Motion in A Straight Line Test

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Motion in A Straight Line Test - 31

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Question 1
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An object is sliding down on an inclined plane. The velocity changes at a constant rate from 10 cm/s to 15 cm/s in 2 seconds. What is its acceleration?

A
$5 cm/s^2$

B
$7.5 cm/s^2$

C
$2.5 cm/s^2$

D
$12.5 cm/s^2$

Solution

$a=\dfrac { v-u }{ t } =\dfrac {15-10}{2} =2.5\ cm/{ s }^{ 2 }$

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Question 2
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Consider two observers moving with respect to each other at a speed $v$ along a straight line. They observe a block of mass $m$ moving a distance $l$ on a rough surface. The following quantities will be same as observed by the two observers.

A
Kinetic energy of the block at time t

B
Work done by friction

C
Total work done on the block

D
Acceleration of the block

Solution

acceleration of the block is same in both frames as they are only moving in different velocities (but constant).
KE is dependent of $v_{rel}$ . Work done on block by forces changes as position vector changes differently in each frame.

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Question 3
1 / -0

If a car at rest accelerates uniformly to a speed of $144\ km/h$ in $20$ second, it covres a distance of :-

A
$20\ m$

B
$400\ m$

C
$1440\ m$

D
$2980\ m$

Solution

speed of car after 20 sec. $=144\quad km/h\quad =\quad 144*\frac { 5 }{ 18 } \quad m/s=40m/s$
$v=u+at\\40=0+20a\\a=2m/s^2$
Distance covered $s=ut+1/2 at^2\\=0*20+\frac 12 *2*{20}^2\\=400 m$

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Question 4
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The unit for the rate of change of velocity is

A
$m s^{-2}$

B
$m s^{-3}$

C
$m s^{-1}$

D
$m^{3} s^{-2}$

Solution

The rate of change of velocity with respect to time is called acceleration.
GIven by
$a=\dfrac{change\quad in\quad velocity}{time}$
Putting units of velocity and time.
$a=\dfrac{m/s}{s}$ = $m/s^2$ or $ms^{-2}$ .Hence this is the unit of acceleration or rate of change of velocity with respect to time.

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Question 5
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A body starts from rest is moving under a constant acceleration up to 20 s . If it moves $S_{1}$ Distance in first 10 s, and $S_{2}$ distance in next 10 s then $S_{2}$ will be equal to:

A
$S_{1}$

B
$2S_{1}$

C
$3S_{1}$

D
$4S_{1}$

Solution

$\displaystyle s_{1}=0+\frac{1}{2}a(10)^{2}=50a.............(i)$
after $10s\;\;\;\;\;\;\;v=10a.....(ii)$
Now $\displaystyle s_{2}= (10a)t+\frac{1}{2}a(10)^{2}$
$\displaystyle s_{2}= (10a)10+\frac{1}{2}a(100)$
$s_{2}=100a +50a=150a=3(50a)\Rightarrow s_{2}=3s_{1}$

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Question 6
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An aeroplane is flying in a horizontal direction at $600 km/hr$ at a height of $6 kms$ and is advancing towards a point which is exactly over a target on earth. At that instant the pilot releases a ball which on descending the earth strike the target. The falling ball appears-

A
To the pilot in the aeroplane, as falling along a parabolic path

B
To a person standing near the target, as falling exactly vertical

C
To a person standing near the target, as describing a parabolic path

D
To the pilot sitting in the aeroplane, as falling in a zigzag path

Solution

Since pilot is moving with same horizontal velocity as the ball so for him it appears to be falling exactly vertical.
To the person near the target it is like any other horizontal projectile from above the ground so it will be parabolic. (C) is correct

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Question 7
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A particle travels $10\ m$ in first $5\ s$ , $10\ m$ in next $3\ s$ . Assuming constant acceleration, what is the distance traveled in the next two seconds.

A
$\displaystyle \frac{20}{3}\ m$

B
$20\ m$

C
$60\ m$

D
$180\ m$

Solution

In the first $5$ sec:
$s = ut+0.5at^2$
$10 = 5u + 0.5a \times 25$

In the next $3$ sec:
$20 = 8u + 0.5a \times 64$

Solving above two equations will give us:
$u = \dfrac{7}{6} , a = \dfrac{1}{3}$

Therefore, in next $2$ sec distance traveled is:
$S = \dfrac{7}{6} \times 10 + 0.5 \times \dfrac{1}{3} \times 100 = \dfrac{170}{6}$

Therefore distance travelled in next $2$ sec is $=\dfrac{170}{6}-20 = \dfrac{20}{3}$

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Question 8
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A body travels $200\ cm$ in the first two seconds and $220\ cm$ in the next $4\ s$ with same acceleration.The velocity of the body at the end of the $7^{th}$ second is

A
$5\ cm/s$

B
$10\ cm/s$

C
$15\ cm/s$

D
$20\ cm/s$

Solution

$200=u\times 2+\dfrac{1}{2}a(2)^{2}$

$100=u+a..........(i)$

$420=u(6)+\dfrac{1}{2}a(6)^{2}$

$70=u+3a..........(ii)$

from $(i)$ and $(ii)$ $2a=-30\Rightarrow a=-15cm/s^{2}$

$\therefore u=100-a=100-(-15)=115\;cm/s$

$u_{7^{th}}=u+at=115+(-15)(7)=115-105=10\;cm/sec$

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Question 9
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A truck of mass $2800 kg$ is moving with a speed of $15\:m/s$ . A frictional retarding force of $1200 N$ are acting on it, then in $10 s$ it shall travel a distance of:

A
$156 m$

B
$122.8 m$

C
$162.5 m$

D
$118 m$

Solution

Acceleration of the truck: $a=\displaystyle \frac{1200-500}{2800}=\frac{1}{4}m/s^{2}$
$\displaystyle S=ut+\frac{1}{2}at^{2}$
$=\displaystyle 15\times 10+\frac{1}{2}\times \frac{1}{4}\times 100=162.5\:m$ .

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Question 10
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Directions For Questions

Comprehension # 2
Two ships, $V$ and $W$ , move with constant velocities $2\:ms^{-1}$ and $4\:ms^{-1}$ along two mutually perpendicular straight tracks toward the intersection point $O$ . At the moment $t=0$ , the ships were located at distances $100\:m$ and $200\:m$ from the point $O$ .

...view full instructions

The distance between them at time $t$ is:-

A
$\;\sqrt{(200)^2+(100)^2}\:m$

B
$\;\sqrt{(200-4t)^2+(100-2t)^2}\:m$

C
$\;[(200-4\,t)+(100-2\,t)]\:m$

D
$\;\sqrt{(200-2\,t)^2+(100-4\,t)^2}\:m$

Solution

Given : $V_V = 2 m/s$ $V_W =4 m/s$
$\therefore$ Distance traveled by ship V in time $t$ , $l_1 = 2t$
Also distance traveled by ship W in time $t$ , $l_2 = 4t$
$\implies$ $OA = 100 - 2t$ and $OB = 200 -4t$

$\therefore$ Distance between them at time $t$ , $AB = L =\sqrt{OA^2 + OB^2}$
$\implies$ $L = \sqrt{(100-2t)^2 + (200 -4t)^2}$ m

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 31

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Motion in A Straight Line Test - 31

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SHARING IS CARING

Question 1

5cm/s25 cm/s^25cm/s2

7.5cm/s27.5 cm/s^27.5cm/s2

2.5cm/s22.5 cm/s^22.5cm/s2

12.5cm/s212.5 cm/s^212.5cm/s2

Solution

Question 2

Kinetic energy of the block at time t

Work done by friction

Total work done on the block

Acceleration of the block

Solution

Question 3

20 m20\ m20 m

400 m400\ m400 m

1440 m1440\ m1440 m

2980 m2980\ m2980 m

Solution

Question 4

ms−2m s^{-2}ms−2

ms−3m s^{-3}ms−3

ms−1m s^{-1}ms−1

m3s−2m^{3} s^{-2}m3s−2

Solution

Question 5

S1S_{1}S1​

2S12S_{1}2S1​

3S13S_{1}3S1​

4S14S_{1}4S1​

Solution

Question 6

To the pilot in the aeroplane, as falling along a parabolic path

To a person standing near the target, as falling exactly vertical

To a person standing near the target, as describing a parabolic path

To the pilot sitting in the aeroplane, as falling in a zigzag path

Solution

Question 7

203 m\displaystyle \frac{20}{3}\ m320​ m

20 m20\ m20 m

60 m60\ m60 m

180 m180\ m180 m

Solution

Question 8

5 cm/s5\ cm/s5 cm/s

10 cm/s10\ cm/s10 cm/s

15 cm/s15\ cm/s15 cm/s

20 cm/s20\ cm/s20 cm/s

Solution

Question 9

156m156 m156m

122.8m122.8 m122.8m

162.5m162.5 m162.5m

118m118 m118m

Solution

Question 10

(200)2+(100)2 m\;\sqrt{(200)^2+(100)^2}\:m(200)2+(100)2​m

(200−4t)2+(100−2t)2 m\;\sqrt{(200-4t)^2+(100-2t)^2}\:m(200−4t)2+(100−2t)2​m

[(200−4 t)+(100−2 t)] m\;[(200-4\,t)+(100-2\,t)]\:m[(200−4t)+(100−2t)]m

(200−2 t)2+(100−4 t)2 m\;\sqrt{(200-2\,t)^2+(100-4\,t)^2}\:m(200−2t)2+(100−4t)2​m

Solution

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$5 cm/s^2$

$7.5 cm/s^2$

$2.5 cm/s^2$

$12.5 cm/s^2$

$20\ m$

$400\ m$

$1440\ m$

$2980\ m$

$m s^{-2}$

$m s^{-3}$

$m s^{-1}$

$m^{3} s^{-2}$

$S_{1}$

$2S_{1}$

$3S_{1}$

$4S_{1}$

$\displaystyle \frac{20}{3}\ m$

$20\ m$

$60\ m$

$180\ m$

$5\ cm/s$

$10\ cm/s$

$15\ cm/s$

$20\ cm/s$

$156 m$

$122.8 m$

$162.5 m$

$118 m$

$\;\sqrt{(200)^2+(100)^2}\:m$

$\;\sqrt{(200-4t)^2+(100-2t)^2}\:m$

$\;[(200-4\,t)+(100-2\,t)]\:m$

$\;\sqrt{(200-2\,t)^2+(100-4\,t)^2}\:m$