**CBSE Board Class IX
Science (Physics)**

**Solutions of NCERT Exercise Questions**

**Chapter 8, MOTION **

*Important Exercise
Questions Solved*

Q.1: 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?

Ans:

Time taken =
2 min 20 sec = 140 sec.

Radius, r =
100 m.

In 40 sec
the athlete complete one round.

So, in 140
sec the athlete will complete = 140 ÷ 40 = 3.5 (three and a half) round.

Or, distance
covered in 140 sec = 2πr x 3.5 = 2 x 22/7 x 100 x 3.5 =
2200 m, Ans.

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

Or,
displacement = diameter = 200 m, Ans.

Q.2: 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?

Ans:

(a)
For motion from A to B:

distance
covered = 300 m; displacement = 300 m.

time
taken = 150 sec.

we
know, average speed = distance covered ÷ time taken = 300 m ÷ 150 sec = 2 ms^{‑1}
Ans.

(b)
For motion from A to C:

distance
covered = 300 + 100 = 400 m.

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

time
taken = 2.5 min + 1 min = 3.5 min = 210 sec.

Average
speed = distance covered ÷ time taken = 400 ÷ 210 = 1.90 ms^{-1} Ans.

Average
velocity = displacement covered ÷ time taken = 200 m ÷ 210 sec = 0.952ms^{-1}.
Ans.

Q.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 30 kmh^{-1}. What
is the average speed of Abul’s trip?

Ans:

Let
one side distance = x km.

Time
taken for forward trip= at a speed of 20 km/h = Distance / Speed = x/20
h.

Time
taken in return trip at a speed of 30 km/h = x/30 h.

Total
time for the whole trip = x/20 + x/30
= 5x/60 h.

Total
distance covered = 2x km.

We
know, average speed = Total distance ÷ total time = 2x
÷ (5x/60) = 24 kmh^{-1 }Ans.

Q.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?

Ans:

Here,
u = 0, a = 3 ms^{-2}, t = 8 s

Since,
s = ut + ½ at^{2} = 0 x 8 + ½ x 3 x 8^{2}
= 96 m. Ans.

Q.5: A driver of a car travelling at 52 kmh^{-1} applies
the brakes and accelerates uniformly in the opposite direction. The car stops
after 5 s. Another driver going at 34 kmh^{-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 two cars. Which of the two cars travelled farther after
the brakes were applied?

Ans:

In
fig AB and CD are the time graphs for the two cars whose initial speeds are 52
km/h and 34 km/h, respectively.

Distance
covered by the first car before coming to rest

=
Area of triangle AOB

= ½ x AO x BO

= ½ x 52 kmh^{-1}
x 5 s

= ½ x (52 x 1000 x 1/3600) ms^{-1} x
5 s = 36.1 m

Distance covered by the second car before coming to rest

= Area of triangle COD

= ½ x CO x DO

= ½ x 34 km h^{-1} x 10 s

= ½ x (34 x 1000 x
1/3600)
ms^{-1} x10 s = 47.2 m

Thus, the second car travels farther than the first car
after they applied the brakes.

Q.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?

Ans:
Here, u = 0, s = 20 m, a = 10 ms^{-2}, v = ?, t = ?

As,
v^{2} - u^{2} = 2as

So,
v^{2} - 0^{2 }= 2 x 10 x 20 = 400

or,
v = 20 ms^{-1}. Ans.

And
t = (v - u) ÷ a = 20 ÷ 10 = 2 s. Ans.

Q.9: State which of the following situations are possible and
give an example of each of the following -

(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.

Ans:

(a)
Yes, a body can have acceleration even when its velocity is zero. When a body
is thrown up, at highest point its velocity is zero but it has acceleration
equal to acceleration due to gravity.

(b)
Yes, an acceleration moving horizontally is acted upon by acceleration due to
gravity that acts vertically downwards.

Q.10: An artificial is moving in a circular orbit of radius
42250 km. Calculate its speed if it takes 24 hrs to revolve around the earth.

Ans:

Here,
r = 42250 km = 42250000 m

T
= 24 h = 24 x 60 x 60 s

Speed,
v = 2πr ÷ T = (2 x 3.14 x 42250000) ÷
(24 x 60 x 60) m/s =
3070.9 m/s = 3.07 km/s Ans.

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