Motion Test

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Motion Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from $$54\ kmph$$ to $$18\ kmph$$ in $$5\ s$$. What is the distance traveled by train during this interval of time?

A
$$52\ m$$

B
$$50\ m$$

C
$$25\ m$$

D
$$80\ m$$

Solution

Initial speed
$$u = 54 \ kmph$$
$$\Rightarrow u= 54\times \dfrac{5}{18}$$
$$\Rightarrow u = 15 \ ms^{-1}$$
Final speed
$$v = 18 \ kmph$$
$$\Rightarrow v =18\times \dfrac{5}{18}$$
$$\Rightarrow v = 5 \ ms^{-1}$$
Time $$t = 5 \ s$$
Using   $$v = u+at$$
$$\therefore$$   $$15 = 5+a\times 5$$
$$\implies \ a = -2 \ m/s^2$$
Using   $$2aS = v^2 - u^2$$
where $$S$$ is the distance travelled.
$$\therefore$$ $$2\times (-2)S = 5^2 - 15^2$$
$$\Rightarrow \ S = 50 \ m$$
Question 2
1 / -0

The speed of a car reduces from $$15 m s^{-1} $$ to $$5 m s^{-1}$$ over a displacement of 10 m. What is the uniform acceleration of the car?

A
$$-10 m s^{-2}$$

B
$$+10 m s^{-2}$$

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

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

Solution

$$\displaystyle a = \frac{v^2 - u^2}{2s} = \frac{5^2 - 15^2}{2 \times 10}$$
$$ \displaystyle = \frac{-20 \times 10}{20}= - 10 m s^{-2}$$
Question 3
1 / -0

A body having zero speed :

A
Is always under rest.

B
Has always zero acceleration.

C
Has always uniform acceleration.

D
Always under motion.

Solution

A body having zero speed. its mean that body is not moving from its location. hence the body will be at rest.
Question 4
1 / -0

A body thrown vertically up reaches a maximum height of 50 m. Another body with double the mass is thrown up with double the initial velocity will reach a maximum height of :

A
100 m

B
200 m

C
400 m

D
50 m
Question 5
1 / -0

A stone is thrown vertically up from the ground. It reaches a maximum height of 50 m in 10 s. After what time will it reach the ground from the maximum height?

A
5 s

B
10 s

C
20 s

D
25 s

Solution

$$t_a = \displaystyle \dfrac{u}{g} = 10 s$$
same amount of time it will take to fall down
Question 6
1 / -0

Two balls are dropped from heights $$h_1$$ and $$h_2$$ respectively. The ratio of their velocities on reaching the ground is __________ .

A
$$\sqrt{h_1} : \sqrt{h_2}$$

B
$$h_1 : h_2$$

C
$$h^2_1 : h^2_2$$

D
none of these

Solution

Initial velocity of ball
$$u = 0$$
Let the final velocity on reaching the ground be $$v$$
Using 3rd equation of motion-
$$v^2 = u^2+ 2gh$$
$$\Rightarrow v^2 = 0+ 2gh$$
$$\Rightarrow \ v = \sqrt{2gh}$$
We get
$$v \propto \sqrt{h}$$
$$\Rightarrow \ v_1:v_2 = \sqrt{h_1}:\sqrt{h_2}$$
Question 7
1 / -0

A stone thrown vertically upwards with an initial velocity u from the top of a tower reaches the ground with a velocity of 3 u. The height of the tower is :

A
$$3 \displaystyle \dfrac{u^2}{g}$$

B
$$4 \displaystyle \dfrac{u^2}{g}$$

C
$$6 \displaystyle \dfrac{u^2}{g}$$

D
$$9 \displaystyle \dfrac{u^2}{g}$$

Solution

$$-h = ut - \displaystyle \dfrac{1}{2} gt^2$$
$$-h = \displaystyle \dfrac{4u^2}{g} - \dfrac{1}{2} g \left ( \dfrac{16u^2}{g^2} \right)$$
$$-h = \displaystyle \dfrac{4u^2}{g} - \frac{8u^2}{g} = \dfrac{-4 u^2}{g}$$
$$h = \displaystyle \dfrac{4u^2}{g}$$
Question 8
1 / -0

A ball is released from the top of height 'h' meters. It takes 't' seconds to reach the ground. Where is the ball at the time $$\displaystyle \dfrac{t}{2} s$$ ?

A
At $$\displaystyle \left ( \dfrac{h}{4} \right) $$ from the ground

B
At $$\displaystyle \left ( \dfrac{h}{2} \right) $$ from the ground

C
At $$\displaystyle \left ( \dfrac{3h}{4} \right) $$ from the ground

D
Depends upon mass and volume of the ball

Solution
Question 9
1 / -0

A car attains a velocity of 10 m s$$^{-1}$$ in 5 s. If initially, it had been at rest, its acceleration must be ___________ .

A
50m s$$^{-2}$$

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

C
5 m s$$^{-2}$$

D
0.6 m s$$^{-2}$$

Solution

Initial velocity of the car   $$u = 0$$
Final velocity   $$v = 10 \ m/s$$
Time taken   $$t = 5 \ s$$
Using   $$v = u+at$$
$$\therefore$$    $$10 = 0+a\times 5$$
$$\implies \ a = 2 \ m/s^2$$
Question 10
1 / -0

A car is moving in a straight line. The position time graph $$(x-t)$$ is as shown in the given figure. Find the average speed of the car for the section- CD.

A
$$1\ ms^{-1}$$

B
$$2.66\ ms^{-1}$$

C
$$3\ ms^{-1}$$

D
$$0.89\ ms^{-1}$$

Solution

$$\underline{\text{For the section-CD}}$$
Total distance travelled is = $$(4-0)=4\ m$$
Total time taken = $$(6-4.5)=1.5\ sec$$

$$\text{Average Speed}=\dfrac {\text{Total distance}}{\text{Total time}} $$
$$\text{Average speed}=\dfrac{4}{1.5}=2.66\ m/s$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 37

Result

Motion Test - 37

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SHARING IS CARING

Question 1

$$52\ m$$

$$50\ m$$

$$25\ m$$

$$80\ m$$

Solution

Question 2

$$-10 m s^{-2}$$

$$+10 m s^{-2}$$

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

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

Solution

Question 3

Is always under rest.

Has always zero acceleration.

Has always uniform acceleration.

Always under motion.

Solution

Question 4

100 m

200 m

400 m

50 m

Question 5

5 s

10 s

20 s

25 s

Solution

Question 6

$$\sqrt{h_1} : \sqrt{h_2}$$

$$h_1 : h_2$$

$$h^2_1 : h^2_2$$

none of these

Solution

Question 7

$$3 \displaystyle \dfrac{u^2}{g}$$

$$4 \displaystyle \dfrac{u^2}{g}$$

$$6 \displaystyle \dfrac{u^2}{g}$$

$$9 \displaystyle \dfrac{u^2}{g}$$

Solution

Question 8

At $$\displaystyle \left ( \dfrac{h}{4} \right) $$ from the ground

At $$\displaystyle \left ( \dfrac{h}{2} \right) $$ from the ground

At $$\displaystyle \left ( \dfrac{3h}{4} \right) $$ from the ground

Depends upon mass and volume of the ball

Solution

Question 9

50m s$$^{-2}$$

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

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

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

Solution

Question 10

$$1\ ms^{-1}$$

$$2.66\ ms^{-1}$$

$$3\ ms^{-1}$$

$$0.89\ ms^{-1}$$

Solution

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