Motion in A Straight Line Test

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Motion in A Straight Line Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Velocity at the top of vertical journey under gravity when a body is projected upward with velocity $$1000m/s$$ is

A
Zero

B
$$10 m/s$$

C
$$100 m/s$$

D
$$1000 m/s$$

Solution

As body goes upwards, due to gravitational force the body will stop at the maximum height and fall towards the ground.
Hence, the velocity at the maximum height is zero.
$$v_m=0m/s$$
The correct option is A.
Question 2
1 / -0

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $$5th$$ sec to that covered in $$5$$ sec is :

A
$$9/25$$

B
$$3/5$$

C
$$25/9$$

D
$$1/25$$

Solution

Given,

$$u=0m/s$$

Distance covered by the $$n^{th}$$ sec,

$$S=ut+a(n-\dfrac{1}{2})$$

For $$5^{th}$$ sec

$$S_1=0+a(5-0.5)$$

$$S_1=4.5a$$. . . . . .. . . . .(1)

The distance covered by the body in 5 sec

$$S_2=ut+\dfrac{1}{2}at^2$$

$$S_2=0+\dfrac{1}{2}a\times 5\times 5$$

$$S_2=12.5a$$. . . . . . . . .(2)

$$\dfrac{S_1}{S_2}=\dfrac{4.5a}{12.5a}$$

$$\dfrac{S_1}{S_2}=\dfrac{9}{25}$$
.
Question 3
1 / -0

A car accelerates from rest at a constant rate $$\alpha$$ for some time, after which it decelerates at a constant rate $$\beta$$ and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by car is

A
$$(\dfrac{\alpha^2+\beta^2}{\alpha \beta})t$$

B
$$(\dfrac{\alpha^2-\beta^2}{\alpha \beta})t$$

C
None of these

D
$$(\dfrac{\alpha \beta r}{\alpha \beta})$$

Solution

Lets consider, $$v=$$ maximum velocity acquired by car
From 1st equation of motion,
$$v=0+\alpha t_1$$
$$t_1=\dfrac{v}{\alpha}$$
Similarly, $$0=v-\beta t_2$$
$$t_2=\dfrac{v}{\beta}$$
$$t_1+t_2=t=\dfrac{v}{\alpha}+\dfrac{v}{\beta}$$
$$t=v(\dfrac{\alpha +\beta}{\alpha \beta})$$
$$v=\dfrac{\alpha \beta}{\alpha +\beta}t$$
Question 4
1 / -0

A particle starts with an initial velocity of $$20 m/s$$ and moves with a constant acceleration on a straight line. If its final velocity is $$60 m/s$$, then velocity of midpoint is

A
$$40 m/s$$

B
$$22\sqrt{5} m/s$$

C
$$20\sqrt{5} m/s$$

D
$$24\sqrt{5} m/s$$

Solution

The following velocity at the mid point is give by the formula,
$$V_{mid \quad point}=\sqrt{\dfrac{(V_{initial})^2+(V_{final})^2}{2}}$$
$$V_{mid \quad point}=\sqrt{\dfrac{20^2+60^2}{2}}=\sqrt{\dfrac{400+3600}{2}}$$
$$=20\sqrt5 m/s$$
Question 5
1 / -0

A block can slide on a smooth inclined plane of inclination '$$q$$' kept on the floor of a lift. When the lift is descending with retardation, '$${a}$$' $$m/s^2$$ the acceleration of the block relative to the incline will be:

A
$$(g-a)\sin q$$

B
$$(g+a)\sin q$$

C
$$g\sin q$$

D
$$g-a$$

Solution
Question 6
1 / -0

A particle travels $$10$$m in first $$5$$ seconds $$10$$ cm in next $$3$$ seconds. Assuming constant acceleration what is the distance travelled in next $$2$$ second ?

A
$$8.3$$ m

B
$$9.3$$ m

C
$$10.3$$ m

D
None of these

Solution

Let assume initial velocity is $$u$$ and constant acceleration $$a$$. using $$s$$=$$ut$$ + $$(0.5)at^2$$,
for first $$5$$ second
$$10$$ = $$5u$$ + $$(0.5)a(25)$$
for next 3 seconds, $$t = 8$$ and $$s = 20$$
$$20 = 8u + (0.5)a(64)$$
Solving,
$$a = {\dfrac{1}{3}}$$ and $$u = {\dfrac{7}{6}}$$
Now for next $$2$$ seconds, $$t = 10$$ and $$S'$$:
$$S' = (10){\dfrac{7}{6} +(0.5){\dfrac{1}{3}}(100)}$$
$$S'$$ = $${\dfrac{170}{6}}$$
Now distance travelled in $$2$$ sec = $${\dfrac{170}{6}} - {20} = {\dfrac{50}{6}}$$ meter
Question 7
1 / -0

A force of $$20N$$ is applied on a body of mass $$5 kg$$ resting on a horizontal plane. The body gains a kinetic energy of $$10 joule$$ after it moves a distance of $$2m$$. The frictional force is:

A
$$10 N$$

B
$$15 N$$

C
$$20 N$$

D
$$30 N$$

Solution

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Question 8
1 / -0

The splash of sound was heard $$6$$s after dropping a stone into a well $$125$$m deep. The velocity of sound is:

A
$$350$$ cm/s

B
$$125$$ m/s

C
$$392$$ cm/s

D
$$0$$ cm/s

Solution

From equation of motion we have
$$s=ut+\dfrac{1}{2}g{t}^{2}$$
where u= $$0$$ m/s;g=acceleration due to gravity=$$10 m/{s}^{2}$$ and s= distance traveled=$$125$$ m

Putting the above-mentioned values in the equation of motion we get

$${t}^{2}=\dfrac{250}{10}$$

$$t=5 s$$
Now as per question the sound of splash is after $$6s$$ of dropping, thus the time taken for the sound to cover a distance of $$125$$ m is $$1$$ second.
Speed of sound $$v=\dfrac{distance}{time}$$=$$125 $$m/s
Question 9
1 / -0

A particle is thrown vertically up with speed u so that distance covered in last second of flight is $$35\ m$$ . If then initial speed of throw is

A
$$20 m/s$$

B
$$26.45m/s$$

C
$$40 m/s$$

D
$$50 m/s$$

Solution

We know from equation of motion we get,
$$v=u-gt$$-------(1)
as velocity of the particle becomes at the very last second of flight $$v=0$$
Thus equation 1 can be written as,
$$t=\dfrac{u}{g}$$-------(A)
Again we know,
$$s=ut-\dfrac{1}{2}g{t}^{2}$$
$$35=\dfrac{{u}^{2}}{g}-\dfrac{1}{2}g{(\dfrac{u}{g})}^{2}$$
$$u=\sqrt{700}=26.45$$
Question 10
1 / -0

Two cars OF SAME LENGTH move in the same direction along parallel roads. One of them is a $$100$$ m long travelling with a velocity of $$7.5$$ $$ms^{-1}$$. How long will it take for the first car to overtake the second car?

A
$$26$$ s

B
$$40$$ s

C
$$60$$ s

D
$$80$$ s

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Motion in A Straight Line Test - 51

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Motion in A Straight Line Test - 51

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SHARING IS CARING

Question 1

Zero

$$10 m/s$$

$$100 m/s$$

$$1000 m/s$$

Solution

Question 2

$$9/25$$

$$3/5$$

$$25/9$$

$$1/25$$

Solution

Question 3

$$(\dfrac{\alpha^2+\beta^2}{\alpha \beta})t$$

$$(\dfrac{\alpha^2-\beta^2}{\alpha \beta})t$$

None of these

$$(\dfrac{\alpha \beta r}{\alpha \beta})$$

Solution

Question 4

$$40 m/s$$

$$22\sqrt{5} m/s$$

$$20\sqrt{5} m/s$$

$$24\sqrt{5} m/s$$

Solution

Question 5

$$(g-a)\sin q$$

$$(g+a)\sin q$$

$$g\sin q$$

$$g-a$$

Solution

Question 6

$$8.3$$ m

$$9.3$$ m

$$10.3$$ m

None of these

Solution

Question 7

$$10 N$$

$$15 N$$

$$20 N$$

$$30 N$$

Solution

Question 8

$$350$$ cm/s

$$125$$ m/s

$$392$$ cm/s

$$0$$ cm/s

Solution

Question 9

$$20 m/s$$

$$26.45m/s$$

$$40 m/s$$

$$50 m/s$$

Solution

Question 10

$$26$$ s

$$40$$ s

$$60$$ s

$$80$$ s

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