Motion Test

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Motion Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

A body falls from rest, its velocity at the end of first second is $$ ( g = 32 \,ft / s^2) $$

A
$$ 16 \, ft / s $$

B
$$ 32 \, ft / s $$

C
$$ 64 \, ft / s $$

D
$$ 24 \, ft / s $$

Solution

Since the body falls from rest, the initial velocity, $$ u=0 ft/s.$$
$$ g= 32\ ft/s^2$$
The velocity attained by body in time $$t = 1\ s$$ is given by :
$$ \nu = u +g t = 0 +32 \times 1 = 32 \, ft/s $$
Option (B) is correct.
Question 2
1 / -0

A body thrown with an initial speed of $$ 96\, ft / sec $$ reaches the ground after $$ ( g = 32 ft / sec^2 )$$

A
$$ 3 \,sec $$

B
$$ 6 \,sec $$

C
$$ 12 \,sec $$

D
$$ 8 \,sec $$

Solution

Time of flight
$$T = \dfrac{2u}{g} $$
$$ T\ = \dfrac{2 \times 96}{32} = 6 \, sec $$
Question 3
1 / -0

The variation of velocity of a particle with time moving along a straight line is shown in the figure. The distance traveled by the particle in $$4\ s$$ is :

A
$$ 60\,m $$

B
$$ 55 \,m $$

C
$$ 25 \,m $$

D
$$ 30\,m $$

Solution

$$Distance$$ = $$Area\ under\ v-t\ graph$$ $$ = A_1 + A_2 + A_3 + A_4 $$
$$ = \dfrac{1}{2} \times 1 \times 20 + (20 \times 1) + \dfrac{1}{2} (20 + 10) \times 1 + (10 \times 1) $$
$$ = 10 + 20 + 15 + 10 = 55 m $$
Question 4
1 / -0

Time taken by an object falling from rest to cover the height of $$ h_1 $$ and $$ h_2 $$ is respectively $$ t_1 $$ and $$ t_2 $$ then the ratio of $$ t_1 $$ to $$ t_2 $$ is

A
$$ h_1 : h_2 $$

B
$$ \sqrt{h_1} : \sqrt{h_2} $$

C
$$ h_1 : 2 h_2 $$

D
$$ 2h : h $$
Solution
By 2nd equation of motion,
$$s=ut+\dfrac{1}{2}at^2$$
Since $$u=0 \ and \ a = g$$ we have:
For $$s=h_1, t = t_1$$
$$h_1=\dfrac{1}{2}gt_1^2 \implies t_1=\sqrt{\dfrac{2h_1}{g}} . . . . . . (i)$$
For $$s=h_2, t = t_2$$
$$h_2=\dfrac{1}{2}gt_2^2 \implies t_2=\sqrt{\dfrac{2h_2}{g}} . . . . . . (ii)$$

By dividing (i) by (ii)
$$ \Rightarrow \, \dfrac{t_1}{t_2} = \sqrt{\dfrac{h_1}{h_2}} \Rightarrow t_1:t_2= \sqrt h_1 : \sqrt h_2$$
Question 5
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The horizontal straight line obtained from the distance-time graph is related to which of the following velocity ?

A
Zero velocity

B
Constant velocity

C
Increasing velocity

D
Decreasing velocity

Solution

According to the given condition in the question,
The Distance-time graph is horizontal and straight line as given in the above image. here the position of object is not changing with respect to time. Hence velocity of an object is equal to zero.
Question 6
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An object moving in upward direction opposite to the gravitational force of earth performs

A
accelerated motion

B
motion with constant velocity

C
oscillations

D
retarded motion

Solution

Answer: Retarded motion.
The motion of an object vertically upwards is retarded motion. Here, the retardation is $$9.8 \ m/s^2.$$ The retardation is nothing but an acceleration which acts opposite direction of velocity. So it has a tendency to slow down the object.
Question 7
1 / -0

Four alternatives are given to each of the following incomplete statements/questions, choose the right answer.
Uniform circular motion is called continuously accelerated motion mainly because

A
Direction of motion changes

B
Speed remains the same

C
Velocity remains the same

D
Direction of motion does not change

Solution

Uniform circular motion is accelerated because the velocity changes due to continuous change in the direction of motion. Thus, for a body moving in a circular path at a constant speed, its velocity changes continuously. The change in velocity gives rise to an acceleration in the moving body.
Question 8
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A lift performs the first part of its ascent with uniform acceleration, $$a$$, and the remaining with uniform retardation, $$2a$$. If, $$t$$, is the time of ascent, the depth of the shaft is

A
$$\displaystyle \dfrac{{at}^{2}}{4}$$

B
$$\displaystyle \dfrac{{at}^{2}}{3}$$

C
$$\displaystyle \dfrac{{at}^{2}}{2}$$

D
$$\displaystyle \dfrac{{at}^{2}}{8}$$

Solution

We have $$t_1=\dfrac{v}{a}$$ and $$t_2=\dfrac{v}{2a}$$
Adding both times, we get
$$t=\dfrac{3v}{2a}$$...... (1)
or
$$h=h_1+h_2$$

$$h=\dfrac{v^2}{2a}+\dfrac{v^2}{4a}$$

$$h=\dfrac{3v^2}{4a}$$...... (2)

From (1) and (2), we get

$$h=\dfrac{at^2}{3}$$
Question 9
1 / -0

Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given
$$V (m\:s^{-1})$$
0

6
12
24
30
36
42
$$t (s)$$
0

2
4
8
10
12
14
What can be concluded about the motion of the body?

A
It moves with uniform speed.

B
It moves with uniform motion

C
It moves with uniform velocity

D
It moves with uniform acceleration

Solution
Question 10
1 / -0

Directions For Questions

A determined student waited to test the law of gravity himself and jumps off the top of CN Tower in Toronto ($$553\ m$$ high) and falls freely. His initial velocity is zero. A rocketeer arrives at the scene $$5.0\ s$$ later and dives off the top of the tower to save him. The rocketeer leaves the tower with an initial speed $${ v }_{ 0 }$$. In order to catch the student and to prevent injury to him, the rocketeer should catch the student at a sufficiently great height above the ground so that the rocketeer and student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by the rocketeer's jet pack, which he turns on when he catches the student, before that rocketeer is in free fall. To prevent discomfort to the student, the magnitude of acceleration should not exceed $$5g$$.

...view full instructions

What is the minimum height above the ground at which the rocketeer should catch the student?

A
$$92.1\ m$$

B
$$460.9\ m$$

C
$$78.8\ m$$

D
$$82.3\ m$$

Solution

$$Let\quad height\quad at\quad which\quad student\quad is\quad caught\quad is\quad h\\ velocity\quad at\quad this\quad point\quad is\quad \sqrt { 2g(553-h) } \\ when\quad the\quad student\quad comes\quad to\quad ground\quad his\quad velocity\quad is\quad zero\\ { v }^{ 2 }-{ u }^{ 2 }=2as\\ 0-2g(553-h)=2(-5g)h\\ \Rightarrow 553-h=5h\Rightarrow h=92.1m$$

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$$V (m\:s^{-1})$$	0		6	12	24	30	36	42
$$t (s)$$	0		2	4	8	10	12	14

Motion Test - 70

Result

Motion Test - 70

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Question 1

$$ 16 \, ft / s $$

$$ 32 \, ft / s $$

$$ 64 \, ft / s $$

$$ 24 \, ft / s $$

Solution

Question 2

$$ 3 \,sec $$

$$ 6 \,sec $$

$$ 12 \,sec $$

$$ 8 \,sec $$

Solution

Question 3

$$ 60\,m $$

$$ 55 \,m $$

$$ 25 \,m $$

$$ 30\,m $$

Solution

Question 4

$$ h_1 : h_2 $$

$$ \sqrt{h_1} : \sqrt{h_2} $$

$$ h_1 : 2 h_2 $$

$$ 2h : h $$

Solution

Question 5

Zero velocity

Constant velocity

Increasing velocity

Decreasing velocity

Solution

Question 6

accelerated motion

motion with constant velocity

oscillations

retarded motion

Solution

Question 7

Direction of motion changes

Speed remains the same

Velocity remains the same

Direction of motion does not change

Solution

Question 8

$$\displaystyle \dfrac{{at}^{2}}{4}$$

$$\displaystyle \dfrac{{at}^{2}}{3}$$

$$\displaystyle \dfrac{{at}^{2}}{2}$$

$$\displaystyle \dfrac{{at}^{2}}{8}$$

Solution

Question 9

It moves with uniform speed.

It moves with uniform motion

It moves with uniform velocity

It moves with uniform acceleration

Solution

Question 10

$$92.1\ m$$

$$460.9\ m$$

$$78.8\ m$$

$$82.3\ m$$

Solution

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