Motion Test

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Motion Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

A body is projected vertically upward with an initial velocity $$u$$. If acceleration due to gravity is $$g$$, the time for which it remains in air, is

A
$$\displaystyle\frac{u}{g}$$

B
$$ug$$

C
$$\displaystyle\frac{2u}{g}$$

D
$$\displaystyle\frac{u}{2g}$$

Solution

$$v=u+gt$$
at highest distance,v$$=$$0,and g acts in opposite direction.
So,
$$0=u-gt$$
$$t= \dfrac{u}{g}$$ since time taken to ascent and descent is same.
So, the total time(t) is $$ \dfrac{2u}{g} $$ for which it remain in air.
Question 2
1 / -0

Give an example of motion in which average speed is not zero, but the average velocity is zero.

A
A car moving with an average speed returns to the initial position

B
A car at rest

C
A car with non zero displacement.

D
All of the above.

Solution

Suppose a car covers a distance of 50 m in 5 seconds and comes back to initial position in 5 seconds. The average speed of car is 10 m/s, but average velocity is zero because displacement is zero.
Question 3
1 / -0

A pebble is thrown vertically upwards with a speed of $$20\ ms^{-1}$$. How high will it be after $$2\ s$$? (Take $$g=10\ ms^{-2}$$)

A
40 m

B
10 m

C
20 m

D
25 m

Solution

Initial velocity , $$u= 20\ m/s$$
Since, the ball is thrown upward, so 2nd equation of motion will be given by,
$$ S = ut - \dfrac{1}{2}gt^{2}$$
$$\Rightarrow S = 20\times 2 - \dfrac{1}{2}\times 10\times 2^{2} = 40-20 =20\ m$$
Question 4
1 / -0

A body is moving vertically upwards. Its velocity changes at a constant rate from $$50 m s^{-1}$$ to $$20 m s^{-1}$$ in 3 s. What is its acceleration ?

A
$$-30 m s^{-2}$$

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

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

D
$$30 m s^{-2}$$

Solution

Acceleration, $$a = \dfrac{Final Velocity-Initial Velocity}{Time} = \dfrac{20 m/s-50 m/s}{3s} = -10 m/s^2 $$
Question 5
1 / -0

A body falls from the top of a building and reaches the ground $$2.5 s$$ later. How high is the building? (Take $$g=10 m {s}^{-2}$$)

A
30.6 m

B
31.25 m

C
30 m

D
25 m

Solution

Height of the building is given by 2nd equation of motion,
$$ S = ut + \dfrac{1}{2}gt^{2}$$
now since initial velocity is zero so 2nd equation of motion, become,
$$ S = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 2.5^{2} = 31.25 m$$
Question 6
1 / -0

A pebble thrown vertically upwards with an initial velocity $$50 \ m s^{-1}$$ comes to a stop in 5 s. Find the retardation assuming displacement in upwards direction to be positive.

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

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

C
$$250 \ m s^{-2}$$

D
$$-250 \ m s^{-2}$$

Solution

Acceleration $$a=\dfrac{v-u}{t}= \dfrac{0-50 m s^{-1}}{5 s}$$
$$= -10 m s ^{-2}$$
$$\text{Retardation}=-\text{Acceleration}$$
$$=10\ ms^{-2}$$
Question 7
1 / -0

The velocity of an object increases at a constant rate from $$20 m s^{-1}$$ to $$50 m s^{-1}$$in 10 s. Find the acceleration.

A
$$-3 \,m \,s^{-2}$$

B
$$-30 \,m \,s^{-2}$$

C
$$30 \,m \,s^{-2}$$

D
$$3 \,m \,s^{-2}$$

Solution

Acceleration $$a=\dfrac {v-u}{t}=\dfrac{50 \ m s^{-1}-20 \ m s^{-1}}{10 s}$$
or $$ a=\dfrac{30 \ m s^{-1}}{10 s}$$
$$= 3 \ m s^{-2}$$
Question 8
1 / -0

A body is dropped from the top of a tower. It acquires a velocity of $$20\ ms^{-1}$$ on reaching the ground. Calculate the height of the tower. (Take $$g=10\ ms^{-2}$$)

A
$$20 m$$

B
$$10 m$$

C
$$40 m$$

D
$$30 m$$

Solution

Final velocity , $$ v= 20 m/s$$
$$g = 10 m/s^{2}$$
By third eqation of motion we have,
$$v^{2} = u^{2} + 2gS$$
Since, $$u =0$$
$$S = \dfrac{v^{2}}{2g}$$
$$\Rightarrow S = \dfrac{20^{2}}{2\times10} = 20m$$
Question 9
1 / -0

A stone is dropped freely from the top of a tower and it reaches the ground in $$4\ s$$. Taking $$g=10\ ms^{-2}$$, calculate the height of the tower.

A
80 m

B
40 m

C
4 m

D
20 m

Solution

$$S = ut + \dfrac{1}{2}gt^{2}$$
as $$ u = 0$$
So, $$S = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 4^{2} = 80 m$$
Question 10
1 / -0

An object falling freely from rest reaches ground in $$2 s$$. If acceleration due to gravity is $$9.8 ms^{-2}$$, the velocity of object on reaching the ground will be

A
$$9.8$$ $$m$$ $${s}^{-1}$$

B
$$4.9$$ $$m$$ $${s}^{-1}$$

C
$$19.6$$ $$m$$ $${s}^{-1}$$

D
zero

Solution

$$v=u+gt$$ , initially u$$=$$0 and g acting in the direction of movement of body.
$$v=0+9.8 \times 2$$
$$v=19.6 m/s$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 42

Result

Motion Test - 42

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SHARING IS CARING

Question 1

$$\displaystyle\frac{u}{g}$$

$$ug$$

$$\displaystyle\frac{2u}{g}$$

$$\displaystyle\frac{u}{2g}$$

Solution

Question 2

A car moving with an average speed returns to the initial position

A car at rest

A car with non zero displacement.

All of the above.

Solution

Question 3

40 m

10 m

20 m

25 m

Solution

Question 4

$$-30 m s^{-2}$$

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

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

$$30 m s^{-2}$$

Solution

Question 5

30.6 m

31.25 m

30 m

25 m

Solution

Question 6

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

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

$$250 \ m s^{-2}$$

$$-250 \ m s^{-2}$$

Solution

Question 7

$$-3 \,m \,s^{-2}$$

$$-30 \,m \,s^{-2}$$

$$30 \,m \,s^{-2}$$

$$3 \,m \,s^{-2}$$

Solution

Question 8

$$20 m$$

$$10 m$$

$$40 m$$

$$30 m$$

Solution

Question 9

80 m

40 m

4 m

20 m

Solution

Question 10

$$9.8$$ $$m$$ $${s}^{-1}$$

$$4.9$$ $$m$$ $${s}^{-1}$$

$$19.6$$ $$m$$ $${s}^{-1}$$

zero

Solution

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