Motion Test

Result

Motion Test - 67

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Question 1
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A particle starts from rest with acceleration $$ 2\ m/s^2 $$, The distance moved by the particle in 5 sec is:

A
22 m

B
28 m

C
25 m

D
13 m

Solution

Given:
Acceleration $$a=2\,m/{{s}^{2}}$$
Time $$t=5\,s$$
Initial velocity $$u=0$$
We know from the equation of motion,
$$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$
$$\Rightarrow s=0+\dfrac{1}{2}\times 2\times 5\times 5 $$
$$ s=25\,m $$
Hence, the distance moved by the particle is $$25\ m$$
Question 2
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A train starting from a railway station and moving with uniform acceleration attains a speed $$40km\ h^{-1}$$ in $$10$$ minutes. Find its acceleration.

A
$$0.0185 m/s^2$$

B
$$0.0195 m/s^2$$

C
$$0.0165 m/s^2$$

D
$$0.0135 m/s^2$$

Solution

Here we have, initial velocity,$$v=40$$km/h$$=11.11$$m/s
Time $$t=10$$minutes$$=60\times 10=600$$secs
We know that $$v=u+at$$
$$\Rightarrow 40\ km{hr}^{-1}=0+a\times 10$$m
$$\Rightarrow 11.11 \ m{s}^{-1}=a\times 600$$secs
$$\Rightarrow \dfrac{11.11}{600}=0.0185\ m{s}^{-2}$$
Question 3
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A body starting from rest moving with uniform acceleration has a displacement of $$16 \ m$$ in first $$4 \ s$$ and $$9 \ m$$ in first $$3 \ s$$. The acceleration of the body is

A
$$1 \ ms^{-1}$$

B
$$2 \ ms^{-1}$$

C
$$3 \ ms^{-1}$$

D
$$4 \ ms^{-1}$$
Question 4
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A stone falls from a balloon that is descending at a uniform rate of $$12\, ms^{-1}$$. The displacement of the stone from the point of release after 10 sec is :-

A
$$490\ m$$

B
$$610\ m$$

C
$$510\ m$$

D
$$725\ m$$

Solution

Given,
$$u=12m/s$$, the velocity of the stone is same as ballon
$$t=10\ sec$$
$$g=9.8\ ms^{-2}$$
From 2nd equation of motion,
$$s=ut+\dfrac{1}{2}gt^2$$
$$\Rightarrow s=12\times 10+\dfrac{1}{2}\times 9.8\times 10\times 10$$
$$\Rightarrow s=120+490$$
$$\Rightarrow s=610m$$
The correct option is B.
Question 5
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$$2nd$$ equation of motion $$(s=ut + \dfrac{1}{2}at^2)$$ can be derived graphically form $$v-t$$ graph by finding

A
Slope

B
Area

C
Either(A) or (B)

D
Insufficient data

Solution

$$2nd$$ equation of the motion
$$S=ut+\cfrac{1}{2}at^2$$
From $$v-t$$ graph-
The area under the $$v-t$$ graph is the displacement of the particle.
The slope represents the acceleration of the particle.
Question 6
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A particle is released from rest from a tower of height 3h.The ratio of times to fall equal height h,i.e., t$$_{1}$$: t$$_{2}$$ :t$$_{3}$$ is

A
$$\sqrt3 :\sqrt2$$: 1

B
3 : 2 : 1

C
9 :4 : 1

D
1 : ($$\sqrt2$$ - 1) :($$\sqrt3 - \sqrt2 -1$$)

Solution

Time in falling first height $$h$$ will be $$t_1=T_1=\sqrt[2]{\dfrac{2h}{g}}$$

Time in falling first height $$2h$$ will be $$T_2=\sqrt[2]{\dfrac{2(2h)}{g}}$$
so the time in falling second $$h$$ will be $$t_2=T_2-T_1=\sqrt[2]{\dfrac{h}{g}}(\sqrt[2]{4}-\sqrt[2]{2})$$

Time in falling in height $$3h$$ will be $$T_3=\sqrt[2]{\dfrac{2(3h)}{g}}$$

so the time in falling third $$h$$ will be $$t_3=T_3-(T_1+T_2)=\sqrt[2]{\dfrac{h}{g}}(\sqrt[2]{6}-\sqrt[2]{4}-\sqrt[2]{2})$$
the ratio will be $$t_1:t_2:t_3=\sqrt[2]{2}:\sqrt[2]{2}(\sqrt[2]{2}-1):\sqrt[2]{2}(\sqrt[2]{3}-\sqrt[2]{2}-1)$$
$$\sqrt[2]{2}$$ is common in all so can be cancelled so the answer is option D.
Question 7
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Directions For Questions

The graph given alongside shows the positions of a body at different times.

...view full instructions

Calculate the speed of the body as if moves from: B to C.

A
$$3\ cm/s$$

B
$$1.5\ cm/s$$

C
$$Zero$$

D
$$-3\ cm/s$$

Solution

$$Speed = \dfrac{change\ in\ distance}{time}$$
At point B, the distance at 5 sec = 3 cm and at point C, the distance at 7 sec = 3cm
Change in distance = (3-3) cm = 0
Hence speed is zero
Question 8
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A helicopter is flying at a height of $$500m$$. If all of sudden its engines stop working, the helicopter will fall on the earth in __ seconds?

A
$$10$$

B
$$12$$

C
$$15$$

D
$$20$$

Solution

Initial velocity, $$u=0$$; only downward velocity considered
acceleration $$=g=10\ ms^{-2}$$
Displacement, $$s=500$$m
From Newton's second equation of motion,
$$s=ut+\dfrac{1}{2}at^2$$
$$\Rightarrow 500=\dfrac{1}{2}\times 10 \times t^2$$
$$\Rightarrow t^2=100$$
$$\Rightarrow t=10s$$
Question 9
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The displacement - time graph of a particle moving in a straight line is shown in the figure. Then

A
Velocity at point $$D$$ is greater than that at point $$B$$

B
Speed of the particle is increasing continuously

C
Acceleration is uniform

D
The particle is first uniformly accelerated and then retarded

Solution

The slope of the displacement - time graph at a point gives us the velocity at that point. From the graph, it is clear that the slope at point $$D$$ is greater than that at point $$B$$. Hence the velocity at point $$D$$ is greater than that at point $$B$$. So, option (A) is correct.
Question 10
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A ball is dropped from a balloon moving upwards with velocity $$10\ m/s$$ at a height of $$20\ m$$ from the ground. The distance traveled by the ball before reaching the ground is: (Take $$g=10\ m/s^2)$$

A
$$40\ m$$

B
$$10\ m$$

C
$$30\ m$$

D
$$20\ m$$

Solution

Initial velocity of the ball, $$u = 10\ m/s $$
Height of the balloon $$h = 20\ m $$
From A to B
Final velocity, $$v = 0\ m/s $$
Let $$x$$ be the distance travelled by the ball till i's velocity $$v = 0$$

Using third equation of motion,
$$2ax = v^{2} - u^{2} $$
$$-2gx = v^{2} - u^{2} $$ ($$a = -g$$ downward)
$$-2 \times 10 \times x = 0 - 10^{2} $$

$$x = \dfrac{100}{20} = 5\ m $$

$$\therefore $$ Total distance $$D$$ travelled by the balloon
$$D = h + x + x = h + 2x $$
$$D = 20 + (2 \times 5)$$
$$D = 30\ m$$