Motion in A Straight Line Test

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Motion in A Straight Line Test - 34

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Question 1
1 / -0

A body is projected horizontally from a certain height, (h) then time of descent is

A
$$\displaystyle { t }_{ d }=\sqrt { \frac { 2h }{ g } } $$

B
$$\displaystyle { t }_{ d }=\sqrt { \frac { h }{ g } } $$

C
$$\displaystyle { t }_{ d }=\sqrt { h g} $$

D
$$\displaystyle { t }_{ d }=\sqrt { h + g} $$

Solution

When a body is projected horizontally, its initial velocity in the vertical direction is zero i.e. $$u_y = 0$$
Displacement of body in vertical direction $$S_y = h$$
Acceleration due to gravity in vertical direction $$a_y = g = 10 \ m/s^2$$
Using $$S_y = u_y t +\dfrac{1}{2}a_yt^2$$
$$\therefore$$ $$h = 0+\dfrac{1}{2}gt_d^2$$
$$\implies \ t_d = \sqrt{\dfrac{2h}{g}}$$
Question 2
1 / -0

The distance travelled by a body starting with a velocity of $$\displaystyle 20\ { ms }^{ -1 }$$, and moving with an acceleration of $$\displaystyle 2\ { ms }^{ -2 }$$, in the $$\displaystyle { 8 }^{ th }$$ second is ____ m.

A
35

B
20

C
120

D
10

Solution

Initial velocity $$u = 20 \ m/s$$
Acceleration   $$a = 2 \ m/s^2$$
Distance covered in $$8^{th}$$ second   $$S_t = u+\dfrac{a}{2}(2t-1)$$
where $$t = 8 \ s$$
Thus,    $$S_8 = 20+\dfrac{2}{2}(2\times 8-1) = 35 \ m$$
Question 3
1 / -0

A stone thrown vertically passes a certain point P at the end of 2 seconds and 8 seconds respectively. Find the maximum height reached by the stone .$$\displaystyle (Take\quad g ={ 10\quad ms }^{ -2 }) $$

A
625 m

B
125 m

C
225 m

D
350 m

Solution

Let initial velocity $$u$$ and point 'P' is at height $$h$$.
At the end of 2 seconds,
Using $$s=ut+\frac{1}{2}at^2$$ $$\Rightarrow h= u\times 2 - \frac{1}{2}g\times 2^2$$ $$\Rightarrow h= 2u-20$$ .......1
At the end of 8 seconds,
$$\Rightarrow h= u\times 8- \frac{1}{2}g\times 8^2$$ $$\Rightarrow h= 8u-320$$ ....2
Solving equations 1 and 2, $$\Rightarrow u=50 {m}/{s}$$,
At maximum height , velocity $$v=0$$,
Using $$v^2=u^2+2as$$ $$\Rightarrow 0= 50^2-2\times 10\times s$$ $$\Rightarrow s=125m$$,
$$\therefore $$ Maximum height $$= 125 m$$
Question 4
1 / -0

A stone is released from a hot air balloon which is rising steadily with a velocity of $$\displaystyle { 4\quad ms }^{ -1 }$$. The velocity of the stone at the end of 3 s after it is released is _____ $$\displaystyle { ms }^{ -1 }$$

A
29.4

B
25.4

C
32.5

D
62.7

Solution

Let the upward direction to be negative.
Initial velocity of stone $$u = -4 \ m/s$$
Acceleration of the stone $$a = g = 9.8 \ m/s^2$$
Velocity of stone $$v = u+at$$
$$\therefore$$ $$v = -4+9.8\times 3 = 25.4 \ m/s$$
Question 5
1 / -0

A body starts moving with an initial velocity of $$\displaystyle 5\ { ms }^{ -1 }$$ and an acceleration of $$\displaystyle 1\ { ms }^{ -2 }$$. The distance travelled by it in the 5th second is

A
$$9.5$$ m

B
$$22.5$$ m

C
$$50$$ cm

D
$$10$$ cm

Solution

Initial velocity $$u = 5 \ m/s$$
Acceleration   $$a = 1 \ m/s^2$$
Distance covered in $$5th$$ second,   $$S_5 = u+\dfrac{1}{2}a(2t-1)$$
where $$t = 5 \ s$$
$$\therefore$$   $$S_5 = 5+\dfrac{1}{2}(2\times 5-1) = 9.5 \ m$$
Question 6
1 / -0

A body starts from rest and moves with uniform acceleration for 2s. If then decelerates uniformly for 3s and stops. If deceleration is $$\displaystyle { 4\quad ms }^{ -2 }$$, the acceleration of the body is ____ $$\displaystyle { ms }^{ -2 }$$

A
10

B
8.7

C
4

D
6

Solution

Given Final Velocity v=0, deceleration $$a=4ms^{-2}$$ and deceleration time $$t=3 s$$,
Let $$u$$ be the velocity before the deceleration is started.
Using $$v=u+at$$ $$\Rightarrow 0=u -4\times 3$$ $$\Rightarrow u=12, ms^{-1}$$,
Now , During acceleration, Initial velocity $$u_1$$=0, time $$t= 2s $$, Final velocity $$u=12ms^{-1}$$,
Let acceleration is $$a$$.
Using $$v=u+at$$ $$\Rightarrow 12=0+a\times 2$$ $$\Rightarrow a= 6 ms^{-2}$$
Question 7
1 / -0

The ratio of the time taken by a body moving with uniform acceleration in reaching two points P and Q along a straight line path is 1 : 2. If the body starts from rest and moves linearly, the ratio of the distance between P and Q from the starting point is:

A
4 : 1

B
1 : 4

C
2 : 3

D
3 : 1

Solution

The ratio of time taken is 1 :2,
Using $$s= ut+ \dfrac{1}{2} gt^2$$,
The ratio of distance $$\dfrac{s_1}{s_2}= \dfrac{\dfrac{1}{2}gt_1^2}{\dfrac{1}{2}gt_2^2}=\dfrac{t_1^2}{t_2^2}$$, ( u=0 given)
$$\Rightarrow \dfrac{s_1}{s_2}=\dfrac{1}{4}$$
Question 8
1 / -0

A hollow iron ball (A) and a solid iron ball (B) and cricket ball (C) are dropped from the same height. Which among the three balls reaches the ground first? Assuming there is no resistance due to air.

A
A

B
B

C
C

D
All the three balls reaches ground simultaneously

Solution

Since all the balls have have zero velocity initially and they experience equal acceleration due to gravity and travel equal distance. Thus all the balls take equal time to reach the ground.
Question 9
1 / -0

A body starts from rest and moves with uniform acceleration for 3s. It then decelerates uniformly for 2s. and stops. If the deceleration is $$\displaystyle 3 \ { ms }^{ -2 }$$ the maximum velocity of the body is ____ $$\displaystyle { ms }^{ -1 }$$

A
zero

B
2

C
6

D
Cannot be determined

Solution

Velocity will be maximum before the deceleration starts.
Given Final Velocity v=0, acceleration $$a=-3ms^{-2}$$ and time $$t=3 s$$,
Let maximum velocity is $$u_{max}$$,
Using $$v=u+at$$ $$\Rightarrow 0=u_{max} -3\times 2$$ $$\Rightarrow u_{max}=6 ms^{-1}$$
Question 10
1 / -0

A body falls from a height of $$45$$ m above the ground. Find the time taken by the body to reach the ground. $$\displaystyle (Take \ g = 10 { ms }^{ -2 })$$

A
$$5$$ s

B
$$2$$ s

C
$$3$$ s

D
$$6$$ s

Solution

Initial speed $$u = 0 \ m/s$$
Acceleration $$a = g = 10\ m/s^2$$
Height $$h = 45 \ m$$
Using $$h = ut+\dfrac{1}{2}at^2$$
Or $$45 = 0+\dfrac{1}{2}(10)t^2$$
$$\implies \ t = 3 \ s$$

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Motion in A Straight Line Test - 34

Result

Motion in A Straight Line Test - 34

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SHARING IS CARING

Question 1

$$\displaystyle { t }_{ d }=\sqrt { \frac { 2h }{ g } } $$

$$\displaystyle { t }_{ d }=\sqrt { \frac { h }{ g } } $$

$$\displaystyle { t }_{ d }=\sqrt { h g} $$

$$\displaystyle { t }_{ d }=\sqrt { h + g} $$

Solution

Question 2

35

20

120

10

Solution

Question 3

625 m

125 m

225 m

350 m

Solution

Question 4

29.4

25.4

32.5

62.7

Solution

Question 5

$$9.5$$ m

$$22.5$$ m

$$50$$ cm

$$10$$ cm

Solution

Question 6

10

8.7

4

6

Solution

Question 7

4 : 1

1 : 4

2 : 3

3 : 1

Solution

Question 8

A

B

C

All the three balls reaches ground simultaneously

Solution

Question 9

zero

2

6

Cannot be determined

Solution

Question 10

$$5$$ s

$$2$$ s

$$3$$ s

$$6$$ s

Solution

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