Motion In One Dimension(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

MOTION IN ONE DIMENSION MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. The initial velocity of a particle is 10 m/sec and its retardation is 2 m/

s e c^{2}

. The distance moved by the particle in

5^{t h}

second of its motion is

1 m
19 m
50 m
75 m

Discuss Question

Answer: Option A. -> 1 m
:
A
Using the formula for the distance travelled in the

n^{t h}

second

S_{n} = u - \frac{a}{2} (2 n - 1) = 10 - \frac{2}{2} (2 \times 5 - 1) = 1

meter

Question 22. A particle located at x = 0, at time t = 0 starts moving along the positive x direction with a velocity v that varies as v =

α

\sqrt{x}

.The displacement of the particle varies with time as

Discuss Question

Answer: Option B. -> t2
:
B
Given, v =

α

\sqrt{x}

\frac{d x}{d t}

= a

x^{\frac{1}{2}}

x^{\frac{- 1}{2}}

dx = adt
integrating above equation on both sides

x^{\frac{1}{2}}

α

at
x

α

t^{2}

Question 23. A train has a speed of 60 km/h for the first one hour and 40 km/h for the next half hour. Its average speed in km/h is

50
53.33
48
70

Discuss Question

Answer: Option B. -> 53.33
:
B
Distance travelled by train in first 1 houris 60 kmand distance in next

\frac{1}{2}

houris 20 km.

v_{a v g} = \frac{60 + 20}{1.5}

= 53.33

Question 24. The initial velocity of a particle is u (at t=0) and the acceleration f is given by

f = a t

where

a

is a constant. Which of the following relation is valid?

v=u+at2
v=u+at22
v=u+at
v = u

Discuss Question

Answer: Option B. -> v=u+at22
:
B
We have

f = \frac{d v}{d t}

On rearrangement and integration

\int_{u}^{v} d v = a \int_{0}^{t} t d t

v - u = a \frac{t^{2}}{2}

v = u + \frac{a t^{2}}{2}

Question 25. A body is moving from rest under constant acceleration and let

S_{1}

be the displacement in the first

(p - 1)

seconds and

S_{2}

be the displacement in the first

p

seconds. The displacement in

(p^{2} - p + 1)^{t h}

second will be

S1+S2
S1S2
S1−S2
S1S2

Discuss Question

Answer: Option A. -> S1+S2
:
A
From

S = u t + \frac{1}{2} a t^{2}

S_{1} = \frac{1}{2} a (p - 1)^{2}

and

S_{2} = \frac{1}{2} a p^{2}

[As u = 0]
We have

S_{1} + S_{2} = \frac{a}{2} [2 p^{2} - 2 p + 1]

Using the expression for distance traveled in

n^{t h}

second

S_{n} = u + \frac{a}{2} (2 n - 1)

S_{(p^{2} - p + 1)^{t h}} = \frac{a}{2} [2 (p^{2} - p + 1) - 1] = \frac{a}{2} [2 p^{2} - 2 p + 1]

It is clear that

S_{(p^{2} - p + 1)^{t h}} = S_{1} + S_{2}

Question 26. A motor car moving with a uniform speed of

20 m s^{- 1}

comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

20m/sec2
−20m/sec2
−40m/sec2
+2m/sec2

Discuss Question

Answer: Option B. -> −20m/sec2
:
B
Using,

v^{2} = u^{2} + 2 a S

\Rightarrow o = u^{2} + 2 a S

\Rightarrow a = - \frac{u^{2}}{2 S}

a = - \frac{(20)^{2}}{2 \times 10}

a = - 20 m s^{- 2}

Question 27. The velocity of a particle moving along a straight line increases according to the linear law v =

v_{0}

+ kx
where k is a constant . Then
1) the accelearation of the particle is k(

v_{0}

+ kx)
2) the particle takes a time

\frac{1}{k}

l o g_{e}

(

\frac{v_{1}}{v_{0}}

) to attain a velocity

v_{1}

3) velocity varies linearly with displacement with slope of velocity displacement curve equal to k
4) data is insufficient to arrive at a conclusion.

1,2,3 are correct
2,3 are correct
all of them are correct
1 and 2 are correct

Discuss Question

Answer: Option A. -> 1,2,3 are correct
:
A
Options (1), (2), (3) are correct.
Accelearation =

\frac{d v}{d t}

=kv

\Rightarrow

a = k (

v_{0}

+ kx)
Further

\Rightarrow

\frac{d v}{d t}

= kv

\Rightarrow

\frac{d v}{v}

= kdt

v_{1} \int v_{0}

\frac{d v}{v}

= k

t \int 0

ft
t =

\frac{1}{k}

l o g_{e}

\frac{v_{1}}{v_{0}}

Since , v =

v_{0}

+ kx . Hence slope of velocity displacement curve is

\frac{d v}{d x}

= k

Question 28. A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone?

12.25 m/s
14.75 m/s
16.23 m/s
17.15 m/s

Discuss Question

Answer: Option A. -> 12.25 m/s
:
A
Time taken by first stone to reach the water surface from the bridge be t,then

h = u t + \frac{1}{2} g t^{2} \Rightarrow 44.1 = 0 \times t + \frac{1}{2} \times 9.8 t^{2}

t = \sqrt{\frac{2 \times 44.1}{9.8}} = 3 s e c

Second stone is thrown 1 sec later and both strikesimultaneously.This means that the time of travel for the second stone = 3 - 1 = 2 sec
Hence, 44.1 =