Motion Test

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Motion Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

A body moving with a constant acceleration travels the distances 3 m and 8 m respectively in 1 s and 2 s. Calculate the initial velocity.

A
$$4\, m\, s^{-1}$$

B
$$0 \, m\, s^{-1}$$

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

D
$$1\, m\, s^{-1}$$

Solution

$$a=ut+\dfrac{1}{2}at^2$$

Putting the values,
$$3=u\times 1+\dfrac{1}{2}a\times 1^2$$
$$a=6-2u..............(i)$$

Similarly, $$8=u\times 2+\dfrac{1}{2}a\times 2^2$$
$$a=4-u.................(ii)$$

Equating (i) and (ii),
$$6-2u=4-u$$
$$u = 2\ m/s$$
Question 2
1 / -0

Figure shows the displacement-time graph for the motion of two boys A and B along a straight road in the same direction. Which of the two has greater velocity?

A
B

B
A

C
Both have same velocity

D
None of the above

Solution

From a graph of displacement vs time, the slope gives the rate of change of displacement or velocity

From the given graph the velocity of $$A$$ is $$\dfrac{20}{4} = 5km/hr$$

$$\&$$

Velocity of $$B$$ is $$\dfrac{20}{2} = 10km/hr$$
Question 3
1 / -0

A car travels with a uniform velocity of $$25\, m\, s^{-1}$$ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find the acceleration.

A
$$5\, m\, s^{-2}$$

B
$$-2.5\, m\, s^{-2}$$

C
$$25\, m\, s^{-2}$$

D
$$10\, m\, s^{-2}$$

Solution

For retardation process, $$u = 25 m/s, v=0, t=10 s$$
$$v=u+at$$
$$0=25+a\times 10$$
$$a = -2.5 m/s^2$$
Question 4
1 / -0

A train starts from rest and accelerates uniformly at a rate of $$2\ ms^{-2}$$ for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s. Find the maximum velocity reached.

A
$$2\, m\, s^{-1}$$

B
$$10 \, m\, s^{-1}$$

C
$$1\, m\, s^{-1}$$

D
$$20\, m\, s^{-1}$$

Solution

Maximum velocity is attained at the end of uniformly accelerated motion.
From, $$v=u+at$$
Maximum velocity, $$v=0+2\times 10 = 20\ m/s$$
Question 5
1 / -0

A car travels a distance 100 m with a constant acceleration and average velocity of $$20\, ms^{-1}$$. The final velocity acquired by the car is $$25\, ms^{-1}$$. Find the acceleration of the car.

A
$$1\, m\, s^{-2}$$

B
$$2\, m\, s^{-2}$$

C
$$0\, m\, s^{-2}$$

D
$$4\, m\, s^{-2}$$

Solution

$$\text{Average velocity} = \dfrac{v+u}{2}$$
$$20 = \dfrac{25+u}{2}$$
$$u = 15\ m/s$$

$$v=u+at$$
$$25=15+a\times 5$$
We get, $$a = 2\ m/s^2$$
Question 6
1 / -0

When brakes are applied to a bus, the retardation produced is $$25\ \ cm\ s^{-2}$$ and the bus takes 20 s to stop. Calculate the initial velocity of the bus.

A
$$500 \, m\, s^{-1}$$

B
$$5\, cm\, s^{-1}$$

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

D
$$12.5 \, m\, s^{-1}$$

Solution

Retardation is $$25 cm/s^2$$, therefore, Acceleration, $$a=-0.25 m/s^2$$
$$t=20 s, v=0$$
Putting the values in $$v=u+at$$
$$0 = u+(-0.25)(20)$$
$$u = 5 m/s$$
Question 7
1 / -0

A ball is thrown vertically upwards with a velocity of $${49 ms^{-1}}$$. Calculate the maximum height to which it rises.

A
122.5 m

B
100 m

C
22.5 m

D
78 m

Solution

$$v^2=u^2+2as$$
$$v=0 m/s$$
$$u=49 m/s$$
$$a=-9.8 m/s^2$$ (for upward direction gravity is -ve)
$$0^2=49^2+2(-9.8)s$$
$$s=\dfrac{49\times49}{2\times9.8}= 122.5m$$
Question 8
1 / -0

Figure below shows the distance-time graph of three objects $$A, B$$ and $$C$$. Study the graph and answer the following questions: which of the three is travelling the fastest?

A
Car $$C$$

B
Car $$A$$

C
Car $$B$$

D
Can not be determined

Solution

The slope distance-time graph determines the velocity of the object.
Greater the slope, greater will be the velocity. Hence, car B is travelling the fastest as its slope is largest among the three.
Question 9
1 / -0

When a ball is thrown up vertically with velocity $$ {v}_{o} $$ it reaches a maximum height of $$h$$. If one wishes to triple the maximum height then the ball should be thrown with velocity.

A
$$\displaystyle \sqrt{3}v_{0}$$

B
$$\displaystyle 3v_{0}$$

C
$$\displaystyle 9v_{0}$$

D
$$\displaystyle \frac{3}{2}v_{0}$$

Solution

Maximum height means final velocity is zero.
Thus $$H=\dfrac{0-{u}^{2}}{-2g}=\dfrac{{u}^{2}}{2g}$$
To triple maximum height, i.e. 3H, we have, $$3H=\dfrac{{u'}^{2}}{2g}$$ or $${u'}^{2}=3{u}^{2}$$ or, $$u'=\sqrt{3} u$$
Question 10
1 / -0

A stone is thrown in vertically upward direction with a velocity of $$5\ ms^{-1}$$. If the acceleration of the stone during its motion is $$10\ ms^{-2}$$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

A
Height$$=1.25 \; m$$ and time $$= 0.5 \;s$$

B
Height$$=2.5 \; m$$ and time $$ =0.5 \;s$$

C
Height$$=2.5 \; m$$ and time $$ =0.05 \;s$$

D
Height$$=1.25 \; m$$ and time $$ =0.05 \;s$$

Solution

Given, initial velocity, $$u=5ms^{−1}$$
Final velocity, $$v=0$$
Since, u is upward & a is downward, it is a retarded motion.
∴$$a=−10ms^{−2}$$
Height attained by stone, $$s=?$$
Time take to attain height, $$ t=?$$
(i) Using the relation, $$v^2−u^2=2as$$, we have
$$s=v^2−u^2/2a$$
$$=(0)^2−(5)^2/2×(−10)=1.25m$$
(ii) Using the relation, $$v=u+at$$
$$0=5+(−10)t$$ or
$$t=5/10=0.5s$$

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Motion Test - 44

Result

Motion Test - 44

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SHARING IS CARING

Question 1

$$4\, m\, s^{-1}$$

$$0 \, m\, s^{-1}$$

$$2\, m\, s^{-1}$$

$$1\, m\, s^{-1}$$

Solution

Question 2

B

A

Both have same velocity

None of the above

Solution

Question 3

$$5\, m\, s^{-2}$$

$$-2.5\, m\, s^{-2}$$

$$25\, m\, s^{-2}$$

$$10\, m\, s^{-2}$$

Solution

Question 4

$$2\, m\, s^{-1}$$

$$10 \, m\, s^{-1}$$

$$1\, m\, s^{-1}$$

$$20\, m\, s^{-1}$$

Solution

Question 5

$$1\, m\, s^{-2}$$

$$2\, m\, s^{-2}$$

$$0\, m\, s^{-2}$$

$$4\, m\, s^{-2}$$

Solution

Question 6

$$500 \, m\, s^{-1}$$

$$5\, cm\, s^{-1}$$

$$5\, m\, s^{-1}$$

$$12.5 \, m\, s^{-1}$$

Solution

Question 7

122.5 m

100 m

22.5 m

78 m

Solution

Question 8

Car $$C$$

Car $$A$$

Car $$B$$

Can not be determined

Solution

Question 9

$$\displaystyle \sqrt{3}v_{0}$$

$$\displaystyle 3v_{0}$$

$$\displaystyle 9v_{0}$$

$$\displaystyle \frac{3}{2}v_{0}$$

Solution

Question 10

Height$$=1.25 \; m$$ and time $$= 0.5 \;s$$

Height$$=2.5 \; m$$ and time $$ =0.5 \;s$$

Height$$=2.5 \; m$$ and time $$ =0.05 \;s$$

Height$$=1.25 \; m$$ and time $$ =0.05 \;s$$

Solution

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