Motion in A Straight Line Test

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Motion in A Straight Line Test - 16

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Weekly Quiz Competition

Question 1
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Identify the wrong statement about acceleration.

A
Acceleration is rate of change of velocity per second

B
Acceleration is rate of change of speed per second in a certain direction

C
Acceleration is a vector quantity

D
Unit of acceleration is $$m{s}^{-1}$$

Solution

Acceleration= rate of change of velocity = change in velocity velocity / time =$$\dfrac{ms^{-1}}{s}= ms^{-2}$$

Unit of acceleration is $$m{s}^{-2}$$.

Hence, option D is correct
Question 2
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Which of the following terms does not go well with the motion of a bus on a crowded road.

A
Uniform velociity

B
Variable velocity

C
Variable acceleration

D
Variable speed

Solution

On a crowded road, depending on traffic conditions, the bus driver has to change its speed and direction of motion many times. The driver will have to frequently apply brakes as well. He might also have to accelerate the bus with different rates. So, the velocity changes both by magnitude and direction. Thus, uniform velocity is not attained in this case.
Question 3
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A body executing non-uniform motion:

A
Undergoes acceleration

B
Never undergoes acceleration

C
May or may not undergo acceleration

D
Never undergoes constant acceleration

Solution

If a body undergoes non-uniform motion, its speed changes and a body with changing speed undergoes acceleration.
This acceleration may or may not be constant.
Acceleration of the particle is given as-
$$\text{acceleration} = \dfrac{ \text{ change in velocity } }{ \text{ time } }$$
Question 4
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A body is moving vertically upwards. Its velocity changes at a constant rate from $$50 m{s}^{-1}$$ to $$20 m{s}^{-1}$$ in $$3 s$$. What is its acceleration?

A
$$10 m{s}^{-2}$$

B
$$- 10 m{s}^{-2}$$

C
$$1 m{s}^{-2}$$

D
$$- 1 m{s}^{-2}$$

Solution

Initial velocity $$= 50 m{s}^{-1}$$
Final velocity $$= 20 m{s}^{-1}$$
Time $$= 3 s$$
Acceleration = $$\dfrac{20-50}{3}$$ m$${s}^{-2}$$ $$= - 10 m{s}^{-2}$$
Question 5
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A scooter increases its speed from 36 km/h to 72 km/h in 10 s. What is its acceleration?

A
$${7.2 m/s^2}$$

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

C
$${3.6 m/s^2}$$

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

Solution

in the given question we have
u=36 km/hr =$$36 \times \frac{5}{18}m/s=10 m/s$$
v= 72 km/hr=$$ 72 \times \frac{5}{18}m/s=20 m/s$$
t=10 s
a=?
we know the formula
v= u +at
20 = 10+ a 10
20-10 =10 a
10 = 10 a
a = 1 $$m/s^{2}$$
Question 6
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A car accelerates at a rate of $$5 m{s}^{-2}$$. Find the increase in its velocity in $$2 s$$.

A
$$10 m{s}^{-1}$$

B
$$2.5 m{s}^{-1}$$

C
$$5 m{s}^{-1}$$

D
$$2 m{s}^{-1}$$

Solution

Acceleration $$= 5 m{s}^{-2}$$
Time $$= 2 s$$
Increase in velocity $$=$$ acceleration $$\times$$ time $$=(5 \times 2)m{s}^{-1}=10m{s}^{-1}$$
Question 7
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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 8
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Directions For Questions

The following options have dot representations. Each dot represents a time interval of one second and the motion of the ball is not necessarily horizontal.

...view full instructions

In which of the following option could represent the ball increasing its speed?

A

B

C

D

E

Solution

Option B; Since the distance between the constant time intervals is increasing, the ball is covering more distance per second each second, which is an increase in speed (acceleration).
Option A: Represents constant speed as same distance covered in each second.
Option C: Represents retardation.
Option D: First accelerate then retardation
Option E: First accelerate then retardation
Question 9
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A car moving is with a particular velocity, which has been measured each second and the data is recorded as per the table below.
Velocity Time
$$0$$ $$0$$
$$2 \ {m}/{s}$$ $$1 \ s$$
$$4 \ {m}/{s}$$ $$2 \ s$$
$$6 \ {m}/{s}$$ $$3 \ s$$
$$8 \ {m}/{s}$$ $$4 \ s$$
Calculate the acceleration of the car ?

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

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

C
$$4 \ {m}/{{s}^{2}}$$

D
$$6 \ {m}/{{s}^{2}}$$

E
$$8 \ {m}/{{s}^{2}}$$

Solution

Acceleration is defined as the ratio of change in velocity to the change in time interval.

Acceleration $$a=\dfrac{\Delta v}{\Delta t}$$

The given data shows that the velocity of car changes by $$2 \ m/s$$ in $$1$$ $$s$$ each.
Thus acceleration of the car $$a = \dfrac{\Delta v}{\Delta t} = \dfrac{2}{1} =2 \ m/s^2$$
Question 10
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Which of the following best define the acceleration of a particle:

A
the rate of change of velocity.

B
only experienced during a change of direction.

C
only experienced during a change of speed.

D
calculated by multiplying speed by velocity.

E
always constant.

Solution

Acceleration is defined as the rate of change of velocity.

acceleration $$a = \dfrac{change \ in \ velocity}{time \ interval}$$

Hence, any change in the speed or direction would cause a change in velocity, and hence will produce acceleration.

So, the correct option is A.