Motion Test

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Motion Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

What does the odometer of an automobile measure?

A
Speed

B
Distance

C
Acceleration

D
All

Solution

An odometer or odograph is an instrument that indicates distance traveled by the automobile
Question 2
1 / -0

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $$3.0\ ms^{-2}$$ for $$8\ s$$. How far does the boat travel during this time?

A
$$96\ m$$

B
$$192\ m$$

C
$$12\ m$$

D
$$72\ m$$

Solution

Given, initial velocity of boat is: $$u=0$$
Acceleration is: $$a=3\ ms^{-2}$$
Time is: $$t=8\ s$$

Using second equation of motion:
$$s=ut+\cfrac {1}{2}at^2$$
$$s=0\times 8+\cfrac {1}{2}\times 3\times 8^2=96\ m$$.
Question 3
1 / -0

A car starts from rest and moves along the x-axis with constant acceleration $$5 ms^{-2}$$ for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?

A
320 m

B
160 m

C
480 m

D
720 m

Solution

$$s_{ 1 }=ut+(1/2)at^2$$
$$=0+\dfrac { 1 }{ 2 } \times 5\times { 8 }^{ 2 }$$
$$ =160\ m$$.
The velocity attained in 8 seconds is given by
$$v=u+at$$
$$=0+5\times 8 = 40\ m/s$$.
The distance traveled in the next 4 seconds is
$$s_{ 2 }=40\times 4 = 160\ m$$.
So, the total distance traveled by the car is $$s_1 + s_2 =160+160= 320\ m$$
Question 4
1 / -0

If the displacement of an object is proportional to square to time, then the object moves with:

A
uniform velocity.

B
uniform acceleration.

C
increasing acceleration.

D
decreasing acceleration.

Solution

Hint:
Motion with uniform acceleration is that in which the acceleration of a body does not change with time.
Step 1: Note all the given values.
It is given that displacement of the body is directly proportional to the square of time. That is,
$$s \propto t^2$$
Or,
$$s = Kt^2$$ ....eq.1
Where $$K$$ is constant.

Step 2: Find velocity and acceleration of object
Now $$\dfrac{ds}{dt} = V = K2t$$
$$\Rightarrow V = (2 K)t$$
$$\dfrac{dv}{dt} = \vec{a} = 2k$$
So acceleration is constant

Thus, if displacement of a body is directly proportional to the square of time then the motion is of uniform acceleration.
Option B is correct.
Question 5
1 / -0

The table below shows the speed of a moving vehicle with respect to time.
Speed (m/s)
2
4
6
8
10
Time (s)
1
2
3
4
5
Find the acceleration of the vehicle.

A
$$2 m/s^2$$

B
$$4 m/s^2$$

C
$$6 m/s^2$$

D
$$8 m/s^2$$

Solution

From the given data, $$v=2t$$
Acceleration, $$a = \dfrac{v_2-v_1}{t_2-t_1}$$
$$= \dfrac{2(t_2-t_1)}{t_2-t_1}\ = 2\ m/s^2$$
Question 6
1 / -0

An object is sliding down on an inclined plane. The velocity changes at a constant rate from 10 cm/s to 15 cm/s in 2 seconds. What is its acceleration?

A
$$5 cm/s^2$$

B
$$7.5 cm/s^2$$

C
$$2.5 cm/s^2$$

D
$$12.5 cm/s^2$$

Solution

$$a=\dfrac { v-u }{ t } =\dfrac {15-10}{2} =2.5\ cm/{ s }^{ 2 }$$
Question 7
1 / -0

Two stones are thrown vertically upwards simultaneously with their initial velocities $$u_1$$ and $$u_2$$ respectively. The heights reached by them would be in the ratio of

A
$$u_1^2:u_2^2$$

B
$$u_1:u_2$$

C
$$u_2^2:u_2^2$$

D
$$u_2:u_1$$

Solution

$$v^2=u^2+2as$$
$$0=(u_1)^2+2(-g)s_1$$
Hence, $$s_1=u_1^2/2g$$
Similarly, $$s_2=u_2^2/2g$$

Therefore, $$\dfrac{s_1}{s_2}=\dfrac{u_1^2}{u_2^2}$$
Question 8
1 / -0

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of $$10 ms^{-2}$$, with what velocity will it strike the ground and time after which it will it strike the ground?

A
20 $$m/s$$, 2 s

B
24 $$m/s$$, $$\sqrt{2}$$ s

C
10 $$m/s$$, 4 s

D
None of the above

Solution

Given, initial velocity of ball, $$u=0$$
Final velocity of ball, $$v=?$$
Distance through which the balls falls, $$s=20 m$$
Acceleration $$a=10 ms^{-2}$$
Time of fall, $$t=?$$
We know
$$v^2-u^2=2as$$
or $$v^2-0=2\times 10\times 20=400$$ or $$v=20 ms^{-1}$$
Now using $$v=u+at$$ we have
$$20=0+10\times t$$ or $$t=2s$$
Question 9
1 / -0

An iron sphere of mass 10 kg is dropped from a height of 80 cm. If the downward acceleration of the sphere is $$10 ms^{-2}$$, calculate the momentum of the sphere when it just strikes the ground

A
100 kg m/s

B
40 kg m/s

C
60 kg m/s

D
75 kg m/s

Solution

Here, initial velocity of sphere, $$u=0$$
Distance travelled, $$s=80 cm=0.8 m$$
Acceleration of sphere, $$a=10 ms^{-2}$$
Final velocity of sphere when it just reaches the ground can be calculated using
$$v^2-u^2=2as$$
$$\therefore v^2-0=2\times 10\times 0.8=16$$
or $$v=\sqrt {16}=4 ms^{-1}$$.
Momentum of the sphere just before it touches the ground $$=mv=10\times 4=40 kg ms^{-1}$$
Question 10
1 / -0

Which of the following relations is correct?

A
Speed = $$\frac { 1 }{ Distance\times Time } $$

B
Speed = $$\frac { Time }{ Distance } $$

C
Speed = $$\frac { Distance }{ Time } $$

D
Speed = Distance $$\times $$ Time

Solution

Speed is defined as the distance travelled by an object per unit time.
$$Speed = \dfrac{Distance}{time}$$

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Motion Test - 45

Result

Motion Test - 45

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Rank

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SHARING IS CARING

Question 1

Speed

Distance

Acceleration

All

Solution

Question 2

$$96\ m$$

$$192\ m$$

$$12\ m$$

$$72\ m$$

Solution

Question 3

320 m

160 m

480 m

720 m

Solution

Question 4

uniform velocity.

uniform acceleration.

increasing acceleration.

decreasing acceleration.

Solution

Question 5

$$2 m/s^2$$

$$4 m/s^2$$

$$6 m/s^2$$

$$8 m/s^2$$

Solution

Question 6

$$5 cm/s^2$$

$$7.5 cm/s^2$$

$$2.5 cm/s^2$$

$$12.5 cm/s^2$$

Solution

Question 7

$$u_1^2:u_2^2$$

$$u_1:u_2$$

$$u_2^2:u_2^2$$

$$u_2:u_1$$

Solution

Question 8

20 $$m/s$$, 2 s

24 $$m/s$$, $$\sqrt{2}$$ s

10 $$m/s$$, 4 s

None of the above

Solution

Question 9

100 kg m/s

40 kg m/s

60 kg m/s

75 kg m/s

Solution

Question 10

Speed = $$\frac { 1 }{ Distance\times Time } $$

Speed = $$\frac { Time }{ Distance } $$

Speed = $$\frac { Distance }{ Time } $$

Speed = Distance $$\times $$ Time

Solution

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