Motion Test

Result

Motion Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following defines displacement correctly?

A
Change in position in a certain direction

B
Change in position

C
Length of path travelled

D
None of these

Solution

Displacement is the linear distance between the initial and final positions. So, it is change in position in a certain direction.
Question 2
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Pick out the wrong statement about retardation.

A
Retardation is negative acceleration

B
Retardation is rate of decrease of velocity with time

C
SI unit of retardation is $$m{s}^{-2}$$

D
Retardation is a scalar quantity

Solution

Retardation, or negative acceleration, is nothing but an acceleration which decreases the speed. Its SI unit is same as that of acceleration. Just like acceleration, it has both magnitude and direction, i.e. it is a vector quantity. So option D is the wrong statement about retardation.
Question 3
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The units of speed and average speed are

A
same across all systems of units

B
same only for SI

C
different across all systems of units

D
different only for SI

Solution

Units for speed and average speed are the same for all systems of units. For example, SI unit is $$m{s}^{-1}$$, practical unit is $$kmph$$ etc.
Formula:
$$speed = \dfrac{distance}{time}$$
$$velocity = \dfrac{displacement}{time}$$
Question 4
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When particles move in a circle at a constant speed then the motion is said to be

A
Uniform circular motion

B
Non uniform motion

C
Projectile motion

D
All

Solution

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
Question 5
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Identify in which of the following graphs velocity of the moving object remains constant?

A

B

C

D

E

Solution

Except graphs A and B, others are Velocity time graphs, As we can see that velocity of particles is changing with time.
Now, Slope of the curve under the position-times graph gives the instantaneous velocity of the object.
As slope of the graph shown in option A is constant, thus velocity of the object is constant in graph of option A.
Question 6
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A jeep is accelerating uniformly in a straight line with $$5 m/s^2$$. Calculate the time required to cover a distance of 200 m if jeep starts from rest :

A
9.0 s

B
10.5 s

C
12.0 s

D
15.5 s

E
20.0 s

Solution

Given : $$S = 200m$$, $$a = 5 m/s^2$$, $$u = 0$$
Using $$2^{nd}$$ equation of motion, $$s = ut + \dfrac{1}{2}at^2$$
$$\therefore$$ $$200 = 0 + \dfrac{1}{2}\times 5t^2$$ $$\Rightarrow t^2 = \cfrac{200\times 2}{5}$$
$$\implies t \approx 9 $$ s
Question 7
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When a body moves in a circular motion :

A
its direction constantly changes

B
its direction remains constant

C
its velocity vector is always perpendicular to the direction of motion

D
all of the above

Solution

As an object moves in a circle, it is constantly changing its direction. In all instances, the object is moving tangent to the circle, and the direction of the velocity vector is the same as the direction of the object's motion.
Question 8
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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.
Question 9
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A car is moving with speed $$2 \ m/s$$ crosses a electric post and starts accelerating at a constant rate of $$2 \ m/s^2$$. How far past the electric post will the car be after $$3 \ s$$?

A
$$8 m$$

B
$$9 m$$

C
$$12 m$$

D
$$15 m$$

E
$$18 m$$

Solution

Given :
Initial velocity $$u =2 $$ m/s
Acceleration $$a = 2$$ $$m/s^2$$
Time $$t = 3$$ s
Using second equation of motion:
$$S = ut + \dfrac{1}{2}at^2$$
Thus the distance of electric post from the car after $$3 \ s$$,
$$S= 2 \times 3 + \dfrac{1}{2}\times 2 \times 3^2 =15m$$
Question 10
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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$${s}^{-2}$$ $$= - 10 m{s}^{-2}$$