Motion Test

Result

Motion Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

A football player initially at rest starts 20 meters from the goal line. He accelerates towards the goal line at 2m/s$$^2$$. Find out the distance between player and goal line after 4s?

A
8 m

B
28 m

C
32 m

D
4 m

E
52 m

Solution

Given : $$u = 0$$ m/s $$t =4$$ s $$a =2 m/s^2$$
Using $$S = ut + \dfrac{1}{2}at^2$$
Thus distance covered in 4 s $$S = 0 + \dfrac{1}{2} \times 2 \times 4^2 = 16$$ m
Thus distance between player and goal $$d =20 -16 = 4$$ m
Question 2
1 / -0

Match the entries in Column-I with those in Column-II.
Column - I Column - II
a Speed 1 cm
b Acceleration 2 s
c Displacement 3 $$m s^{-1}$$
d Time 4 $$km h^{-2}$$

A
a-2, b-1, c-4, d-3

B
a-3, b-4, c-1, d-2

C
a-4, b-2, c-3, d-1

D
a-3, b-2, c-2, d-1

Solution

In a given table, Column-I has given Physical quantity and Column-II has given units.
Displacement: Minimum distance between two points and its C.G.S unit is $$cm.$$
Speed: It is defined as the ratio of distance over time and its S.I unit is $$m/s.$$
Time: It is defined as the duration in which all things happen and it unit is seconds (s).
Acceleration: it is equal to velocity over time and its S.I unit is $$m/s^2$$, its unit can also be written as $$kmh^{-2}$$

So correct match is,
Column-I   Column-II
(a) Speed (3)  $$m/s$$
(b) Acceleration (4)   $$kmh^{-2}$$
(c) Displacement (1) $$cm.$$
(d) Time (2)  $$s$$
Question 3
1 / -0

Choose the correct option.
Odometer: Distance :: ______:______

A
Weighing scale : Weight

B
Length : Width

C
Inch : Foot

D
Mileage : Speed

Solution

An odometer is an instrument used in a vehicle to measure the distance covered by it.
Similarly, the weighing scale is an instrument used to measure the weight of an object.
Question 4
1 / -0

The accelerated motion of a body changes due to change in:

A
Speed

B
Direction of motion.

C
Velocity.

D
All of the above

Solution

The accelerated motion of a body changes due to a change in speed, direction of motion, velocity.
As acceleration posses magnitude and direction. Its magnitude changes by a change in speed, velocity, and direction can be changed by the direction of motion and velocity.
Question 5
1 / -0

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12 m $$s^{-1}$$, and decreases at the rate of 0.5 m $$s^{-2}$$. Find the time it will take to come to rest.

A
6 s

B
12 s

C
24 s

D
48 s

Solution

$$v=u+at$$
$$\therefore t=\dfrac{v-u}{a}$$
$$v=0, u=12ms^{-1}, a=-0.5ms^{-2}$$
$$t=\dfrac{0-12}{-0.05}=24s$$
Question 6
1 / -0

Uniform circular motion is called continuously accelerated motion mainly because its :

A
direction of motion changes

B
speed remains the same

C
velocity remains the same

D
direction of motion does not change

Solution

An accelerating body is an object that is changing its velocity. And since velocity is a vector that has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity.
Hence a uniform circular motion is a accelerated motion because direction of motion keeps on changing
Question 7
1 / -0

A truck and a car are moving with equal velocity. On applying brakes, if both decelerate at the same rate,what happens?

A
The truck will cover less distance before stopping.

B
The car will cover less distance before stopping.

C
Both will cover equal distance.

D
None of the above

Solution

Consider, the car has a mass ‘m’ and truck has a mass ‘M’.

where ,$$ M>>m$$

Momentum = (mass of the body) x (velocity of the same body)

As, the velocities of the car and truck are the same….its the mass which decides the distance which will be covered after applying brakes by each of them.

As the mass of truck is much higher than that of a car, the truck has a larger momentum value.
HENCE, THE TRUCK TRAVELS A LARGER DISTANCE AFTER APPLYING BRAKES.
Question 8
1 / -0

A car starts from rest and accelerates uniformly over a time of $$5.21$$ seconds for a distance of $$110$$ m. Determine the acceleration of the car.

A
$$8.10 m/s/s$$

B
$$8.1 m/s/s$$

C
$$8.0 m/s/s$$

D
$$7.10 m/s/s$$

Solution

Time taken $$t=5.21 s$$
Initial velocity $$u=0$$ and distance traveled $$s=110 m$$
Using, $$s=ut+at^2/2$$
where $$a$$ is the acceleration.
$$ \Rightarrow 110=0(5.21)+a(5.21)^2/2$$
We get $$a=8.10 m/s^2$$
Question 9
1 / -0

In a uniform linear motion which of the following quantities remains zero?

A
Speed

B
Velocity

C
Acceleration

D
None of these

Solution

In uniform linear motion, velocity is constant which means acceleration is zero.
$$\text{acceleration} = \dfrac{ \text{ change in velocity } }{ \text{ time } }$$
Question 10
1 / -0

Speed can be defined as :

A
$$\dfrac{Distance }{time}$$

B
$$\dfrac{Distance }{{Time}^2}$$

C
$$\dfrac{Time}{Distance}$$

D
$$None\ of\ the\ above$$

Solution

Speed of the body is defined as the distance travelled by the body in unit time.
Mathematically, $$Speed = \dfrac{Distance}{time}$$

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Re-Attempt Weekly Quiz Competition

	Column - I		Column - II
a	Speed	1	cm
b	Acceleration	2	s
c	Displacement	3	$$m s^{-1}$$
d	Time	4	$$km h^{-2}$$

Column-I	Column-II
(a) Speed	(3) $$m/s$$
(b) Acceleration	(4) $$kmh^{-2}$$
(c) Displacement	(1) $$cm.$$
(d) Time	(2) $$s$$

Motion Test - 28

Result

Motion Test - 28

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Rank

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Question 1

8 m

28 m

32 m

4 m

52 m

Solution

Question 2

a-2, b-1, c-4, d-3

a-3, b-4, c-1, d-2

a-4, b-2, c-3, d-1

a-3, b-2, c-2, d-1

Solution

Question 3

Weighing scale : Weight

Length : Width

Inch : Foot

Mileage : Speed

Solution

Question 4

Speed

Direction of motion.

Velocity.

All of the above

Solution

Question 5

6 s

12 s

24 s

48 s

Solution

Question 6

direction of motion changes

speed remains the same

velocity remains the same

direction of motion does not change

Solution

Question 7

The truck will cover less distance before stopping.

The car will cover less distance before stopping.

Both will cover equal distance.

None of the above

Solution

Question 8

$$8.10 m/s/s$$

$$8.1 m/s/s$$

$$8.0 m/s/s$$

$$7.10 m/s/s$$

Solution

Question 9

Speed

Velocity

Acceleration

None of these

Solution

Question 10

$$\dfrac{Distance }{time}$$

$$\dfrac{Distance }{{Time}^2}$$

$$\dfrac{Time}{Distance}$$

$$None\ of\ the\ above$$

Solution

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