Motion(9th Grade > Physics ) Questions and answers for exam Preparation

9th Grade > Physics

MOTION MCQs

Total Questions : 58 | Page 5 of 6 pages

An express train going towards Delhi is travelling at a speed of $90 k m p h$ . Brakes are applied so as to produce a uniform acceleration of $- 0.5 m s^{- 2}$ . Find how far the train will go before it is brought to rest.

$600 m$
$725 m$
$625 m$
$500 m$

Discuss Question

Answer: Option C. ->

625 m

:
C

Given,
initial velocity, $u = 90 k m p h$ = $90 \times \frac{5}{18}$ $= 25 m s^{- 1}$
acceleration, $a = - 0.5 m s^{- 2}$
final velocity, $v = 0$ (brakes are applied)
Let 's' be the distance travelled.
From the third equation of motion,
$v^{2} = u^{2} + 2 a s$
$0 = 25^{2} + (2 \times - 0.5 \times s)$
$\Rightarrow s = 625 m$
Hence the distance covered is $625 m$ .

Question 42.

A $60 m$ long train moving on a straight level track passes a pole in $5 s$ . Find the speed of the train and the time it will take to cross a $540 m$ long bridge.

$13 m s^{- 1}$ , $40 s$
$25 m s^{- 1}$ , $30 s$
$12 m s^{- 1}$ , $50 s$
$15 m s^{- 1}$ , $65 s$

Discuss Question

Answer: Option C. ->

12 m s^{- 1}

50 s

:
C
Given:
Length of the train,

D = 60 m

,
Time taken to cross the pole,

T = 5 s

,
Length of the bridge

= 540 m

Speed of the train, s = \frac{D}{T}

Speed of the train, s = \frac{60}{5} = 12 m s^{- 1}

Now, distance to be covered to cross a

540 m

long bridge is sum of Length of the train and Length of the bridge.

d = 60 + 540 = 600 m

∴ Time taken = \frac{d}{s}

Time taken = \frac{600}{12} = 50 s

∴

Speed of the train is

12 m s^{- 1}

and the time it will take to cross the bridge is

50 s

Question 43.

In the given figure, the velocity of the body at A is zero.

True
False
$625 m$
$500 m$

Discuss Question

Answer: Option A. -> True
:
A

The slope of a displacement-time graph gives us the velocity. The slope at any point in a curve can be found by drawing a tangent on it. In the figure, a tangent has been drawn at point A.

Slope = \frac{Change in y-coordinate}{Change in x-coordinate}

At point A, clearly, there is no change in the y-coordinate. Hence, the slope is zero. This implies that the velocity at point A is zero.

Question 44.

An auto travels the first half time with a uniform speed $u$ and the next half time with a uniform speed $v$ . Find its average speed.

$\frac{u + v}{2}$
$\frac{u v}{2}$
$2 u v$
$u v$

Discuss Question

Answer: Option A. ->

\frac{u + v}{2}

:
A

Given, speed of the auto during the first half of the journey is $u$ and the next half of the journey is $v$
Let total time taken for the journey be $T$
$∴$ Time taken for the first half of the journey be $\frac{T}{2}$ and the next half of the journey be $\frac{T}{2}$
Let the average speed be $V_{a v}$
We know, $s p e e d = \frac{D i s t a n c e}{t i m e}$
$∴$ $D i s t a n c e = s p e e d \times t i m e$
During the first half of the journey, distance, $d_{1} = u \times \frac{T}{2}$
During the second half of the journey, distance, $d_{2} = v \times \frac{T}{2}$
Total distance $= d = d_{1} + d_{2}$
$= u \times$ $\frac{T}{2} + v \times \frac{T}{2}$
$\Rightarrow d = \frac{T}{2} (u + v)$
Average speed, $V_{a v}$ =   $\frac{T o t a l d i s t a n c e}{T o t a l t i m e t a k e n}$
   $= \frac{d}{T}$
$= \frac{\frac{T}{2} (u + v)}{T}$
$= \frac{u + v}{2}$

Question 45.

An artificial satellite is moving in a circular orbit of radius $4200 k m$ around the earth. What is its speed if it takes one day to complete one revolution?

$4099 kmph$
$3099 kmph$
$1099 kmph$
$2099 kmph$

Discuss Question

Answer: Option C. ->

1099 kmph

:
C
Given, the radius of the orbit,

R = 4200 k m

The time taken to revolve around the Earth,

T = 24 h r (1 d a y)

The speed of the satellite,

v = \frac{Circumference of the orbit}{Time taken}

= \frac{2 π R}{T}

= \frac{2 \times 3.14 \times 4200 km}{24 hr}

= 1099 km {hr}^{- 1}

= 1099 kmph

Question 46.

The velocity - time graph of a body with a constant acceleration is:

Straight line parallel to velocity axis
Parabola
Hyperbola
Straight line inclined at some angle with respect to time axis

Discuss Question

Answer: Option D. -> Straight line inclined at some angle with respect to time axis
:
D

Since acceleration is constant, the slope of the velocity-time graph should be a straight line with constant slope. Thus, the graph which shows a straight line inclined at some angle with respect to time axis represents the velocity-time graph of a body with a constant acceleration.

Question 47.

Motion of a satellite in a circular orbit is an example of

non-uniform circular motion
uniform circular motion
uniform linear motion
non-uniform linear motion

Discuss Question

Answer: Option B. -> uniform circular motion
:
B

The motion of a satellite in a circular orbit is uniform circular motion.In uniform circular motion, a body moves in a circle with constant speed but the velocity changes due to a constant change in direction.

Question 48.

A car covers a distance of 300 m in 20 seconds. Calculate the speed of the car.

$25 m s^{- 1}$
$20 m s^{- 1}$
$10 m s^{- 1}$
$15 m s^{- 1}$

Discuss Question

Answer: Option D. ->

15 m s^{- 1}

:
D
Given, distance travelled by the car

= 300 m

and the time taken

= 20 s

.
We know,

s p e e d = \frac{d i s t a n c e}{t i m e}

\Rightarrow s p e e d = \frac{300}{20}

15 m s^{- 1}

∴

Speed of car is

15 m s^{- 1}

Question 49.

The area below velocity – time graph gives:

acceleration
displacement
average speed
average velocity

Discuss Question

Answer: Option B. -> displacement
:
B

The area under a velocity-time graph gives the displacement of the object.
The area under the curve is velocity x time = displacement.

$∵ V e l o c i t y = \frac{Displacement}{Time}$

In the graph, the area under the curve is $d i s p l a c e m e n t$ = Area of triangle ADE + Area of rectangle ADBC

Question 50.

Brakes are applied to a truck to produce an acceleration of $- 10 m s^{- 2}$ . If the truck takes $5 s$ to stop after applying the brakes, then find the distance covered by the truck before coming to rest.

$250 m$
$125 m$
$- 125 m$
$- 250 m$

Discuss Question

Answer: Option B. ->

125 m

:
B

Given, final velocity, $v = 0$ , acceleration, $a = - 10 m s^{- 2}$ , time taken, $t = 5 s$ .
Let the initial velocity be $u$ .
From the first equation of motion, $v = u + a t$ ,
$0 = u + (- 10 \times 5)$
$\Rightarrow u = 50 m s^{- 1}$
From third equation of motion, $v^{2}$ = $u^{2} + 2 a s$ ,
$0 = 50^{2} + 2 \times - 10 \times s$
$\Rightarrow s = 125 m$
Therefore, the truck will stop after $125 m$ .