Motion Test

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Motion Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

A car is travelling at 30 m/s. If its velocity increases to 50 m/s while accelerating at the rate of 2.5 m/s$$\displaystyle ^{2}$$ find the time taken to reach that velocity.

A
10 s

B
8 s

C
4 s

D
5 s

Solution

Given Final velocity (v) = 50 m/s
Initial velocity (u) = 30 m/s
Acceleration (a) = 2.5 m/s$$\displaystyle ^{2}$$
We know v = u + at
$$\displaystyle \therefore $$ 50 = 30 + 2.5 t
$$\displaystyle \Rightarrow $$ 50 - 30 = 2.5 t
$$\displaystyle \Rightarrow $$ $$\displaystyle \frac{20}{2.5}$$ = t or 8 = t
$$\displaystyle \therefore $$ Time taken to accelerate = 8 seconds
Question 2
1 / -0

The quantity which tells the distance an object travels in a certain time is called -

A
acceleration

B
speed

C
velocity

D
none of these

Solution

The quantity which tells the distance an object travels in a certain time is called speed.
$$Speed = \dfrac{Distance}{Time}$$
Question 3
1 / -0

A stone dropped from the top of a tower reaches the ground with a velocity of 58.8 m/s in 6 seconds. Calculate the value of acceleration due to gravity.

A
10 m/s$$\displaystyle ^{2}$$

B
19.6 m/s$$\displaystyle ^{2}$$

C
8.2 m/s$$\displaystyle ^{2}$$

D
9.8 m/s$$\displaystyle ^{2}$$

Solution

Here Initial velocity (u) = 0
Final velocity (v) = 58.8 m/s
Time (t) = 6 seconds
We know $$v = u +at$$
$$58.8 = a\times6$$
$$\displaystyle \Rightarrow 6 a = 58.8$$
$$\displaystyle \Rightarrow a = \frac{58.8}{6}$$
$$\displaystyle \therefore a = 9.8 m/s^{2}$$
$$\displaystyle \therefore$$ The value of acceleration due to gravity = 9.8 m/s$$\displaystyle ^{2}$$
Question 4
1 / -0

If two bodies are moving with the same speed but in different directions they will have -

A
same velocities

B
same acceleration

C
different velocities

D
none of these

Solution

Since velocity is a vector quantity as it has both magnitude as well as direction. So the two bodies moving with same speed but in different directions have different velocities.
Question 5
1 / -0

A boy cycles a distance of 640 m in 2 minutes and 40 seconds. What is the speed of the cycle?

A
4 m/s

B
8 m/s

C
12 m/s

D
16 m/s

Solution

Distance covered = 640 m
Time taken = 2 min 40 sec
= $$2$$ $$\displaystyle \times 60 + 40 = 160s$$ (because 1 minute =60seconds)
$$\displaystyle \therefore Speed = \frac{Distance}{Time}= \frac{640}{160}=4m/s$$
Question 6
1 / -0

The velocity of a bullet is reduced from $$200\;\text{m/s}$$ to $$100\;\text{m/s}$$ while travelling through a wooden block of thickness $$10\;\text{cm}$$. Assuming it to be uniform, the retardation will be:

A
$$15\times10^4\;\text{m/s}^2$$

B
$$10\times10^4\;\text{m/s}^2$$

C
$$12\times10^4\;\text{m/s}^2$$

D
$$14.5\;\text{m/s}^2$$

Solution

Using the formula, $$\text{v}^2=\text{u}^2+2\text{as}$$

$$(100)^2=(200)^2-2\text{a}\times\displaystyle\frac{10}{100}$$

$$2\text{a}\times\displaystyle\frac{10}{100}=(200)^2-(100)^2=300\times100$$

$$\text{a}=\displaystyle\frac{3\times10^5}{2}=15\times10^4\;\text{m/s}^2$$
Question 7
1 / -0

A ball thrown vertically upwards with a velocity of 15 m/s reaches its highest point at 11.25 m in 1.5 sec. Find the total distance travelled by the ball and its position after 2 sec respectively. Take $$g=10 \ m/s^2$$

A
Total distance = 12 m, After 2 sec = 9.5 m.

B
Total distance = 17.5 m, After 2 sec =7.5 m.

C
Total distance = 14 m, After 2 sec = 9 m.

D
Total distance = 12.50 m, After 2 sec = 10 m.

Solution

The ball is thrown upwards.
$$\therefore$$ Acceleration due to gravity $$= -10 \ m/s^2$$
Final velocity $$=\ 0m/s$$
Initial velocity $$= 15\ m/s$$
Distance $$= 11.25\ m$$
Time $$= 1.5\ sec$$
To find total distance traveled:
the distance traveled by the ball in upward and downward motion will be the sum of the height reached upto 1.5 sec plus height descended in next 0.5 sec.

Height descended,
$$h=ut+\dfrac{1}{2} at^2$$
$$h=0+\dfrac{1}{2} (-10)(0.5)^2$$
$$h=1.25 m$$

Hence, the total distance will be $$11.25\ m + 1.25\ m = 12.50\ m$$
To find position after $$2\ sec$$
Using the formula:
$$s=ut+\dfrac{1}{2} at^2$$
$$s=15\times 2+\dfrac{1}{2}\times -10 \times (2)^2$$
$$s=30+(-40)$$
$$s=-10 m$$

So the option D is correct.
Question 8
1 / -0

A runner running along a circular track at a constant speed has _______ velocity.

A
Variable

B
Uniform

C
Potential

D
Parallel

Solution

the magnitude may remain same but the direction of velocity keeps on changing, so velocity is varriable
Question 9
1 / -0

Can a moving object cover zero distance?

A
Always

B
Sometimes

C
Never

D
Can't be said from given condition

Solution

A moving object always changes its position. Then, it moves along a path with some length.
So, if the body is moving, it will always cover some path that can never have zero length. Thus distance traveled by a moving object can never be zero.
Question 10
1 / -0

A body, initially at rest, starts moving with a constant acceleration $$2 m{s}^{-2}$$. Calculate the distance travelled in 5 s.

A
$$10m$$

B
$$15m$$

C
$$25m$$

D
$$50m$$

Solution

Initial velocity, $$u=0$$

Acceleration, $$a=2ms^{-2}$$

Time, $$t=5s$$

Distance traveled, $$s=ut+\dfrac{1}{2}at^2$$

$$s=(0\times 5)+(\dfrac{1}{2}\times 2\times 5^2)=25m$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 54

Result

Motion Test - 54

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Rank

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SHARING IS CARING

Question 1

10 s

8 s

4 s

5 s

Solution

Question 2

acceleration

speed

velocity

none of these

Solution

Question 3

10 m/s$$\displaystyle ^{2}$$

19.6 m/s$$\displaystyle ^{2}$$

8.2 m/s$$\displaystyle ^{2}$$

9.8 m/s$$\displaystyle ^{2}$$

Solution

Question 4

same velocities

same acceleration

different velocities

none of these

Solution

Question 5

4 m/s

8 m/s

12 m/s

16 m/s

Solution

Question 6

$$15\times10^4\;\text{m/s}^2$$

$$10\times10^4\;\text{m/s}^2$$

$$12\times10^4\;\text{m/s}^2$$

$$14.5\;\text{m/s}^2$$

Solution

Question 7

Total distance = 12 m, After 2 sec = 9.5 m.

Total distance = 17.5 m, After 2 sec =7.5 m.

Total distance = 14 m, After 2 sec = 9 m.

Total distance = 12.50 m, After 2 sec = 10 m.

Solution

Question 8

Variable

Uniform

Potential

Parallel

Solution

Question 9

Always

Sometimes

Never

Can't be said from given condition

Solution

Question 10

$$10m$$

$$15m$$

$$25m$$

$$50m$$

Solution

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