Motion in A Straight Line Test

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Motion in A Straight Line Test - 58

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Question 1
1 / -0

A ball is thrown up under gravity $$\left( g = 10 \mathrm { m } / \mathrm { sec } ^ { 2 } \right) .$$ Find its velocity after $$1.0 sec $$ at a height of 10$$\mathrm { m }$$

A
$$5$$ $$\mathrm { m } / \mathrm { sec } ^ { 2 }$$

B
$$5$$ $$\mathrm { m } / \mathrm { sec }$$

C
$$10$$ $$\mathrm { m } / \mathrm { sec }$$

D
$$15$$ $$\mathrm { m } / \mathrm { sec }$$
Question 2
1 / -0

A stone drop from height 'h' on Earth surface fall in 1 sec. If the same stone taken to Moon and drop freely then it will reaches from the surface of the Moon in the time (The 'g' of Moon is 1$$/ 6$$ times ofEarth):-

A
$$\sqrt { 6 }$$ second

B
9 second

C
$$\sqrt { 3 }$$ second

D
6 second

Solution
Question 3
1 / -0

In case of a freely falling body, the ratio of kinetic energy at the end of the third second increase in kinetic energy in the next three seconds is

A
$$1:1$$

B
$$1:2$$

C
$$1:3$$

D
$$1:9$$

Solution
Question 4
1 / -0

The figure represents the velocity-time graph of body moving in a straight line. How much distance does it travel during the last 10 seconds?

A
20 m

B
100 m

C
50 m

D
200 m

Solution

Distance travelled in last $$10$$ sec, is $$d = velocity \times time$$, which is nothing but area under the curve between the limit $$t = 10s$$ to $$t = 20s$$
So, area of triangle is $$= \dfrac {base \times height}{2} = \dfrac {(20 - 10)\times 20}{2} = 100\ m$$
Hence, correct option is (B).
Question 5
1 / -0

A body released from the top of a tower of height h takes time 't' to reach the ground. After time t/2, its height from the ground is

A
$$h/2$$

B
$$h/4$$

C
$$3h/4$$

D
$$h/3$$

Solution
Question 6
1 / -0

A particle moving with constant acceleration describes in the last second of its motion $$\dfrac { 5 }{ 9 } $$ of the total distance. If it stats from rest it is in motion for a time.

A
1s

B
2s

C
3s

D
4s

Solution
Question 7
1 / -0

A body moving with uniform acceleration covers 48m in the $${ 10 }^{ th }$$ second and 56m in the $${ 12 }^{ th }$$. Distance moved in the $${ 11 }^{ th }$$ second is

A
50m

B
51m

C
52m

D
53m
Question 8
1 / -0

How long will a train, running at a speed of 45 kmph cross a standing man, given the length of the train is 450m?

A
100 sec

B
150 sec

C
50 sec

D
36 sec

Solution
Question 9
1 / -0

A 640- N hunter gets a rope around a 3200-N polar bear. They are stationary, 30 m apart, on frictionless ice. When the hunter pulls the polar bear to meet him, the polar bear will move:

A
24 m

B
15 m

C
20 m

D
5 m

Solution
Question 10
1 / -0

A particle starts from rest with a uniform acceleration, It travels a distance X is first two second & a distance y in the next two seconds.Then:-

A
Y = x

B
Y = 2x

C
Y = 3x

D
Y = 4 x

Solution

$$ \begin{array}{l} \text { Correct option is (c) } \\ \text { Let acceleration = a } \\ \text { As we know, } \\ \text { distance in first two seconds } \\ \qquad \begin{array}{l} s_{2}=v t+\frac{1}{2} a t^{2} \\ s_{2}=\frac{1}{2} x a \times(2)^{2} \end{array} \\ s_{2}=2 a=x \end{array} $$

$$ \begin{array}{l} \text { distance in next two seconds = } \\ \qquad \begin{aligned} y &=s_{4}-s_{2} \\ &=\frac{1}{2} x a x(4)^{2}-2 a \\ \frac{y}{y} &=8 a-2 a \\ \text { so, } y &=3 x \end{aligned} \end{array} $$