Motion in A Straight Line Test

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Motion in A Straight Line Test - 37

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Question 1
1 / -0

An object thrown vertically upwards with a velocity of 25 m/s takes 4 sec to reach the thrower. What is displacement of the object?

A
100 m

B
180 m

C
0 m

D
120 m

Solution

Displacement is the minimum distance between the initial and final position of an object If a ball thrown from point A reaches point B and then return to point A, then a displacement of the ball becomes zero as it returns to its initial position.
Question 2
1 / -0

A ball is thrown upwards with a velocity of $$25 m/s$$. What is the time taken by the ball to return to the thrower ($$g=10\:m/s^2$$)

A
$$5 sec$$

B
$$2.5 sec$$

C
$$3 sec$$

D
$$4.2 sec$$

Solution

Initial velocity of the ball final velocity $$= 0 m/s$$
acceleration due to gravity $$(g) = -10 m/s$$ $$^2$$ (upward motion)
By using the formula,
$$v=u+at$$
$$O=25+(-10)\times t$$
$$10t=25$$
$$t=2.5 sec.$$
The time taken for upward journey is equal to the time taken for downward journey.
Therefore, total time taken by the ball $$= 2.5 + 2.5=5 sec.$$
Question 3
1 / -0

Directions For Questions

These questions consist of two statements each, printed as assertion and reason. While answering these questions you are required to choose anyone of the following four responses.

...view full instructions

Assertion: The driver in a vehicle moving with a constant speed on a straight road is an inertial frame of reference.
Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.

A
If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

B
If both Assertion and Reason are true but the Reason is not the correct explanation of the Assertion

C
If Assertion is true statement but Reason is false.

D
If both Assertion and Reason are false statements.

Solution

Inertial frames are non accelerated frames i.e they move with constant velocity.
As the driver in a vehicle is moving with a constant speed, thus the reference frame is an inertial frame.
Also the Newtonian mechanics (law's of motion) are applicable only in inertial frame.
Thus the assertion is true but the reason is false.
Question 4
1 / -0

A car accelerates steadily so that it goes from a velocity of 20 m/s to a velocity of 40 m/s in 4 seconds. What is its acceleration?

A
0.2 $$m/s^2$$

B
4 $$m/s^2$$

C
5 $$m/s^2$$

D
10 $$m/s^2$$

E
80 $$m/s^2$$

Solution

Given,
$$u=20\ m/s\\v=40\ m/s\\t=4\ sec$$

Acceleration is a measure of the change in velocity over time.
Change in velocity is $$\Delta v=v-u$$
$$\Delta v=40m/s-20m/s=20m/s$$
Since this change in velocity takes place over 4 seconds,
Car acceleration is $$a=\dfrac{\Delta v}{\Delta t}$$

$$a=\dfrac{20m/s}{4s}=5m/s^2$$
Question 5
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Which of the following statement must always be true?
I.If an objects acceleration is zero, then its speed must remain constant.
II. If an objects acceleration is constant, then it must move in a straight line.
III. If an objects speed remains constant, then its acceleration must be zero.

A
I and II only

B
I and III only

C
I only

D
III only

E
II and II only

Solution

Acceleration is the rate of change of speed of the object. Thus when acceleration is zero, the speed of object remains constant.

Acceleration of an object moving in a circular path is $$\dfrac{v^2}{R}$$. Thus an object with constant acceleration may not move in a straight line.

Again in case of circular path, the speed remains same, but acceleration is finite.
Question 6
1 / -0

The velocity-time graph below represents the velocity of a toy train as it moves north and south with velocity near the middle of the vertical axis. During which, Interval(s) is the toy train speeding up?

A
$$0\ to\ A$$ only

B
$$0\ to\ A$$ and $$D\ to\ E$$

C
$$A\ to\ B$$

D
$$B\ to\ D$$ only

E
$$A\ to\ B$$ and $$D\ to\ E$$

Solution

The toy train will speed up if the rate of change of velocity will increases with respect to time.
Therefore, the train will speed up in the intervals $$0\ to A$$ and $$D\ to\ E$$.
Question 7
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A sprinter starts from rest and accelerates at a steady rate for the first $$50 m $$ of a $$100 m$$ race, and then continues at a constant velocity for the second $$50 m$$ of the race. If the sprinter runs the $$100 m$$ in a time of $$10 s$$, what is his instantaneous velocity when he crosses the finish line?

A
$$5 m/s$$

B
$$10 m/s$$

C
$$12 m/s$$

D
$$15 m/s$$

E
$$20 m/s$$

Solution

We know the total distance the sprinter covers, and we know the total time. However, since the acceleration isnt uniform, we cant calculate the velocity quite so simply. Rather, we need two equations, one for the first 5 0 meters of the race, and another for the second 5 0 meters. In the first 50 meters, the sprinter accelerates from an initial velocity of $$v_0=0$$ to a final velocity of v in an amount of time, $$t_1$$. We can express this relationship using the kinematic equation that leaves out velocity, and then solve for t:
$$x=x_0+\dfrac{1}{2}(r+ v_0)t_1$$
$$50 m=\dfrac{1}{2}vt_1$$
$$t_1-\dfrac{100m}{v}$$
In the last 5 0 meters of the race, the sprinter runs with a constant velocity of v , covering a distance of x = 50 m in a time $$t_2$$, Solving for, $$t_2$$ we find:
$$t_2=\dfrac{50m}{v}$$
We know that the total time of the race, $$t_1 +t_2=10 s$$. With this in mind, we can add the two sprint times together and solve for v :
$$10 s=\dfrac{100m }{v}+\dfrac{50 m}{v}=\dfrac{150 m}{v}$$
$$v=\dfrac{150 m}{10 s}=15 m/s$$
Question 8
1 / -0

Which of the following vehicles is undergoing a deacceleration?

A
A car driving straight to the east on a road at a constant speed

B
A truck rounding a corner at a constant speed

C
A van slowing down as it approaches a stop sign

D
None of these

Solution

A object is said to have an acceleration if it changes its velocity either by increasing its speed, decreasing its speed or changing the direction of its velocity.
Since the car and the truck move with constant speed, thus they have zero acceleration.
But the van is slowing down its speed, thus it has deacceleration.
Hence option C is correct.
Question 9
1 / -0

A girl is 13m away from a tree. An apple falls from the tree at height of 2.8 m. If she wants to catch the apple before it hits the ground then how fast does she need to run?

A
$$9.83{m}{s^{-1}}$$

B
$$0.755 \frac{m}{s}$$

C
More information is needed to solve

D
$$17.22 {m}{s^{-1}}$$

E
$$ 1.33 \frac{m}{s}$$

Solution

Height of the tree $$h = 2.8$$ m
Distance of girl from tree $$d = 13$$ m
Time taken by apple to reach the ground $$t = \sqrt{\dfrac{2h}{g}} =\sqrt{\dfrac{2 \times 2.8}{9.8}} =0.76$$ s
Thus the speed of the girl $$v = \dfrac{d}{t} =\dfrac{13}{0.76} =17.11 \simeq 17.22$$ m/s
Question 10
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By applying brakes, a car moving at $$1 m{s}^{-1}$$ is brought to rest in 3 s. If it moves at $$2 m{s}^{-1}$$, how long will it take to come to rest?

A
$$6s$$

B
$$12s$$

C
$$3s$$

D
$$1s$$

Solution

$$u=1\ ms^{-1}$$
$$v=0$$
$$t=3\ s$$
$$\therefore$$ Retardation, $$a=\dfrac{u-v}{t}=\dfrac{1-0}{3}ms^{-2}=\dfrac{1}{3}ms^{-2}$$
Now, $$u=2ms^{-1}$$$$,v=0, a=-\dfrac{1}{3}ms^{-2}$$
Let the time be $$t$$.
$$v=u+at$$
$$0=2-\dfrac{1}{3}t$$
$$t=6s$$

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Motion in A Straight Line Test - 37

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Motion in A Straight Line Test - 37

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Question 1

100 m

180 m

0 m

120 m

Solution

Question 2

$$5 sec$$

$$2.5 sec$$

$$3 sec$$

$$4.2 sec$$

Solution

Question 3

If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

If both Assertion and Reason are true but the Reason is not the correct explanation of the Assertion

If Assertion is true statement but Reason is false.

If both Assertion and Reason are false statements.

Solution

Question 4

0.2 $$m/s^2$$

4 $$m/s^2$$

5 $$m/s^2$$

10 $$m/s^2$$

80 $$m/s^2$$

Solution

Question 5

I and II only

I and III only

I only

III only

II and II only

Solution

Question 6

$$0\ to\ A$$ only

$$0\ to\ A$$ and $$D\ to\ E$$

$$A\ to\ B$$

$$B\ to\ D$$ only

$$A\ to\ B$$ and $$D\ to\ E$$

Solution

Question 7

$$5 m/s$$

$$10 m/s$$

$$12 m/s$$

$$15 m/s$$

$$20 m/s$$

Solution

Question 8

A car driving straight to the east on a road at a constant speed

A truck rounding a corner at a constant speed

A van slowing down as it approaches a stop sign

None of these

Solution

Question 9

$$9.83{m}{s^{-1}}$$

$$0.755 \frac{m}{s}$$

More information is needed to solve

$$17.22 {m}{s^{-1}}$$

$$ 1.33 \frac{m}{s}$$

Solution

Question 10

$$6s$$

$$12s$$

$$3s$$

$$1s$$

Solution

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