Motion Test

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Motion Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

What is the acceleration of the race car that moves at constant velocity of 300 km/hr ?

A
73.32 $$m/sec^2$$.

B
0 $$m/sec^2$$

C
63.33 $$m/sec^2$$.

D
53.33 $$m/sec^2$$.

Solution

Since the car is moving with a constant velocity, so change in the velocity of car is zero i.e. $$\Delta v = 0$$
Acceleration of car $$a = \dfrac{\Delta v}{\Delta t} = 0 \ m/s^2$$
Question 2
1 / -0

If a bus travelling at $$20\ m/s$$ is subject to a steady deceleration of $$5\ m/s^{2}$$, how long will it take to come rest?

A
400 s

B
0.4 s

C
40 s

D
4 s

Solution

Using first equation of motion
$$ v= u+at$$, we have
$$ 0=20 +(-5)t$$
$$\Rightarrow 5t=20$$
$$\Rightarrow t=4s$$
Question 3
1 / -0

The velocity of a body moving with a uniform acceleration of $$2 m/sec^2$$ is $$ 10 m/sec$$. Its velocity after an interval of 4 sec is

A
12 m/sec

B
14 m/sec

C
16 m/sec

D
18 m/sec

Solution

By the first eqn of motion
$$v=u+at$$ Put the given values
$$v=10+2\times 4$$
$$u=18\ m/sec$$
Question 4
1 / -0

A heavy stone is thrown from a cliff of height $$h$$ with speed $$v$$. The stone will hit the ground with maximum speed if it is thrown:

A
vertically downward

B
vertically upward

C
horizontally

D
the speed does not depend on the inital direction

Solution

Correct Answer: Option (D)

According to the Equation $$v^{2}-u^{2}=2gh$$,Where v is final Velocity, u is initial velocity, g is acceleration due to gravity and h is the height of the cliff

As 2gh and u is constant, the final velocity(v) is constant irrespective of the direction.
Question 5
1 / -0

The rate of change of velocity is:

A
Force

B
Momentum

C
Acceleration

D
Displacement

Solution

C. Acceleration
Question 6
1 / -0

Ball A is dropped freely while another ball B is thrown vertically downward with an initial velocity $$v$$ from the same point simultaneously. After $$t\ seconds$$ they are separated by a distance of: (Given that $$t$$ is less than the time required by ball B to reach the ground.)

A
$$\dfrac{vt}{2}$$

B
$$\dfrac{1}{2}gt^{2}$$

C
$$vt$$

D
$$vt+\dfrac{1}{2}gt^{2}$$

Solution

First case (for ball A):
$$s_{1}=0(t)+\dfrac{1}{2}gt^{2}$$
Second case (for ball B):
$$s_{2}=v(t)+\dfrac{1}{2}gt^{2}$$
after $$t\ seconds$$, distance between them is $$s_{2}-s_{1}$$
$$\Rightarrow s_{2}-s_{1}=vt$$
Note: $$\text{After ball B has landed on ground the distance between A and B decreases and becomes zero.}$$
So, C is correct.
Question 7
1 / -0

A body starting with a velocity $$v$$ returns to its initial position after $$t$$ second with the same speed along the same line. Acceleration of the particle is :

A
$$\dfrac{-2v}{t}$$

B
$$\dfrac{2v}{t}$$

C
$$\dfrac{v}{2t}$$

D
$$\dfrac{t}{2v}$$

Solution
Question 8
1 / -0

A body dropped from the top of a tower reaches the ground in $$4\;s$$. Height of the tower is (Take $$g=9.8 \ m/s^2$$):

A
$$39.2\ m$$

B
$$44.1\ m$$

C
$$58.8\ m$$

D
$$78.4\ m$$

Solution

We have, $$s = ut + \frac {1}{2} at^2$$
Given,
$$t = 4\ sec$$
$$a = g$$
$$u = 0$$
$$\Rightarrow s = \dfrac {1}{2} gt^2$$
$$s = \dfrac {1}{2} g(4)^2 = 8g = 8 \times 9.8 = 78.4\ m$$
Height of the tower = 78.4 m
Question 9
1 / -0

The displacement - time graphs of two bodies A and B are OP and OQ respectively. If $$\angle $$POX is $$60^{0}$$ and $$\angle $$QOX is $$45^{0}$$, the ratio of the velocity of A to that of B is

A
$$\sqrt{3}:\sqrt{2}$$

B
$$\sqrt{3}:1$$

C
$$1: \sqrt{3}$$

D
$$3:1$$

Solution

$$V_{A}= \tan 60^{0}$$
$$V_{B}= \tan 45^{0}$$

So, $$ \dfrac{V_{A}}{V_{B}} = \dfrac{\tan 60^o}{\tan 45^o} = \dfrac{\sqrt{3}}{1} = \sqrt{3}: 1$$
Question 10
1 / -0

A ball dropped freely takes $$0.2\;s$$ to travel the last $$6\;m$$ distance before hitting the ground. Total time of fall is $$(g=10\;ms^{-2})$$:

A
$$2.9\ s$$

B
$$3.1\ s$$

C
$$2.7\ s$$

D
$$0.2\ s$$

Solution

Let $$T=$$ total time taken in second
$$S=$$ total distance traveled in meter
$$V=$$ velocity of particle in m/sec
So in this question the condition is of free fall so acceleration of the particle will remain constant through out the process.
$${T}={T}_{1}+{T}_{2}$$
$${S}={S}_{1}+{S}_{2}$$
Given that: $${T}_{1}=0.2 \;s$$ and $${S}_{1}=6 \;m$$
Thus, $$6=u(0.2)+\dfrac{1}{2}10(0.2)^2$$
Thus, we get $$u=29 m/s$$
Now using:
$$u= 0 +gt $$(as the ball is dropped with initial velocity as zero)

$$t=\dfrac{29}{10}=2.9 s$$

Thus we get total time as $$2.9+0.2=3.1 s$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 29

Result

Motion Test - 29

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SHARING IS CARING

Question 1

73.32 $$m/sec^2$$.

0 $$m/sec^2$$

63.33 $$m/sec^2$$.

53.33 $$m/sec^2$$.

Solution

Question 2

400 s

0.4 s

40 s

4 s

Solution

Question 3

12 m/sec

14 m/sec

16 m/sec

18 m/sec

Solution

Question 4

vertically downward

vertically upward

horizontally

the speed does not depend on the inital direction

Solution

Question 5

Force

Momentum

Acceleration

Displacement

Solution

Question 6

$$\dfrac{vt}{2}$$

$$\dfrac{1}{2}gt^{2}$$

$$vt$$

$$vt+\dfrac{1}{2}gt^{2}$$

Solution

Question 7

$$\dfrac{-2v}{t}$$

$$\dfrac{2v}{t}$$

$$\dfrac{v}{2t}$$

$$\dfrac{t}{2v}$$

Solution

Question 8

$$39.2\ m$$

$$44.1\ m$$

$$58.8\ m$$

$$78.4\ m$$

Solution

Question 9

$$\sqrt{3}:\sqrt{2}$$

$$\sqrt{3}:1$$

$$1: \sqrt{3}$$

$$3:1$$

Solution

Question 10

$$2.9\ s$$

$$3.1\ s$$

$$2.7\ s$$

$$0.2\ s$$

Solution

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