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

Question 1. A particle moving in a straight line has velocity and displacement equation as V = 4

\sqrt{1 + S}

, where V in m/s and s is in m.
The initial velocity of the particle in m/s is ______

7
8
9
4

Explanation:-

Answer: Option D. -> 4
:
D

v = \frac{d s}{d t} = 4 \sqrt{1 + s} \Rightarrow \int_{o}^{s} \frac{d s}{\sqrt{1 + s}} = 4 \int_{o}^{t} d t \Rightarrow 4 t = 2 [\sqrt{1 + s} - 1] \Rightarrow 2 t + 1 = \sqrt{1 + s} \Rightarrow 1 + s = 4 t^{2} + 1 + 4 t \Rightarrow s = 4 t^{2} + 4 t A t t = 0 s, s = 0 m \Rightarrow v |_{a t t = 0 s} = 4 \sqrt{1 + 0} = 4 m s^{- 1}

Question 2. A particle is moving along the path given by y =

\frac{c}{6}

t^{6}

(where c is positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at t=5 sec. Is

5a=v
a = √v
a=5v
a=v

Explanation:-

Answer: Option D. -> a=v
:
D
Differentiate y we get

v = c (t^{5})

Differentiate v we get

a = 5 c (t^{4})

In expressions for v and a, substitute t = 5 s and take ratio of the two quantities.
We get, v=a

Question 3. The position of a particle travelling along x - axis is given by

x_{t}

=

t^{3}

- 9

t^{2}

+ 6t where

x_{t}

is in m and t is in second . Then
1) the particle does not come to rest at all.
2) the particle comes to rest firstly at (3 -

\sqrt{7}

) s and then at (3 +

\sqrt{7}

) s
3) the speed of the particle at t= 2s is 18 m

s^{- 1}

4) the acceleration of the particle at t= 2s is 6 m

s^{- 2}

2,3, 4 are correct
2,3 are correct
All of them are correct
1 and 2 are correct

Explanation:-

Answer: Option A. -> 2,3, 4 are correct
:
A

x_{t}

=

t^{3}

- 9

t^{2}

+ 6t
v=

\frac{d x_{t}}{d t}

= (3

t^{2}

- 18t + 6) m

s^{- 1}

So, the particlecomes to rest since this equation has two valid roots.

\Rightarrow

t_{1}

=(3 -

\sqrt{7}

) s and

\Rightarrow

t_{2}

=(3 -

\sqrt{7}

) s
Substituting the value t = 2s in the equation for velocity, we get, v= = - 18m

s^{- 1}

The acceleration of the particle is a=6t - 18m

s^{- 2}

Substituting the value t = 2s in the equation for acceleration, we get, a= = - 6m

s^{- 2}

Question 4. A frictionless wire AB is fixed on a circular frame of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is

2√gRgcosθ
2√gRcosθg
2√Rg
gR√gcosθ

Explanation:-

Answer: Option C. -> 2√Rg
:
C

∠ A B C = 90^{0} Acceleration along AB is a = g c o s θ Distance travelled is A B = 2 R C o s θ We have 2 R C o s θ = \frac{1}{2} \times g c o s θ t^{2} \Rightarrow t = 2 \sqrt{\frac{R}{g}}

Question 5. A Body moves 6 m north. 8 m east and 10m vertically upwards, what is the magnitude of its resultant displacement from initial position?

10√2m
10m
10√2m
10×2m

Explanation:-

Answer: Option A. -> 10√2m
:
A

\to r = x ˆ i + y ˆ j + z ˆ k ∴ r = \sqrt{x^{2} + y^{2} + z^{2}}

r = \sqrt{6^{2} + 8^{2} + 10^{2}} = 10 \sqrt{2} m

Question 6. A particle moves according to the equation

\frac{d v}{d t}

=

α

-

β

v , where

α

and

β

are constants. Find the velocity as a funtion of time. Assume body starts from rest.

v = (βα) (1 - e−βt)
v = (βα) (e−βt)
v = (αβ) (1 - e−βt)
v = (αβ) (e−βt)

Explanation:-

Answer: Option C. -> v = (αβ) (1 - e−βt)
:
C
Take the relation given for acceleration and integrate with the given limits to obtain the desired function for velocity.

\frac{d v}{d t}

=

α

-

β

v

\Rightarrow

v \int 1

\frac{d v}{α - β v}

=

t \int 0

dt
v = (

\frac{α}{β}

) (1 -

e^{- β t}

)

Question 7. A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is

4.9 m
9.8 m
19.6 m
24.5 m

Explanation:-

Answer: Option D. -> 24.5 m
:
D
The first body would have travelled for 3s. Since both of them start from rest,the
separation between the two bodies, two seconds after the release of second body is

x = \frac{1}{2} \times 9.8 [(3)^{2} - (2)^{2}] = 24.5 m

Question 8. A body A is projected upwards with a velocity of 98 m/s. The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

6 sec
8 sec
10 sec
12 sec

Explanation:-

Answer: Option D. -> 12 sec
:
D
Let tbe the time of flight of the first body when they meet.
Then the time of flight of the second body will be(t - 4)sec.
Since the displacementof both the bodies from the ground will be the same,

h_{1} = h_{2}

∴ 98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g (t - 4)^{2}

On solving, we get t = 12 seconds.

Question 9. A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time

\frac{t}{2}

s.

At h2 from the ground
At h4 from the ground
Depends upon mass and volume of the body
At 3h4 from the ground

Explanation:-

Answer: Option D. -> At 3h4 from the ground
:
D
Let the body after time

\frac{t}{2}

fall a distancex from the top. Since it starts from rest, we have

x = \frac{1}{2} g \frac{t^{2}}{4} = \frac{g t^{2}}{8}

.............(i)
It reaches the ground in t seconds, we have

h = \frac{1}{2} g t^{2}

..............(ii)
Eliminate t from (i) and (ii), we get x =

\frac{h}{4}

∴

Height of the body from the ground
=

h - \frac{h}{4} = \frac{3 h}{4}

Question 10. A stone falls from a balloon that is descending at a uniform rate of 12 m/s. The magnitude of the displacement of the stone from the point of release after 10 sec is

490 m
510 m
610 m
725 m

Explanation:-

Answer: Option C. -> 610 m
:
C
u = -12m/s, g = - 9.8 m/

s e c^{2}

, t = 10sec
Displacement =

u t + \frac{1}{2} g t^{2}

=

- 12 \times 10 + \frac{1}{2} \times - 9.8 \times 100 = - 610 m

(since downwards)
The magnitudeis 610m.