Motion in A Straight Line Test

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Motion in A Straight Line Test - 33

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Question 1
1 / -0

$$A$$ and $$B$$ are arguing about uniform acceleration. $$A$$ states that acceleration means "the longer you go." $$B$$ states that acceleration means "the further you go." Who is right?

A
$$A$$

B
$$B$$

C
Both $$A$$ and $$B$$

D
None of these

Solution

Acceleration of the object means how fast the object is changing its velocity or at what rate its velocity is changing. Acceleration does not mean distance covered by the object or time taken to cover that distance. So, none of them is right.
Question 2
1 / -0

A ball is thrown upwards in a train which is accelerating. The ball will fall ______________.

A
ahead of the thrower

B
behind the thrower

C
in the hands of thrower

D
data inadequate

Solution

When train have some initial velocity and acceleration, and if ball if thrown in air it will carry only velocity of train and will behave as a freely falling body.
Now initial horizontal velocity of both the bodies is same and train have some acceleration in horizontal velocity but ball dont have any acceleration in horizontal.
So displacement of train is more in horizontal as compare to ball, so ball will land behind the thrower.
So best possible amswer is option B.
Question 3
1 / -0

A body, when dropped from a certain height, falls to the ground in 4 s. What is the time taken by the body to cover the last 100 cm? $$\displaystyle (Take\quad g={ 10\quad ms }^{ -2 })$$

A
$$0.65 s$$

B
$$0.15 s$$

C
$$0.125 s$$

D
$$0.025 s$$

Solution

Let Height $$h$$ from a body is dropped.
Using $$s=ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= 0\times 4 -\frac{1}{2}g\times 4^2$$ $$\Rightarrow h= 80 m$$,
Let time $$t$$ to cover the 79$$m$$.
Using $$s=ut+\frac{1}{2}at^2$$ $$\Rightarrow -79= 0\times t -\frac{1}{2}g\times t^2$$ $$\Rightarrow t=\sqrt{\frac{79}{5}}$$= 3.975 $$s$$
$$\therefore $$ Required time= $$4-3.975= 0.025$$ $$s$$
Question 4
1 / -0

A body under uniformly accelerated motion covers $$20 m$$ and $$60 m$$ in first two seconds of time from the beginning. The body starts with an initial velocity

A
Zero

B
$$5{m}/{s}$$

C
$$10{m}/{s}$$

D
$$15{m}/{s}$$

Solution

Here $$t = 1, {S}_{1} = 20$$ and
$$t = 2$$ and $${S}_{2} = 60 + 20 = 80$$
$$\because {S}_{1} = ut + \displaystyle\frac{1}{2}a{t}^{2}$$
$$\therefore 20 = u + \displaystyle\frac{1}{2}a$$
or, $$2u + a = 40$$ .....(1)
and also $${S}_{2} = ut + \displaystyle\frac{1}{2}a{t}^{2}$$
$$\therefore 80 = 2u + \displaystyle\frac{1}{2}a \times 4 = 2u + 2a$$
or, $$2u + a = 40$$ ......(2)
From equations (1) and (2) we get
$$u = 0$$.
Question 5
1 / -0

Which one of the following represents the displacement-time graph of two objects a and b moving with zero relative speed ?

A

B

C

D
None of these

Solution

For zero relative velocity the velocity of both body must be same. $$v=v_a-v_b=0$$
The slope of displacement vs time graph must be same hence option B is correct.
Question 6
1 / -0

A ball is thrown vertically upwards with a velocity $$u$$. What is the maximum height to which it will rise before falling back?

A
$${u^2/2g}$$

B
$$g/u$$

C
$$2g/u$$

D
$$u/2g$$

Solution

Answer is A.

If we know the initial velocity, from the equation, $${ v }^{ 2 }-{ u }^{ 2 }=2as$$ with s = h and v = 0 at the top of the path, initial velocity is given as 'u' and a = -g then:
h = $$\frac { { u }^{ 2 } }{ 2g } $$.
Therefore, the maximum height reached = $$\frac { { u }^{ 2 } }{ 2g } $$.
Question 7
1 / -0

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m.If the ball is in contact with the floor for 0.01 sec, then average acceleration during contact is

A
$$\displaystyle 2100{ m }/{ { s }^{ 2 } }$$

B
$$\displaystyle 1400{ m }/{ { s }^{ 2 } }$$

C
$$\displaystyle 700{ m }/{ { s }^{ 2 } }$$

D
$$\displaystyle 400{ m }/{ { s }^{ 2 } }$$

Solution

Let u be the velocity with which the ball hits the ground, then
$$\displaystyle { u }^{ 2 }=2gh$$
$$\displaystyle =2\times 9.8\times 10=196$$
$$\displaystyle \therefore \quad u=14 m/sec$$
If v be the velocity with which it rebounds, then
$$\displaystyle { v }^{ 2 }=2\times 9.8\times 2.5=49$$
$$\displaystyle \Rightarrow v=7m/sec$$
$$\displaystyle \therefore \quad \Delta v=\left( v-u \right) $$
$$\displaystyle =\left( 7m/sec \right) -\left( -14m/sec \right) $$
$$\displaystyle =21m/sec$$
$$\displaystyle \therefore \quad a=\frac { \Delta v }{ \Delta t } =\frac { 21 }{ 0.01 } $$
$$\displaystyle =2100{ m }/{ { s }^{ 2 } }$$
Question 8
1 / -0

If a car at rest accelerates uniformly and attains a speed of 72 km/hr in 10 s, then it covers a distance of:

A
50 m

B
100 m

C
200 m

D
400 m

Solution

Given
u=0
v=72 km/h=20 m/s
t=2s
$$a=\dfrac{v-u}{t}$$
$$a=2$$
$$s=ut+\dfrac{at^2}{2}$$
putting values
$$s=100m$$
Question 9
1 / -0

An automobile travelling with a speed of $$\displaystyle 72 km {h}^ {-1} $$ , can be stopped within a distance of 30 m, by applying brakes. What will be the stopping distance, if the automobile speed is increased to $$\displaystyle \sqrt { 3 } $$ times and the same braking force is applied?

A
30 m

B
90 m

C
60 m

D
120 m

Solution

Given Initial Velocity $$u=72 kmh^{-1}$$= 20$$ms^{-1}$$, Distance $$s= 30 m$$, Final Velocity $$v=0$$,
Let acceleration is $$a$$.
Using $$v^2= u^2-2as$$ $$\Rightarrow 0= 20^2 +2\times a\times 30$$ $$\Rightarrow a= -\dfrac{20}{3}$$,
Now initial velocity is increases by $$\sqrt3$$ times and the same braking force is applied.
Let distance covered is $$s$$.
Using $$v^2= u^2-2as$$ $$\Rightarrow 0= (20\sqrt 3)^2+2\times (-\dfrac{20}{3}) \times s$$ $$\Rightarrow s=90 m$$
Distance covered is 90 m.
Question 10
1 / -0

A stone dropped from the top of a building taken 5 s to reach the ground. If it is stopped momentarily 4 s after it is dropped and then released again, how much time would it take from the moment it is released again to reach the ground? $$\displaystyle (Take\quad g = 10 {ms} ^{-2})$$

A
1 s

B
3 s

C
4 s

D
5 s

Solution

First Case-
Let the distance covered by the stone in 5 s is $$h$$.
Using $$s=ut+\frac{1}{2}gt^2$$ $$\Rightarrow h= 0\times 5+\frac{1}{2} \times 10\times 5^2$$ ( $$\because u=0$$)
$$\Rightarrow h= 125 m$$,

Second Case-
Let the distance covered by the stone in 4 s is $$h_1$$.
Using $$s=ut+\frac{1}{2}gt^2$$ $$\Rightarrow h= 0\times 4+\frac{1}{2} \times 10\times 4^2$$ ( $$\because u=0$$)
$$\Rightarrow h_1=80 m$$
At this moment stone is stopped and then released,
Let it take $$t$$ seconds to cover the remaining distance.
Remaining distance = $$h-h_1$$= 125-80=45 $$m$$,
Using $$s=ut+\frac{1}{2}gt^2$$ $$\Rightarrow 45= 0\times t+\frac{1}{2} \times 10\times t^2$$ ( $$\because u=0$$)
$$\Rightarrow t=3 s$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 33

Result

Motion in A Straight Line Test - 33

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SHARING IS CARING

Question 1

$$A$$

$$B$$

Both $$A$$ and $$B$$

None of these

Solution

Question 2

ahead of the thrower

behind the thrower

in the hands of thrower

data inadequate

Solution

Question 3

$$0.65 s$$

$$0.15 s$$

$$0.125 s$$

$$0.025 s$$

Solution

Question 4

Zero

$$5{m}/{s}$$

$$10{m}/{s}$$

$$15{m}/{s}$$

Solution

Question 5

None of these

Solution

Question 6

$${u^2/2g}$$

$$g/u$$

$$2g/u$$

$$u/2g$$

Solution

Question 7

$$\displaystyle 2100{ m }/{ { s }^{ 2 } }$$

$$\displaystyle 1400{ m }/{ { s }^{ 2 } }$$

$$\displaystyle 700{ m }/{ { s }^{ 2 } }$$

$$\displaystyle 400{ m }/{ { s }^{ 2 } }$$

Solution

Question 8

50 m

100 m

200 m

400 m

Solution

Question 9

30 m

90 m

60 m

120 m

Solution

Question 10

1 s

3 s

4 s

5 s

Solution

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