Motion in A Straight Line Test

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Motion in A Straight Line Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

A car travels a distance $$100 m$$ with a constant acceleration and average velocity of $$20\ ms^{-1}$$. The final velocity acquired by the car is $$25\ ms^{-1}$$. Find the initial velocity.

A
$$15 \,m \,s^{-1}$$

B
$$10 \,m \,s^{-1}$$

C
$$30 \,m \,s^{-1}$$

D
$$0 \,m \,s^{-1}$$

Solution

For uniformly accelerated motion, $$\text{Average velocity} = \dfrac{v+u}{2}$$
$$20 = \frac{25+u}{2}$$
$$u=15\ m/s$$
Question 2
1 / -0

A ball is thrown vertically upwards with an initial velocity of $$49 m {s}^{-1}$$. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take $$g=9.8 m {s}^{-2}$$)

A
(i) 122.5 m, (ii) 10 s

B
(i) 120.5 m, (ii) 10 s

C
(i) 122.5 m, (ii) 20 s

D
(i) 120 m, (ii) 10 s

Solution

(i)
$$v^2=u^2+2as$$
$$0=49^2-2(9.8)(s)$$
$$s=122.5\ m/s$$

(ii)
$$v=u+at$$
$$0=49-9.8t$$
$$t=5\ s$$
total time$$=10\ s$$
Question 3
1 / -0

A ball thrown up vertically returns to the thrower after 6 s Find it's initial velocity.

A
20 m/s

B
40 m/s

C
35 m/s

D
30 m/s

Solution

The ball returns to the thrower in $$6$$ s,thus the time for its upward journey$$=$$$${6/2}$$$$=$$ 3s
For the upward motion of ball.
Initial velocity $$u=$$?;final velocity $$v=$$0
($${\because }$$ Ball comes to rest)
Time ,$$t=3$$s
Acceleration due to gravity ,$${g=-10 ms^{-2}}$$
[In upward direction ,$$g$$ is taken -ve]
$$v$$$$=$$$$u+gt$$
or $${0=u-10 \times 3,or -u=-30}$$
or $${u=30 ms^{-1}}$$
$${\therefore}$$ Initial velocity of ball is $${30 ms ^{-1}}$$
Question 4
1 / -0

A boy on a cliff $$49 m$$ high drops a stone.One second later,he throws a second after the first .They both hit the ground at the same time.With what speed did he throw the second stone?

A
$$12.1 m/s$$

B
$$10 m/s$$

C
$$21.1 m/s$$

D
$$15.1 m/s$$

Solution

If time for first stone is$$ t$$, time for second stone is $$t-1$$
$$s=49m$$ is the same for both stones, $$u=0$$ for first stone
$$v^2=u^2+2as$$
$$v^2=0+2(9.8)(49)$$
$$v=31 m/s$$
$$v=u+at$$
$$31=0+9.8t$$
$$t=3.16 s$$
$$t'=2.16 s$$
$$s=ut+0.5at^2$$
$$49=u'(2.16)+0.5(9.8)(2.16)^2$$
$$u'=12.1 m/s$$
Question 5
1 / -0

According to the following graph, what happens to the distance covered by the body from 0 -10 minutes?

A
It goes on increasing

B
It goes on decreasing

C
It first increases and then decreases

D
It first decreases and then increases

Solution

We know that distance traveled by an object is the area under it speed time graph.
Now, in this case, as the area under the speed-time graph is increasing from 0-10 minutes. So, the distance will keep on increasing from 0-10 minutes.

Hence, correct answer is option $$A$$
Question 6
1 / -0

The table below shows the speed of a moving vehicle with respect to time.
Speed (m/s)
2
4
6
8
10
Time (s)
1
2
3
4
5
Find the acceleration of the vehicle.

A
$$2 m/s^2$$

B
$$4 m/s^2$$

C
$$6 m/s^2$$

D
$$8 m/s^2$$

Solution

From the given data, $$v=2t$$
Acceleration, $$a = \dfrac{v_2-v_1}{t_2-t_1}$$
$$= \dfrac{2(t_2-t_1)}{t_2-t_1}\ = 2\ m/s^2$$
Question 7
1 / -0

A stone is thrown vertically upward with an initial velocity of $${40 ms^{-1}}$$,Taking g=$${10 ms^{-2}}$$, find the maximum height reached by the stone.

A
$$100 m$$

B
$$60 m$$

C
$$80 m$$

D
$$120 m$$

Solution

$$v^2=u^2+2as$$

$$0=40^2+2(-10)s$$

$$s=\dfrac{40\times40}{2\times10}$$

$$s=80 m$$
Question 8
1 / -0

A ball is thrown vertically upwards with a velocity of $${20 ms^{-1}}$$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take $${g= 10 ms^{-2}}$$.

A
t=5s

B
t=10s

C
t=15s

D
t=20s

Solution

Time to reach maximum height can be obtained from $$v=u+at$$
$$0=20+(-10)t$$
$$t=2 s$$
$$s=ut+0.5at^2=20(2)+0.5(-10)(2)^2= 20 m$$
Thus, total distance for maximum height is 45 m
$$s=ut+0.5at^2$$
$$45=0+0.5(10)(t')^2$$
$$t'=3 s$$
Total time= 3+2= 5s
Question 9
1 / -0

A car starts from rest and uniformly accelerates in a straight line. In the first second the car covers a distance of $$2$$ m. The velocity of the car at the end of $$2^{nd}$$ $$second$$ will be

A
$$\displaystyle 4.0\ ms^{-1}$$

B
$$\displaystyle 8.0\ ms^{-1}$$

C
$$\displaystyle 6\ ms^{-1}$$

D
None of these

Solution

distance covered by car in t seconds$$ =0*t+\frac { 1 }{ 2 } a{ t }^{ 2 }$$
for $$t=1$$ distance is 2m.
$$2 = \frac { 1 }{ 2 } a*{ 1 }^{ 2 }\\ a=4 m/s^2$$
velocity after 2 seconds.
$$v= u +at\\=0+4*2 = 8m/s$$
Question 10
1 / -0

A stone is allowed to fall from the top of a tower $$100 m$$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $${25 ms ^{-1}}$$.Calculate where the two stones will meet.(Take $${g=10 m/s^2}$$).

A
10 m

B
40 m

C
20 m

D
50 m

Solution

For stone moving downward,acceleration due to gravity ,$${g=10 ms^{-2}}$$.
Initial velocity (u)$$=$$o;distance, s$$=$$100-x;time, t$$=$$?
$${s=ut+\dfrac{1}{2}gt^2}$$, or (100-x)=$${0\times t+\dfrac{1}{2}\times 10 t^2}$$
$${\Rightarrow 100 -x=5t^2}$$ .....(1)
For the stone moving vertically upwards,
For the stone moving vertically upwards,initial velocity, $${u=25 ms ^{-1}}$$;time (t)=?;
acceleration due to gravity $${g=-10 ms^{-1}}$$
[In upward direction ,g is taken -ve]
Distance, $$s=x$$
we know $${s-ut+\dfrac{1}{2} gt^2}$$ or
$${x=25\times t +\dfrac{1}{2}\times (-10 t^2)}$$
$${\Rightarrow x=25 t-5t^2}$$ .....(2)
Substituting the value of x from (2) in (1) we get,
$${100-(25t-5t^2)=5t^2}$$
$${100-25 t+5t^2=5t^2}$$
$$25 t=100$$ or $$t=4s$$
Put the value of tin (1) $${\Rightarrow 100 -x =5(4)^2}$$
$${\Rightarrow 100-x=80}$$ or x$$=$$20 m
$${\therefore}$$ The stones will meet at a height of 20m

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Motion in A Straight Line Test - 30

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Motion in A Straight Line Test - 30

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SHARING IS CARING

Question 1

$$15 \,m \,s^{-1}$$

$$10 \,m \,s^{-1}$$

$$30 \,m \,s^{-1}$$

$$0 \,m \,s^{-1}$$

Solution

Question 2

(i) 122.5 m, (ii) 10 s

(i) 120.5 m, (ii) 10 s

(i) 122.5 m, (ii) 20 s

(i) 120 m, (ii) 10 s

Solution

Question 3

20 m/s

40 m/s

35 m/s

30 m/s

Solution

Question 4

$$12.1 m/s$$

$$10 m/s$$

$$21.1 m/s$$

$$15.1 m/s$$

Solution

Question 5

It goes on increasing

It goes on decreasing

It first increases and then decreases

It first decreases and then increases

Solution

Question 6

$$2 m/s^2$$

$$4 m/s^2$$

$$6 m/s^2$$

$$8 m/s^2$$

Solution

Question 7

$$100 m$$

$$60 m$$

$$80 m$$

$$120 m$$

Solution

Question 8

t=5s

t=10s

t=15s

t=20s

Solution

Question 9

$$\displaystyle 4.0\ ms^{-1}$$

$$\displaystyle 8.0\ ms^{-1}$$

$$\displaystyle 6\ ms^{-1}$$

None of these

Solution

Question 10

10 m

40 m

20 m

50 m

Solution

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