home v 2av = 30 km h-1

Also we know Time taken from Home to School t 1 =S / v 1av

Similarly Time taken form School to Home t 2 =S/v 2av

Total distance from home to school and backward = 2 S

Total time taken from home to school and backward (T) = S/20+

S/30

S/30

Therefore, Average speed (V av) for covering total distance (2S) = Total

Distance/Total Time

Distance/Total Time

= 2S / (S/20 +S/30)

= 2S / [(30S+20S)/600]

= 1200S / 50S

= 24 kmh -1

**4) A motorboat starting from**

rest on a lake accelerates in a straight line at a constant rate of 3.0 m s

rest on a lake accelerates in a straight line at a constant rate of 3.0 m s

**−**

**2 for 8.0 s. How far the boat**

travels during this time?

travels during this time?

**ANSWER**:-Given Initial velocity of motorboat,

u = 0

Acceleration of motorboat, a = 3.0

m s -2

m s -2

Time under consideration, t =8.0 s

We know that Distance, s = ut+ (1/2) at 2

Therefore, The distance travel by motorboat = 0

x 8 + (1/2)3.0 x 8 2

x 8 + (1/2)3.0 x 8 2

= (1/2) x 3 x 8 x 8 m

= 96 m

**5) A driver of a car**

travelling at 52 km h

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

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?

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:**–

Distance Travelled by first car before coming to rest =Area of

△ OPR

= (1/2) x OR x OP

= (1/2) x 5 s x 52 kmh-1

= (1/2) x 5 x (52 x 1000) / 3600) m

= (1/2) x 5x (130 / 9) m

= 325 / 9 m

= 36.11 m

Distance Travelled by second car before coming to rest

=Area of △ OSQ

= (1/2) x OQ x OS

= (1/2) x 10 s x 3 kmh-1

= (1/2) x 10 x (3 x 1000) /

3600) m

= (1/2) x 10 x (5/6) m

= 5 x (5/6) m

= 25/6 m

= 4.16 m

**6) Fig 8.11 shows the**

distance-time graph of three objects A, B and C. Study the graph and answer the

following

distance-time graph of three objects A, B and C. Study the graph and answer the

following

**Questions:**

(a) Which of the three is travelling the

fastest?

fastest?

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

road?

road?

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

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

C?

C?

**ANSWER**🙁 a) Object B

(b) No

(c) 5.714 km

(d) 5.143 km

Speed = slope of the graph

Since slope of object B is greater than objects A and C, it is

travelling the fastest.

travelling the fastest.

(b) All three objects A, B and C never meet at a single point.

Thus, they were never at the same point on road.

Thus, they were never at the same point on road.

**7) A ball is gently dropped**

from a height of 20 m. If its velocity increases uniformly at the rate of 10 m

s

from a height of 20 m. If its velocity increases uniformly at the rate of 10 m

s

**−**

**2, with what velocity will it strike the ground? After what time**

will it strike the ground?

will it strike the ground?

**ANSWER**:-Let us assume, the final velocity

with which ball will strike the ground be ‘v’ and time it takes to strike the

ground be’t’

Initial Velocity of ball, u =0

Distance or height of fall, s =20 m

Downward acceleration, a =10m s -2

As we know, 2as =v 2-u2

v 2 = 2as+ u2

= 2 x 10 x 20 + 0

= 400

∴ Final velocity of

ball, v = 20ms-1

ball, v = 20ms-1

t = (v-u)/a

∴Time taken by the

ball to

strike = (20-0)/10

ball to

strike = (20-0)/10

= 20/10

= 2 seconds

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

(a) Find out 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.

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?

uniform motion of the car?

a)

**ANSWER:**–

The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents

the distance travelled by the car in the first 4 s.

the distance travelled by the car in the first 4 s.

b) The part of the graph between time 6 s to 10 s represents uniform motion of the car.

**10) State which of the**

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

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

**(a) An object with a constant**

acceleration but with zero velocity.

acceleration but with zero velocity.

**(b) An object moving in a**

certain direction with acceleration in the perpendicular direction.

certain direction with acceleration in the perpendicular direction.

**ANSWER**:-(a) Possible When a ball is

thrown up at maximum height, it has zero velocity, although it will have

constant acceleration due to gravity, which is equal to 9.8 m/s2.

(b) Possible When a car is moving

in a circular track, its acceleration is perpendicular to its direction.

in a circular track, its acceleration is perpendicular to its direction.

**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

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 of the circular orbit, r=

42250 km

Time taken to revolve around the earth, t= 24 h

Speed of a circular moving object, v= (2π r)/t

= [2× (22/7) ×42250 × 1000] / (24 × 60 × 60)

= (2×22×42250×1000) / (7 ×24× 60 × 60) m s -1

=3073.74 m s -1