Motion Test

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Motion Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Velocity is the (distance or displacement) per unit time.

A
Displacement

B
Both Displacement and Distance

C
None

D
Sum of both

Solution

Velocity of an object is speed of that object plus direction in which object is moving.
It is given by
$$Velocity=\dfrac{Displacement}{Time}$$
Hence option A is correct.
Question 2
1 / -0

The unit for the rate of change of velocity is

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

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

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

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

Solution

The rate of change of velocity with respect to time is called acceleration.
GIven by
$$a=\dfrac{change\quad in\quad velocity}{time}$$
Putting units of velocity and time.
$$a=\dfrac{m/s}{s}$$=$$m/s^2$$ or $$ms^{-2}$$.Hence this is the unit of acceleration or rate of change of velocity with respect to time.
Question 3
1 / -0

The cheetah is the fastest land animal that can attain a maximum speed of $$112~ km/h$$. Express this speed in $$m/s$$.

A
$$31 ~m/s$$ (approx.)

B
$$30 ~m/s$$ (approx.)

C
$$32 ~m/s$$ (approx.)

D
$$29 ~m/s$$ (approx.)

Solution

Given,
The maximum speed by the cheetah, $$v=112\ km/h$$
We know,
$$1\ km=1000\ m$$
$$1\ h=60\ min$$
$$1\min=60\ s$$
$$\therefore 1\ h=60\times 60=3600\ s$$

$$v=\dfrac{112\times 1000}{3600}$$

$$v=31.1\ m/s$$
Question 4
1 / -0

What distance was covered by the bus at 9.45 a.m.?

A
70 m

B
80 m

C
90

D
Can't be determined

Solution
Question 5
1 / -0

Acceleration is defined as

A
the rate of change of velocity with respect to time.

B
the rate of change of position with respect to time.

C
the rate of change of momentum with respect to time.

D
None of the above

Solution

Acceleration is defined as the rate of change of velocity with respect to time.
$$acceleration = \dfrac{change\ in\ velocity}{time}$$
Question 6
1 / -0

If a train travelling at $$72$$ km/h is to be brought to rest at a distance of $$200$$ m then its retardation should be :

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

B
$$2 \ s^{2}$$

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

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

Solution

$$\displaystyle u=72\frac{km}{h}=72\times\frac{5}{18}=20\;m/s$$

Now $$v^{2}=u^{2}+2as$$

$$0=(20)^{2}+2a(200)$$

$$0=400+400a$$
$$400a=-400$$

retardation $$=-a=1\;m/s^{2}$$
Question 7
1 / -0

The rate of change of velocity of an object with respect to time is called ..........

A
Momentum

B
Displacement

C
Acceleration

D
Impulse

Solution

Acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

$$\text{acceleration} = \dfrac{\text{Change in the velociity}}{\text{time}}$$
Question 8
1 / -0

What does the graph represents:

A
uniform speed

B
non-uniform speed

C
both A and B

D
none of these

Solution

$$\textbf{Step 1-Equation of distance and time}$$
Relationship between distance and time is linear. So, its equation will be in form of:

$$D = k t$$, where k is any constant, D is distance and t is time.
$$\textbf{Step 2-Calculation of velocity}$$

On differentiating this equation with respect to time, at left hand side it will be speed as differentiation of distance with respect to time ..

$$V = k$$

So, speed for the given graph is constant or uniform.

Answer:

Hence, option A is the correct answer.
Question 9
1 / -0

A particle moving rectilinearly with constant acceleration is having initial velocity of 10 m/s. After some time, its velocity becomes 30 m/s. Find out velocity of the particle at the mid point of its path?

A
$$10 \sqrt{5}$$ m/s

B
$$50 \sqrt{5}$$ m/s

C
$$60 \sqrt{5}$$ m/s

D
$$5 \sqrt{5}$$ m/s

Solution

Let total distance be $$2x$$
$$v^2= u^2 + 2as$$
$$30^2 = 10^2 + 2 \times a \times (2x)$$
$$4ax = 800\ ............(i)$$

At mid point,
$$v_{mid}^2 = 10^2 + 2 \times a \times x$$
$$4ax = 2 v_{mid}^2\ - 200 .............(ii)$$

Subtracting (ii) from (i),
$$0 = 1000 - 2 v_{mid}^2$$
$$v_{mid}^2 = 500$$
$$v_{mid} = 10 \sqrt 5\ m/s$$
Question 10
1 / -0

A person walks along an east-west street and a graph of his displacement from home is shown in figure. His average velocity for the whole time interval is

A
$$0\;ms^{-1}$$

B
$$23\;ms^{-1}$$

C
$$8.4\;ms^{-1}$$

D
None of above

Solution

Net displacement of the particle is zero i.e $$S = 0$$ over the whole time.
$$\therefore$$ Average velocity $$v_{avg} = \dfrac{S}{t} = 0$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 47

Result

Motion Test - 47

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SHARING IS CARING

Question 1

Displacement

Both Displacement and Distance

None

Sum of both

Solution

Question 2

$$m s^{-2}$$

$$m s^{-3}$$

$$m s^{-1}$$

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

Solution

Question 3

$$31 ~m/s$$ (approx.)

$$30 ~m/s$$ (approx.)

$$32 ~m/s$$ (approx.)

$$29 ~m/s$$ (approx.)

Solution

Question 4

70 m

80 m

90

Can't be determined

Solution

Question 5

the rate of change of velocity with respect to time.

the rate of change of position with respect to time.

the rate of change of momentum with respect to time.

None of the above

Solution

Question 6

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

$$2 \ s^{2}$$

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

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

Solution

Question 7

Momentum

Displacement

Acceleration

Impulse

Solution

Question 8

uniform speed

non-uniform speed

both A and B

none of these

Solution

Question 9

$$10 \sqrt{5}$$ m/s

$$50 \sqrt{5}$$ m/s

$$60 \sqrt{5}$$ m/s

$$5 \sqrt{5}$$ m/s

Solution

Question 10

$$0\;ms^{-1}$$

$$23\;ms^{-1}$$

$$8.4\;ms^{-1}$$

None of above

Solution

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