Motion Test

Result

Motion Test - 56

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

A train starts from rest and accelerates uniformly at a rate of $$2 \ m{s}^{-2}$$ for $$10 \ s$$. It then maintains a constant speed for $$200 \ s$$. The brakes are then applied and the train is uniformly retarded and comes to rest in $$50 \ s$$.

...view full instructions

Find the total distance travelled.

A
$$460 \ m$$

B
$$4600 \ m$$

C
$$46 \ km$$

D
$$460 \ km$$

Solution

Given, initially the train accelerates at a rate $$a_1$$ for time $$t_1$$
$$u_1=0$$
$$a_1=2 \ ms^{-2}$$
$$t_1=10 \ s$$

Using the second equation of motion , displacement is given by,
$$ s_1=u_1t_1+\dfrac{1}{2}a_1t_1^2=(0\times 10)+(\dfrac{1}{2}\times 2\times 10^2)=100 \ m$$

Calculating the velocity $$v_1$$ of the train after time $$t_1$$
Using the first equation of motion, to calculate the final velocity,
$$v_1=u_1+a_1t_1= 0+2 \times 10 = 20 \ ms^{-1}$$

Now, the train moves at constant speed $$v_1$$ for time $$t_2=200s$$
Hence, displacement $$s_2$$ is given by
$$\therefore s_2=v_1\times t_2=(20\times 200)=4000 \ m$$

Now, brakes are applied so that train comes to rest in time $$t_3$$
final velocity $$w=0$$
time taken $$t_3=50 \ s$$
acceleration is given by,
$$\therefore a_2=\dfrac{w-v_1}{t_3}=-0.4 \ ms^{-2}$$

Using the second equation of motion, to calculate the distance travelled,
$$s_3=(v_1\times t_3)+(\dfrac{1}{2}\times a_2\times t_3^2)=(20\times 50) -(\dfrac{1}{2}\times 0.4\times 50^2)$$
$$s_3=500 / m$$

Hence, total distance , $$s=s_1+s_2+s_3=(100+4000+500) \ m=4600 \ m$$
Question 2
1 / -0

A ball is initially moving with a velocity $$0.5 m{s}^{-1}$$. Its velocity decreases at a rate of $$0.05 m{s}^{-2}$$. How much time will it take to stop?

A
$$5s$$

B
$$10s$$

C
$$15s$$

D
$$20s$$

Solution

Given,
$$u=0.5\ ms^{-1}$$
$$a=-0.05\ ms^{-2}$$ and $$v=0$$

Now, $$v=u+at$$
$$0=0.5+(-0.05\times t)$$
$$t=10\ s$$
Question 3
1 / -0

Directions For Questions

A car travels with a uniform velocity of 25 m$${s}^{-1}$$ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.

...view full instructions

Find the distance which the car travels before the brakes are applied.

A
$$75m$$

B
$$100m$$

C
$$120m$$

D
$$125m$$

Solution

Before brakes are applied, the car was moving with uniforms velocity.
Distance = uniform velocity $$\times$$ time $$=(25\times 5)m=125m$$
Question 4
1 / -0

A particle starts to move in a straight line from a point with velocity $$10 m{s}^{-1}$$ and acceleration $$- 2.0 m{s}^{-2}$$. Find the position of the particle at $$t = 5 s$$

A
$$25m$$

B
$$-25m$$

C
$$50m$$

D
$$-50m$$

Solution

Initial Velocity, $$u=10\ ms^{-1}$$
Acceleration, $$a=-2\ ms^{-1}$$
Time, $$t=5\ s$$
Displacement, $$s$$

Using Newton's second equation of motion,
$$s=ut+\dfrac{1}{2}at^2$$
$$s=(10\times 5)-(\dfrac{1}{2}\times 2\times 5^2)$$
$$s=25\ m$$
Question 5
1 / -0

Larger the slope of a displacement-time graph

A
lesser the velocity

B
higher the velocity

C
lesser the acceleration

D
higher the acceleration

Solution

The slope of a displacement-time graph with time axis measures velocity. Larger slope (steeper graph) means the displacement is changing with a faster rate with respect to the time. Faster rate of change of change of distance with respect to time simply means the body has a larger velocity.

So, larger slope indicates higher velocity.
Question 6
1 / -0

A train is moving with a velocity of $$90\ kmph$$. When brakes are applied, the retardation produced is $$0.5\ m{s}^{-2}$$. Find the velocity after $$10 s$$.

A
$$20 m{s}^{-1}$$

B
$$25 m{s}^{-1}$$

C
$$30 m{s}^{-1}$$

D
$$40 m{s}^{-1}$$

Solution

Initial Velocity, $$u=90\ kmph=\dfrac{90\times 1000}{60\times 60}ms^{-1}=25\ ms^{-1}$$

Retardation, $$a=-0.5\ ms^{-2}$$

Time, $$t=10\ s$$

Using equation of motion
$$v=u+at$$

$$v=25-(0.5\times 10)=20\ ms^{-1}$$
Question 7
1 / -0

A train is moving with a velocity of $$90\ kmph$$. When brakes are applied, the retardation produced is $$0.5 m{s}^{-2}$$. Find the time taken by the train to come to rest.

A
$$20s$$

B
$$30s$$

C
$$40s$$

D
$$50s$$

Solution

Initial Velocity, $$u=90\ kmph=25\ ms^{-1}$$

Retardation, $$a=-0.5\ ms^{-2}$$

Final velocity,$$v=0$$

Let time $$= t$$

$$v=u+at$$

$$0=25-(0.5\times t)$$

$$t=50\ s$$
Question 8
1 / -0

If a displacement-time graph of an object is a straight line parallel to the time axis, the object is

A
stationary

B
moving away from the reference point

C
moving towards the reference point

D
moving with an infinite velocity

Solution

Displacement-time graph being parallel to the time axis, its slope is zero.
Slope of a displacement-time graph measures velocity.
So, velocity of the object is zero.
The object is stationary.
Question 9
1 / -0

A vehicle moving with a constant acceleration from $$A$$ to $$B$$ in a straight line $$AB$$, has velocities $$u$$ and $$v$$ at $$A$$ and $$B$$ respectively. $$C$$ is the mid point of $$AB$$. If time taken to travel from $$A$$ to $$C$$ is twice the time to travel from $$C$$ to $$B$$ then the velocity of the vehicle $$v$$ at $$B$$ is:

A
$$5u$$

B
$$6u$$

C
$$7u$$

D
$$8u$$

Solution

Let the velocity at A be $$u$$ and acceleration be $$a$$.
Thus . using the third equation of motion , $$v^2=u^2+2as$$
where $$s$$ is the distance between A and B.

From the first equation of motion,

Time taken to travel from A to C is $$\dfrac{v'-u}{a}$$, where $$v'$$ is the velocity of vehicle at $$C$$.
Similarly time taken to travel from C to B is $$\dfrac{v-v'}{a}$$
Thus $$\dfrac{v'-u}{a}=2\dfrac{v-v'}{a}$$
$$\implies v'=\dfrac{u+2v}{3}$$
Also $$v'^2-u^2=2a\dfrac{s}{2}=as$$
Thus $$v^2=u^2+2(v'^2-u^2)=2v'^2-u^2$$
$$=2(\dfrac{u+2v}{3})^2-u^2$$
On simplification, we get,
$$v^2-8uv+7u^2=0$$
$$v^2-uv-7uv+7u^2=0$$
$$(v-u)(v-7u)=0$$
$$v=u,7u$$
But , since the body is moving with constant acceleration , $$v>u$$
$$\implies v=7u$$
Question 10
1 / -0

When a body moves in an uniform circular motion, changes in velocity and speed will be

A
velocity changes but speed remains constant

B
velocity remains constant but speed changes

C
both remains constant

D
both changes

Solution

When a body moves in a circular path its direction of motion changes continuously. So, even when the body moves at a constant speed, its velocity changes.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Motion Test - 56

Result

Motion Test - 56

Score

-

Rank

-

SHARING IS CARING

Question 1

$$460 \ m$$

$$4600 \ m$$

$$46 \ km$$

$$460 \ km$$

Solution

Question 2

$$5s$$

$$10s$$

$$15s$$

$$20s$$

Solution

Question 3

$$75m$$

$$100m$$

$$120m$$

$$125m$$

Solution

Question 4

$$25m$$

$$-25m$$

$$50m$$

$$-50m$$

Solution

Question 5

lesser the velocity

higher the velocity

lesser the acceleration

higher the acceleration

Solution

Question 6

$$20 m{s}^{-1}$$

$$25 m{s}^{-1}$$

$$30 m{s}^{-1}$$

$$40 m{s}^{-1}$$

Solution

Question 7

$$20s$$

$$30s$$

$$40s$$

$$50s$$

Solution

Question 8

stationary

moving away from the reference point

moving towards the reference point

moving with an infinite velocity

Solution

Question 9

$$5u$$

$$6u$$

$$7u$$

$$8u$$

Solution

Question 10

velocity changes but speed remains constant

velocity remains constant but speed changes

both remains constant

both changes

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.