Motion Test

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Motion Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

A driver has to apply sudden brakes to stop his vehicle moving at $$72 \ {km}/{hr}$$ and stops within $$2 \ seconds$$. The retardation produced is then

A
$$20 \ {m}/{{s}^{2}}$$

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

C
$$15 \ {m}/{{s}^{2}}$$

D
$$18 \ {m}/{{s}^{2}}$$

Solution

Using the first equation of motion, $$v = u + at$$
Here, $$v = 0$$
$$u = 72 \times \dfrac{1000}{3600} = 20 \ m/s$$
$$t = 2s$$

Hence, $$0 = 20 + a\times 2$$

$$a = -\displaystyle\frac{20}{2} = -10 \ {m}/{{s}^{2}}$$

Since, sign of $$a$$ is negative. Hence, retardation with magnitude $$10 \ m/s^2$$ is produced.
Question 2
1 / -0

If a trolley starts from rest with an acceleration of 2 $$m/s^2$$, the velocity of the body after 4 s would be

A
2 m/s

B
8 m/s

C
4 m/s

D
6 m/s

Solution

Given:
initial velocity $$u=0$$ (Since the body is starting from rest)
acceleration $$a=2\ m/s^2$$
time taken $$t=4\ s$$
final velocity $$v=?$$

According to the first equation of motion $$v = u + at$$
$$v = 0 +2\times4 = 8\ m/s$$
Question 3
1 / -0

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

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

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

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

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

Solution

Initial velocity is $$u$$ and final velocity is $$v=3u$$ and acceleration is $$g$$.
So height is $$h=(v^2-u^2)/2g=((3u)^2-u^2)/2g=(9u^2-u^2)/2g=4u^2/g$$
Question 4
1 / -0

A driver is driving his car along a road is shown in Fig. The driver makes sure that the speedometer reads exactly 40 km/h. What happens to the velocity of the car from P to Q?

A
Velocity remains constant

B
Velocity first increases then decreases

C
Velocity first decreases then increases

D
Nothing can be decided

Solution

As said in the question speedometer reads 40. So the magnitude of the velocity is not changing. The path PQ is a straight line. So the direction of velocity is also not changing. So it is clear that the velocity of the car is constant.
Question 5
1 / -0

Figure shows the displacement-time graphs (a) and (b) for a body moving in a straight path drawn on the same scales. Then

A
slope of line in (a) is greater than the slope of line in (b)

B
slope of line in (b) is greater than the slope of line in (a)

C
slope of line in (a) is equal to the slope of line in (b)

D
nothing can be said about the slopes

Solution

slope of any graph is given by $$dy/dx=Tan\theta $$
means change of function y with respect to function x.
$$\theta $$ is angle made by the graph with x axis.
as angle for a is greater then that of b so as $$Tan\theta$$ is increasing between 0 to 90, so $$Tan\theta _a$$>$$Tan\theta _b$$
so slope of a is greater then that of b.
beat possible answer is optionA.
Question 6
1 / -0

A body moving along a circular path may have

A
A constant speed

B
A constant velocity

C
No tangential velocity

D
No radial acceleration

Solution

A body moving along a circular path may have a constant speed.

The magnitude of the velocity of the body is constant but the direction is constantly changing. This means that, even though the speed is not changing, the velocity is changing.
Question 7
1 / -0

The SI unit of speed is:

A
$$m/s$$

B
$$km/s$$

C
$$cm/s$$

D
None of these

Solution

The distance moved by an object in a unit time is known as speed.
The SI unit of speed is $$m/s$$.
Question 8
1 / -0

What does the speed of a body describes about its motion ?

A
Direction

B
State

C
Type

D
Rapidity

Solution

$$speed =\frac{distance}{time}$$ Speed tells us how fast or slow the distance is being covered by the moving body .Hence it denotes it's rapidity(Means how rapid the moving object is).
Question 9
1 / -0

The velocity of a particle increases from u to v in a time t during which it covers a distances S. If the particle has a uniform acceleration, which one of the following equations does not apply to the motion?

A
2 S = (v + u) t

B
$$a = \displaystyle \frac{v - u}{t}$$

C
$$v^2 = u^2 - 2 aS$$

D
$$S= \displaystyle \left[ u + \frac{1}{2} at \right] t$$

Solution

$$\dfrac{dv}{dt}=a$$
by integrating once we get $$v-u=at$$ i.e. option B.
also $$\dfrac{ds}{dt}=v$$ so integrating once more we get $$S=ut+\frac{1}{2}at^2$$ i.e. option D
by using $$v=u+at $$ and $$ expression of S can be converted to A.
now eleminating t from equation in option A and option B we will get
and expression $$v^2-u^2=2as$$ but in option C where $$ v^2-u^2=-2aS$$ so expression in option C is not correct so answer is option C.
Question 10
1 / -0

A stone projected vertically upwards reaches to the level of window $$10\ m$$ from the ground. Find the magnitude of the velocity of the stone at the time of its projection.

A
$$\displaystyle { 7\ ms }^{ -1 }$$

B
$$\displaystyle { 36\ ms }^{ -1 }$$

C
$$\displaystyle { 28\ ms }^{ -1 }$$

D
$$\displaystyle { 14\ ms }^{ -1 }$$

Solution

Height attained by the stone thrown vertically upwards
$$h = 10 \ m$$
Velocity of projection
$$u = \sqrt{2gh}$$
$$u$$: Initial velocity
$$g$$: Acceleration due to gravity
$$h$$: Height attained by the object
$$\Rightarrow \ u = \sqrt{2\times 10\times 9.8}$$
$$\Rightarrow u = 14\ m/s$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 51

Result

Motion Test - 51

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SHARING IS CARING

Question 1

$$20 \ {m}/{{s}^{2}}$$

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

$$15 \ {m}/{{s}^{2}}$$

$$18 \ {m}/{{s}^{2}}$$

Solution

Question 2

2 m/s

8 m/s

4 m/s

6 m/s

Solution

Question 3

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

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

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

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

Solution

Question 4

Velocity remains constant

Velocity first increases then decreases

Velocity first decreases then increases

Nothing can be decided

Solution

Question 5

slope of line in (a) is greater than the slope of line in (b)

slope of line in (b) is greater than the slope of line in (a)

slope of line in (a) is equal to the slope of line in (b)

nothing can be said about the slopes

Solution

Question 6

A constant speed

A constant velocity

No tangential velocity

No radial acceleration

Solution

Question 7

$$m/s$$

$$km/s$$

$$cm/s$$

None of these

Solution

Question 8

Direction

State

Type

Rapidity

Solution

Question 9

2 S = (v + u) t

$$a = \displaystyle \frac{v - u}{t}$$

$$v^2 = u^2 - 2 aS$$

$$S= \displaystyle \left[ u + \frac{1}{2} at \right] t$$

Solution

Question 10

$$\displaystyle { 7\ ms }^{ -1 }$$

$$\displaystyle { 36\ ms }^{ -1 }$$

$$\displaystyle { 28\ ms }^{ -1 }$$

$$\displaystyle { 14\ ms }^{ -1 }$$

Solution

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