Motion Test

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Motion Test - 27

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Question 1
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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 2
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A car moving with speed $$2 \ m/s$$ crosses an electric post and starts accelerating at a constant rate of $$2 \ m/s^2$$. What will be the speed of the car after $$3 \ s$$?

A
$$8 \ m/s$$

B
$$9 \ m/s$$

C
$$12 \ m/s$$

D
$$15 \ m/s$$

E
$$18 \ m/s$$

Solution

Given : initial speed $$u =2 \ m/s $$
acceleration $$a = 2\ m/s^2$$
Calculate speed after time $$t = 3 \ s$$

Using the first equation of motion $$v = u+at$$
$$\therefore$$ $$v =2 + 2\times 3 =8 \ m/s$$
Question 3
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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 4
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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 5
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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
6.10 m $$s^{-2}$$

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

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

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

Solution

using second eq of motion $$s=ut+1/2at^2$$
as u=0, $$a=\dfrac{2s}{t^2}=\dfrac{2*110}{5.21^2}=8.10m/s^2$$
Question 6
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The position-time graphs above represent the motions of cars 1 to 5. How do they rank according to their speeds (greatest first)?

A
$$1,\ 2,\ 3, 4,\ 5$$

B
$$1,\ 2,\ 3\ and\ 5\ tie,\ 4$$

C
$$1,\ 2,\ 4,\ 3\ and\ 5\ tie$$

D
$$3\ and\ 5\ tie,\ 4,\ 2,\ 1$$

E
$$5,\ 4,\ 3,\ 2,\ 1$$

Solution

We know,
$$speed=\dfrac{distance}{time}$$

If time is constant, then speed will depend on distance travelled.

Consider a time interval from $$t_1$$ to $$t_2$$.
The distance travelled by $$car\ 1$$ is greatest, followed by $$car\ 2$$ and $$car\ 4$$.
The distance travelled by $$car\ 3$$ and $$car\ 4$$ is $$0$$.
$$\therefore$$ The descending order of their speed is $$1 \gt 2\gt 4\gt 3=5$$
Question 7
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The graph above shows the position versus time for three different cars 1,2, and 3. Rank these cars according to the magnitudes of their velocities, greatest first.

A
1, 2, 3

B
3, 2, 1

C
2, 1, 3

D
2, 3, 1

E
all tie

Solution

Slope of position- time graph gives the velocity of the car.
From figure, slopes of all the three curves are zero.(Slope of horizontal line is zero)
Thus velocity of each car is zero.
Hence option E is correct.
Question 8
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A ball of mass $$m$$ is thrown straight upward from the top of a multi-storey building with an initial velocity of $$15 \ m/s$$.
Find out the time taken by the ball to reach its maximum height?

A
$$25 \ s$$

B
$$15 \ s$$

C
$$10 \ s$$

D
$$5 \ s$$

E
$$1.5 \ s$$

Solution

Given : initial velocity $$u = 15 \ m/s$$
acceleration $$a = -g = -10 m/s^2$$ (taking upward direction to be positive)
Final velocity of the ball at the highest point $$v = 0 \ m/s$$

Using the first equation of motion $$v = u + at$$
$$\therefore $$ $$0 = 15 + (-10) t$$
$$\implies t = 1.5 \ s$$
Question 9
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A car accelerates from rest at a rate of $$4 m/s^2$$ for $$10 \ s$$. Calculate the speed of the car at the end of $$10 \ s$$?
[Unless otherwise stated, use $$g=10m/s^2$$ and neglect air resistance ]

A
14 m/s

B
6 m/s

C
2.5 m/s

D
0.4 m/s

E
40 m/s

Solution

Given :
Initial velocity $$u = 0$$ m/s
Acceleration $$a = 4$$ $$m/s^2$$
Time $$t = 10$$ s
Using first equation of motion
$$v = u+at$$
$$\therefore$$ $$v = 0 + 4 (10) = 40 \ m/s$$
Question 10
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The above graph shows position as a function of time for an object moving along a straight line. During which time(s) is the object at rest?
I. $$0.5$$ seconds
II. From $$1$$ to $$2$$ seconds
III. $$2.5$$ seconds

A
I only

B
II only

C
III only

D
I and III only

Solution

Slope of the curve under position-time graph gives the instantaneous velocity of the object.
Slope of the curve is zero only in the time interval $$1<t<2$$ s.
Thus the object is at rest (or velocity is zero) only from $$1$$ to $$2$$ s.
Hence option B is correct.