Motion Test

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Motion Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

A car was moving at $$1 m{s}^{-1}$$ when brakes were applied. The brakes offered a retardation of $$0.5 m{s}^{-2}$$. If the brakes are applied for $$2 s$$, after $$2 s$$

A
the car moves at a reduced speed of $$0.5 m{s}^{-1}$$

B
the car moves at a reduced speed of $$0.25 m{s}^{-1}$$

C
the car stops

D
the car moves with an increased speed of $$2 m{s}^{-1}$$

Solution

Initial velocity, $$u = 1 m{s}^{-1}$$
Retardation, $$a = 0.5 m{s}^{-2}$$
Time, $$t = 2 s$$
Final velocity $$= v$$
Using equation of motion,
$$a = \dfrac{u-v}{t}$$
$$0.5 = \dfrac{1 - v}{2}$$
$$v = 1 - 1 = 0$$
So, final velocity is $$0$$.
After $$2 s$$, the body stops.
Question 2
1 / -0

Which of the following is not an unit of displacement?

A
$$m$$

B
$$cm$$

C
$$km$$

D
$$ms^{-1}$$

Solution

$$m$$, $$cm$$ and $$km$$ are respectively the SI, cgs and practical units of displacement.
$$m{s}^{-1}$$ is the SI unit of speed/velocity.
Question 3
1 / -0

A car travelling at $$2 m{s}^{-1}$$ undergoes an acceleration of $$1 m{s}^{-2}$$. How long will it take to double its velocity?

A
$$1s$$

B
$$2s$$

C
$$3s$$

D
$$4s$$

Solution

$$u=2\ ms^{-1}$$
$$a=1\ ms^{-2}$$
$$v=2u=4\ ms^{-1}$$
$$\therefore t=\dfrac{v-u}{a}=\dfrac{4-2}{1}s=2\ s$$
Question 4
1 / -0

Can, a body in motion, have zero displacement?

A
Yes, always

B
Yes, provided its initial and final positions are same

C
Never

D
Yes, only if its initial and final positions are not same

Solution

Displacement is the shortest distance between the initial and final positions of a body.
So, if a body in motion comes back to its initial position, it will have zero displacement.
Question 5
1 / -0

Directions For Questions

A vehicle is accelerating on a straight road. Its velocity at a certain point of time is $$30 kmph$$. After $$2 s$$, its velocity is $$33.6 kmph$$ and, after further $$2 s$$, it is $$37.2 kmph$$.

...view full instructions

Find the acceleration of the vehicle.

A
$$0.1\ A m{s}^{-2}$$

B
$$0.3\ m{s}^{-2}$$

C
$$0.5\ m{s}^{-2}$$

D
$$1\ m{s}^{-2}$$

Solution

Given,
Initial velocity $$u=30\ kmph=\dfrac{30\times 1000}{60\times 60}ms^{-1}=\dfrac{25}{3}ms^{-1}$$

After 2 sec, its velocity $$v=33.6\ kmph=\dfrac{33.6\times 1000}{60\times 60}ms^{-1} = \dfrac{28}{3}ms^{-1}$$
$$t_1= 2s$$

$$\therefore a=\dfrac{v-u}{t_1}=\dfrac{\frac{28}{3}-\frac{25}{3}}{2}ms^{-2}=0.5ms^{-2}$$
Question 6
1 / -0

A car moving at $$2.5 m{s}^{-1}$$ doubles its velocity with an acceleration of $$0.5 m{s}^{-2}$$ in some time. If the same car travels at $$1.5 m{s}^{-1}$$, what will be its final velocity if same acceleration acts on it for same time?

A
$$2 m{s}^{-1}$$

B
$$3 m{s}^{-1}$$

C
$$4 m{s}^{-1}$$

D
$$5 m{s}^{-1}$$

Solution

$$u=2.5\ ms^{-1}$$
$$v=2u=5\ ms^{-1}$$
$$a=0.5ms^{-2}$$
Let us use: $$v=u+at$$
or, $$5=2.5+(0.5\times t)$$
or, $$t=\dfrac{2.5}{0.5}=5\ s$$
So, using this value, but different initial velocity $$u$$ =$$1.5ms^{-1}$$ we get:
$$v=u+at=1.5+(0.5\times 5)=4ms^{-1}$$
Question 7
1 / -0

Distance is

A
the shortest path between the initial and final positions

B
the longest possible path between the initial and final positions

C
the actual path travelled by the body between the initial and final positions

D
none of these

Solution

Distance is defined as the length of the actual path travelled by a moving body between its initial and final positions.
Question 8
1 / -0

Length of the straight line joining the initial to the final positions of a moving body is known as its

A
distance

B
displacement

C
position

D
none of these

Solution

Displacement is the shortest distance joining initial and final points through a straight line.
Question 9
1 / -0

A body with an initial velocity of $$18\ kmph$$ accelerates uniformly at $$9\ cm{s}^{-2}$$ over a distance of $$200\ m$$. Find its final velocity in $$m{s}^{-1}$$.

A
$$\sqrt{61}$$

B
$$\sqrt{78}$$

C
$$\sqrt{51}$$

D
None of these

Solution

Given,
Acceleration, $$a=9\ cms^{-2}=0.09\ ms^{-2}$$
Displacement, $$s=200m$$
Initial velocity, $$u=18 kmph$$
$$u=\dfrac{18\times 1000}{60 \times 60}ms^{-1}=5ms^{-1}$$

Using Newton's third equation of motion,
$$2as=v^2-u^2$$
$$\implies v^2=u^2+2as$$
$$v^2=(5)^2+(2\times 0.09\times 200)$$
$$v^2=61$$
$$v=\sqrt{61}\ ms^{-1}$$
Question 10
1 / -0

When brakes are applied to a bus, the retardation produced is $$25 cm{s}^{-2}$$ and the bus takes $$20s$$ to stop. Calculate the initial velocity of the bus.

A
$$5 m{s}^{-1}$$

B
$$10 m{s}^{-1}$$

C
$$15 m{s}^{-1}$$

D
$$20 m{s}^{-1}$$

Solution

Final Velocity, $$v=0$$

Retardation, $$a=25\ cm s^{-2}=0.25\ ms^{-2}$$

Time, $$t=20s$$

Let, initial velocity $$=u$$
According to newton first law
$$v=u-at$$
$$0=u-(0.25\times 20)$$
$$u=5\ ms^{-1}$$

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Motion Test - 55

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Motion Test - 55

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Question 1

the car moves at a reduced speed of $$0.5 m{s}^{-1}$$

the car moves at a reduced speed of $$0.25 m{s}^{-1}$$

the car stops

the car moves with an increased speed of $$2 m{s}^{-1}$$

Solution

Question 2

$$m$$

$$cm$$

$$km$$

$$ms^{-1}$$

Solution

Question 3

$$1s$$

$$2s$$

$$3s$$

$$4s$$

Solution

Question 4

Yes, always

Yes, provided its initial and final positions are same

Never

Yes, only if its initial and final positions are not same

Solution

Question 5

$$0.1\ A m{s}^{-2}$$

$$0.3\ m{s}^{-2}$$

$$0.5\ m{s}^{-2}$$

$$1\ m{s}^{-2}$$

Solution

Question 6

$$2 m{s}^{-1}$$

$$3 m{s}^{-1}$$

$$4 m{s}^{-1}$$

$$5 m{s}^{-1}$$

Solution

Question 7

the shortest path between the initial and final positions

the longest possible path between the initial and final positions

the actual path travelled by the body between the initial and final positions

none of these

Solution

Question 8

distance

displacement

position

none of these

Solution

Question 9

$$\sqrt{61}$$

$$\sqrt{78}$$

$$\sqrt{51}$$

None of these

Solution

Question 10

$$5 m{s}^{-1}$$

$$10 m{s}^{-1}$$

$$15 m{s}^{-1}$$

$$20 m{s}^{-1}$$

Solution

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