Motion in A Straight Line Test

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Motion in A Straight Line Test - 60

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Weekly Quiz Competition

Question 1
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The following figure gives the movement of an object. Select the correct statement from the given choices.

A
The total distance travelled by the object is $$975\ m$$

B
The maximum acceleration of the object is $$2\ ms^{-2}$$

C
The maximum deceleration happened between $$25th$$ and $$85th$$ seconds

D
The object was at rest between $$10th$$ and $$15th$$ seconds

E
At $$40th$$ second, the speed of object was decelerating

Solution
Question 2
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A body starts from rest and is uniformly accelerated for $$30s$$. The distance travelled in the first $$10s$$ is $$x_{ 1 }$$ next $$10s$$ is $$x_{ 2 }$$ and the last $$10s$$ is $$x_{ 3 }$$.Then $$x_{ 1 }:x_{ 2 }:x_{ 3 }$$ is the same as :-

A
$$1:2:4$$

B
$$!:2:5$$

C
$$1:3:5$$

D
$$1:3:9$$

Solution

$$u = 0 $$
$$s = ut + \dfrac{1}{2}at^2 \\$$
$$s = o*t + \dfrac{1}{2}at^2$$
$$s = \dfrac{1}{2}at^2$$
In first 10 sec, travelled distance $$x_1 = \dfrac{1}{2}a(10)^2 \\ x_1 = 50a$$
After next 10 sec, travelled distance $$x_2 = \dfrac{1}{2}a(10+10)^2 - x_1\\ x_2 = \dfrac{1}{2}a*(400) - 50 a \\ x_2 = 150a$$
In last 10 sec, travelled distance $$x_3 = \dfrac{1}{2}a(10+10+10)^2 - x_1 - x_2 \\ x_3 = 450a - 50a - 150a \\x_3 = 250a \\$$
$$x_1:x_2:x_3 = 50a : 150a: 250a $$
$$x_1:x_2:x_3 = 1 : 3: 5 $$
Question 3
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A parachutist drops first freely from an aeroplane for 10 s and then his parachutist opens out. Now he descends with a net retardation of $$2.5ms^{-2}$$. If he bails out of the plane at a height of 2495 m and $$g = 10 ms^{-2}$$, his velocity on reaching the ground will be

A
$$5 ms^{-1}$$

B
$$10 ms^{-1}$$

C
$$15 ms^{-1}$$

D
$$20 ms^{-1}$$

Solution

The velocity v acquired by the parachutist after 10 s:
$$v=u+gt=0+10\times 10=100ms^{-1}$$
Then, $$s_1=ut+\dfrac{1}{2}gt^2=0+\dfrac{1}{2}\times 10\times 10^2=500m$$
The distance travelled by the parachutist under retardation is
$$s_2=2495-500=1995m$$
Let $$v_g$$ be the velocity on reaching the ground. Then $$v_{ g }^{ 2 }-v^{ 2 }=2as_{ 2 }$$
or $$v_{ g }^{ 2 }-(100)^2=2\times(-2.5)\times1995$$ or $$v_g=5ms^{-1}$$
Question 4
1 / -0

A person moves 30 m north and then 20 m towards east and finally $$30\sqrt { 2 } $$ m in south-west direction. The displacement of the person from the origin will be

A
10 m along north

B
10 m along south

C
10 m along west

D
Zero

Solution

$$AB=30 m, BC=20 m$$,
$$CD=30\sqrt { 2 } m$$
$$DBCE$$ is isosceles (Fig.), so
$$BE=BC=20m$$
$$AE=AB-BE=30-20=10 m$$
$$EC=20\sqrt { 2 } m$$
$$ED=CD-EC=30\sqrt { 2 } -20\sqrt { 2 } =10\sqrt { 2 } m$$
$$DADE$$ is isosceles, so$$ AD=AE=10 m$$
so the displacement is $$10m$$ due west
Question 5
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Directions For Questions

A car accelerates from rest at a constant rate $$\alpha $$ for some time and then decelerates at a constant rate $$\beta $$ to come to rest. The total time elapsed is t.

...view full instructions

The maximum velocity attained by the car is

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

B
$$\dfrac { \alpha \beta }{ \alpha +\beta } t$$

C
$$\dfrac { 2\alpha \beta }{ \alpha +\beta } t$$

D
$$\dfrac { 4\alpha \beta }{ \alpha +\beta } t$$

Solution

From A to B, applying $$v = u + at$$, we get
$$v_0=0+at\Rightarrow t_1=v_0/a$$
From B to C, again applying v = u + at, we get
$$0=v_0-bt_2\Rightarrow t_2=v_0/b$$
Given $$t_1+t_2=t\Rightarrow$$ $$\dfrac{v_0}{\alpha }+\dfrac{v_0}{\beta }$$ $$=t\Rightarrow v_0=\dfrac{\alpha \beta t}{\alpha+\beta}$$
$$v_0$$ is the maximum velocity attained.
Question 6
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Directions For Questions

The displacement of a body is given by $$4s=M+2Nt^4,$$ where M and N are constants.

...view full instructions

The velocity of the body at any instant is

A
$$\dfrac { M+{ 2 }Nt^{ 4 } }{ 4 } $$

B
$$2N$$

C
$$\dfrac { M+{ 2 }N }{ 4 } $$

D
$$2Nt^3$$

Solution

$$4s=m+2nt^{4}$$
$$s=\dfrac{m+2nt^{4}}{4}$$
$$s=\dfrac{m}{4}+\dfrac{2nt^{4}}{4}$$
$$v=\dfrac{ds}{dt}=\dfrac{dm}{dt^{4}}+\dfrac{2nt^{4}}{4dt}$$
$$\dfrac{ds}{dt}=0+\dfrac{4.2nt^{3}}{4}$$ Differentiation of any constant is 0
$$v=\dfrac{ds}{dt}=2nt^{3}$$
Question 7
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A point moves with uniform acceleration and $$v_1,v_2,$$ and $$v_3$$ denote the average velocities in the three successive intervals of time $$t_1,t_2,$$ and $$t_3$$. Which of the following relations is correct?

A
$$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2+t_3)$$

B
$$(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$$

C
$$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_1-t_3)$$

D
$$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2-t_3)$$

Solution

Suppose u be the initial velocity.
Velocity after time $$t_1:v_{11}=u+at_1$$
Velocity after time $$t_1+t_2:v_{22}=u+a(t_1+t_2)$$
Velocity after time $$t_1+t_2+t_3:$$ $$v_33=u+a(t_1+t_2+t_3)$$
Now $$v_1=\dfrac { u+v_{ 11 } }{ 2 } =\dfrac { u+u+at_{ 1 } }{ 2 } =u+\dfrac { 1 }{ 2 } at_{ 1 }$$
$$v_2=\dfrac{v_{11}+v_{22}}{2}=u+at_1+\dfrac{1}{2}at^2$$
$$v_3=\dfrac{v_{22}+v_{33}}{2}=u+at_1+at_2\dfrac{1}{2}at^3$$
So $$v_1-v_2=-\dfrac{1}{2}a(t_1+t_2)$$
$$v_2-v_3=--\dfrac{1}{2}a(t_2+t_3)$$
$$(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$$
Question 8
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Directions For Questions

A body is moving with uniform velocity of 8$$ms^{-1}$$. When the body just crossed another body, the second one starts and moves with uniform acceleration of 4$$ms^{-2}$$.

...view full instructions

The distance covered by the second body when they meet is

A
8 m

B
16 m

C
24 m

D
32 m

Solution

Let at time t, both bodies will meet.
distance traveled by body A = distance traveled by body B
$$u_1* t = u_2t + \dfrac{1}{2} at^2$$
$$8* t = 0 * t + \dfrac{1}{2}* 4 *t^2$$
$$ 8t = 2 t^2$$
$$ t = 4 s$$
Distance covered by body B $$= u_2t^2 + \dfrac{1}{2}at^2 \\ = 0*t + \dfrac{1}{2} * 4 *(4)^2 \\ = 32 m$$
Option 'D' is correct.
Question 9
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Directions For Questions

The displacement of a body is given by $$4s=M+2Nt^4,$$ where M and N are constants.

...view full instructions

The velocity of the body at the end of 1 s from the start is

A
$$2N$$

B
$$\dfrac{M+2N}{4}$$

C
$$2(M+N)$$

D
$$\dfrac{2M+N}{4}$$

Solution

$$s=\dfrac{m+2Nt^4}{4}$$ $$\Rightarrow v=\dfrac{ds}{dt}=2Nt^3$$
Putting $$t=1 s$$, we get $$v = 2N$$
Question 10
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Directions For Questions

A bus starts moving with acceleration 2$$ms^{-2}$$. A cyclist 96 m behind the bus starts simultaneously towards the bus at a constant speed of 20 $$ms^{-1}.$$

...view full instructions

After what time will he be able to overtake the bus?

A
4 s

B
8 s

C
12 s

D
16 s

Solution

Let at time t, the cyclist overtake the bus, then $$96+$$ (distance travelled by bus in time t) = (distance travelled by cyclist in time t)
$$\Rightarrow\dfrac{1}{2}\times2\times t^2+96=20\times t \Rightarrow t^2-20t+96=0$$ This gives $$t =8$$ s or $$12 s$$. Hence, the cyclist will overtake the bus at $$8 s.$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 60

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Motion in A Straight Line Test - 60

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SHARING IS CARING

Question 1

The total distance travelled by the object is $$975\ m$$

The maximum acceleration of the object is $$2\ ms^{-2}$$

The maximum deceleration happened between $$25th$$ and $$85th$$ seconds

The object was at rest between $$10th$$ and $$15th$$ seconds

At $$40th$$ second, the speed of object was decelerating

Solution

Question 2

$$1:2:4$$

$$!:2:5$$

$$1:3:5$$

$$1:3:9$$

Solution

Question 3

$$5 ms^{-1}$$

$$10 ms^{-1}$$

$$15 ms^{-1}$$

$$20 ms^{-1}$$

Solution

Question 4

10 m along north

10 m along south

10 m along west

Zero

Solution

Question 5

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

$$\dfrac { \alpha \beta }{ \alpha +\beta } t$$

$$\dfrac { 2\alpha \beta }{ \alpha +\beta } t$$

$$\dfrac { 4\alpha \beta }{ \alpha +\beta } t$$

Solution

Question 6

$$\dfrac { M+{ 2 }Nt^{ 4 } }{ 4 } $$

$$2N$$

$$\dfrac { M+{ 2 }N }{ 4 } $$

$$2Nt^3$$

Solution

Question 7

$$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2+t_3)$$

$$(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$$

$$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_1-t_3)$$

$$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2-t_3)$$

Solution

Question 8

8 m

16 m

24 m

32 m

Solution

Question 9

$$2N$$

$$\dfrac{M+2N}{4}$$

$$2(M+N)$$

$$\dfrac{2M+N}{4}$$

Solution

Question 10

4 s

8 s

12 s

16 s

Solution

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