Motion Test

Result

Motion Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

The displacement-time graph of a moving particle is as shown in the figure. The instantaneous velocity of the particle is negative at the point

A
C

B
D

C
E

D
F

Solution

$$\textbf{Explanation}$$
The velocity is
$$v = dx / dt$$

The instantaneous velocity is given by the slope of the displacement time graph.

Since slope is negative at point $$E$$ , instantaneous velocity is negative at $$E$$

Hence, Option C is correct
Question 2
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Figure shows the displacement (x) -time (t) graph of the particle moving on the x-axis.

A
The particle is at rest

B
The particle is continuously going along x-direction

C
The velocity of the particle increases upto time $$t_0$$ and then becomes constant.

D
The particle moves at a constant velocity up to a time $$t_0$$ and then stops.

Solution

The slope of the Displacement- time graph provides the value of the velocity of a moving object. From the graph, it is visible that, till time $$t_{0}$$ object is having a constant slope and hence its velocity will be constant. Whereas after $$t_{0}$$ slope is zero and hence the object is not moving at all. Thus option D is most suitable and hence is the correct option.

The correct option is D.
Question 3
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Which of the following statements is incorrect?

A
Path length is a scalar quantity whereas displacement is a vector quantity.

B
The magnitude of displacement is always equal to the path length traversed by an object over a given time interval.

C
The displacement depends only on the end points whereas path length depends on the actual path followed.

D
The path length is always positive whereas displacement can be positive, negative and zero.

Solution

Displacement is the smallest distance between initial and final is a vector quantity it may be $$+ve$$, $$-ve$$, zero according to direction. (in figure displacement is $$d$$) But path length is a scalar quantity and this is equal to total distance cover, it is always positive in figure path length $$=x_{1}+x_{2}+x_{3}$$
Question 4
1 / -0

Which of the following graphs represents the position-time graph of a particle moving with negative velocity?

A

B

C

D

Solution

$$velocity = \frac{\left | Distance x \right |}{time t}=\frac{\mathrm{d} x}{\mathrm{d} t}$$

$$\int \mathrm{d} x=-v\mathrm{d} t$$
$$x=-vt+c$$
Hence graph shown in B is the correct option.
Question 5
1 / -0

A particle is released from rest from a tower of height 3h. The ratio of the intervals of time to cover three equal heights h is:

A
$$t_1 : t_2 : t_3$$ = 3 : 2 : 1

B
$$t_1 : t_2 : t_3$$ = $$ 1 : (\sqrt2 - 1) : (\sqrt3 -2)$$

C
$$t_1 : t_2 : t_3$$ = $$\sqrt3 : sqrt2 : 1$$

D
$$t_1 : t_2 : t_3$$ = $$ 1 : (\sqrt2 - 1) : (\sqrt3 - \sqrt2)$$

Solution

For $$A$$ to $$B-$$
$$h=(a)(t_1)+\cfrac{1}{2}gt^2_1\quad\quad\quad [2^{nd} \text{ equation of motion.}]$$
$$t_1=\sqrt{\cfrac{2h}{g}}$$..................(1)
From $$A$$ to $$C-$$
$$2h=(0)(t_2)+\cfrac{1}{2}gt^2_2$$
$$t_2=\sqrt{\cfrac{4h}{g}}$$ ..................(2)
Similarly, $$t_3=\sqrt{\cfrac{6h}{g}}$$ ........(3
)
Times taken from $$A$$ to $$B=t_1=\sqrt{\cfrac{2h}{g}}$$

Times taken from $$B$$ to $$C=t_2-t_1=\sqrt{\cfrac{4h}{g}}-\sqrt{\cfrac{2h}{g}}= \ \, \sqrt{\cfrac{2h}{g}}(\sqrt{2}-1)$$

Time taken from $$C$$ to $$D=t_3-t_2=\sqrt{\cfrac{6h}{g}}-\sqrt{\cfrac{4h}{g}}=\sqrt{\cfrac{2h}{g}}(\sqrt{3}-\sqrt{2})$$

$$\therefore$$ Ration $$=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$$

Option- D
Question 6
1 / -0

A football is kicked into the air vertically upwards with velocity $$u$$. The velocity of the ball at the highest point is:

A
$$u$$

B
$$2u$$

C
$$zero$$

D
$$4u$$

Solution

When football is kicked upwards in air
net acceleration of the football is given by
$$F=mg$$
$$a=\dfrac{F}{m}$$

$$a=\dfrac{mg}{m}=g=9.8m/s^{2}$$

now as it moves up its velocity will decrease due to gravity which acts opposite to the motion.now as the ball moves high velocity continuously decreases and finally velocity will become $$ZERO$$ as it will reach the highest point.
so the velocity of football at the top position will be $$ZERO$$
Question 7
1 / -0

Which parameters shown below are common between uniform circular motion and uniform linear motion

A
Velocity

B
Speed

C
Displacement

D
Acceleration

Solution

In uniform circular motion, speed remains constant and even in a uniform linear motion, speed remains constant. The direction of velocity in a uniform circular motion keeps changing as the particle moves along the curve

Hence option b is the correct answer
Question 8
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A body dropped from a height $$h$$ with initial velocity zero, strikes the ground with a velocity $$3m/s$$. Another body of same mass dropped from the same height $$h$$ with an initial velocity of $$4m/s$$. The final velocity of second mass, with which it strikes the ground is

A
$$5m/s$$

B
$$12m/s$$

C
$$3m/s$$

D
$$4m/s$$

Solution

First body initial speed $$u=0$$
Final speed $$=3 m/s$$
Second body initial speed$$ =4 m/s$$
Body is dropped from height $$h$$
So $$a= g= 10m/{s}^{2}$$
$${v}^{2}={u}^{2}+2gh$$
$${3}^{2}=0+20 h$$
$$9=20h$$
$$h=\frac{9}{20}$$
Second object is also dropped from same height.
therefore, $$h=\frac{20}{9}$$
$$u =4 m/s$$
$${v}^{2}={u}^{2}+2gh$$
$$v=5 m/s$$
Question 9
1 / -0

For the V-t graph shown displacement during 15 seconds is

A
25 m

B
50 m

C
zero

D
75m

Solution

The total displacement of a particle is equal to the area under the velocity-time graph.
Net Displacement =Area of triangle
$$\Rightarrow d =\dfrac {1}{2}base \times hight$$
$$\Rightarrow d = \dfrac12(15 \times 10)$$
$$\Rightarrow d =75\ m$$
Question 10
1 / -0

A force of $$5\ N$$ acts on a $$15\ kg$$ body initially at rest. The work done by the force during the first second of motion of the body is

A
$$5\ J$$

B
$$6\ J$$

C
$$\dfrac {5}{6}\ J$$

D
$$75\ J$$

Solution

Initial speed   $$u = 0$$
Acceleration $$a = \dfrac{F}{m} = \dfrac{5}{15} = \dfrac{1}{3} \ m/s^2$$
Distance covered in one second   $$S = ut+\dfrac{1}{2}at^2$$
where $$t = 1 \ s$$
Or,   $$S = 0+\dfrac{1}{2}(\dfrac{1}{3})(1)^2 = \dfrac{1}{6} \ m$$
Work done $$W = FS = 5\times \dfrac{1}{6} = \dfrac{5}{6} \ J$$