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Motion in One Dimensions

141768 Distance travelled by a body falling from rest in the first, second and third second is in the ratio of

1

1 : 2 : 3

2

1 : 3 : 5

3

1 : 4 : 9

4 None of the above

Explanation:

B Given, initial velocity $(u) = 0$
Distance travelled in first second
$s_{1} = \frac{1}{2} {at}^{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 1 = 5 m$
$S_{2} = u + \frac{1}{2} g (2 t - 1)$
$= 0 + \frac{1}{2} \times 10 (2 \times 2 - 1)$
$S_{2} = 15 m \dots \dots .$ . (ii)
$S_{3} = u + \frac{1}{2} g (2 t - 1)$
$S_{3} = 25 m$
from eqn (i), (ii) and (iii)
Now, $s_{1} : s_{2} : s_{3} = 1 : 3 : 5$

SRMJEE - 2013

Motion in One Dimensions

141769 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first 3 seconds will be :

1

9 m

2

16 m

3

27 m

4

36 m

Explanation:

A Given,
Variation of velocity with $t$ is given as $v = kt$
$\Rightarrow \frac{dx}{dt} = kt$
$\Rightarrow dx = ktdt$
Distance covered in first 3 seconds will be
$\int_{0}^{x} dx = k \int_{0}^{3} tdt$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9 (k = 2 m / s^{2})$
$∴ x = 9 m$

SRMJEE - 2013

Motion in One Dimensions

141770 A body moving with uniform acceleration 6 $m / s^{2}$ starts from rest. The distance covered by it in $4^{th}$ second will be

1

21 m

2

35 m

3

53 m

4

12 m

Explanation:

A Given, initial velocity $(u) = 0$
Acceleration of the body, $a = 6 m / s^{2}$
Velocity acquired in first three seconds of the motion $v = u + at = 6 \times 3 m / s = 18 m / s$
This will be the initial velocity for the last one second of the motion
$∴$ The distance covered in $4^{th}$ second
$s = 18 \times 1 + \frac{1}{2} \times 6 \times 1^{2} = 18 + 3 = 21 m$

SRMJEE - 2014

Motion in One Dimensions

141772 A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 {ms}^{- 1}$ to $20 {ms}^{- 1}$ while passing through a distance $135 m$ in $t$ sec. The value of $t$ is

1 10

2 1.8

3 12

4 9

Explanation:

D Given, $u = 10 m / s, v = 20 m / s, s = 135 m$
We know that, $v = u +$ at
$20 = 10 + a t$
$a = \frac{10}{t}$
Now, $s = ut + \frac{1}{2} {at}^{2}$
$135 = 10 \times t + \frac{1}{2} \times \frac{10}{t} (t)^{2}$
$135 = 10 t + 5 t \Rightarrow 135 = 15 t$
$t = 9 \sec$

AIPMT 2008

Motion in One Dimensions

141768 Distance travelled by a body falling from rest in the first, second and third second is in the ratio of

1

1 : 2 : 3

2

1 : 3 : 5

3

1 : 4 : 9

4 None of the above

Explanation:

B Given, initial velocity $(u) = 0$
Distance travelled in first second
$s_{1} = \frac{1}{2} {at}^{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 1 = 5 m$
$S_{2} = u + \frac{1}{2} g (2 t - 1)$
$= 0 + \frac{1}{2} \times 10 (2 \times 2 - 1)$
$S_{2} = 15 m \dots \dots .$ . (ii)
$S_{3} = u + \frac{1}{2} g (2 t - 1)$
$S_{3} = 25 m$
from eqn (i), (ii) and (iii)
Now, $s_{1} : s_{2} : s_{3} = 1 : 3 : 5$

SRMJEE - 2013

Motion in One Dimensions

141769 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first 3 seconds will be :

1

9 m

2

16 m

3

27 m

4

36 m

Explanation:

A Given,
Variation of velocity with $t$ is given as $v = kt$
$\Rightarrow \frac{dx}{dt} = kt$
$\Rightarrow dx = ktdt$
Distance covered in first 3 seconds will be
$\int_{0}^{x} dx = k \int_{0}^{3} tdt$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9 (k = 2 m / s^{2})$
$∴ x = 9 m$

SRMJEE - 2013

Motion in One Dimensions

141770 A body moving with uniform acceleration 6 $m / s^{2}$ starts from rest. The distance covered by it in $4^{th}$ second will be

1

21 m

2

35 m

3

53 m

4

12 m

Explanation:

A Given, initial velocity $(u) = 0$
Acceleration of the body, $a = 6 m / s^{2}$
Velocity acquired in first three seconds of the motion $v = u + at = 6 \times 3 m / s = 18 m / s$
This will be the initial velocity for the last one second of the motion
$∴$ The distance covered in $4^{th}$ second
$s = 18 \times 1 + \frac{1}{2} \times 6 \times 1^{2} = 18 + 3 = 21 m$

SRMJEE - 2014

Motion in One Dimensions

141772 A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 {ms}^{- 1}$ to $20 {ms}^{- 1}$ while passing through a distance $135 m$ in $t$ sec. The value of $t$ is

1 10

2 1.8

3 12

4 9

Explanation:

D Given, $u = 10 m / s, v = 20 m / s, s = 135 m$
We know that, $v = u +$ at
$20 = 10 + a t$
$a = \frac{10}{t}$
Now, $s = ut + \frac{1}{2} {at}^{2}$
$135 = 10 \times t + \frac{1}{2} \times \frac{10}{t} (t)^{2}$
$135 = 10 t + 5 t \Rightarrow 135 = 15 t$
$t = 9 \sec$

AIPMT 2008

Motion in One Dimensions

141768 Distance travelled by a body falling from rest in the first, second and third second is in the ratio of

1

1 : 2 : 3

2

1 : 3 : 5

3

1 : 4 : 9

4 None of the above

Explanation:

B Given, initial velocity $(u) = 0$
Distance travelled in first second
$s_{1} = \frac{1}{2} {at}^{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 1 = 5 m$
$S_{2} = u + \frac{1}{2} g (2 t - 1)$
$= 0 + \frac{1}{2} \times 10 (2 \times 2 - 1)$
$S_{2} = 15 m \dots \dots .$ . (ii)
$S_{3} = u + \frac{1}{2} g (2 t - 1)$
$S_{3} = 25 m$
from eqn (i), (ii) and (iii)
Now, $s_{1} : s_{2} : s_{3} = 1 : 3 : 5$

SRMJEE - 2013

Motion in One Dimensions

141769 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first 3 seconds will be :

1

9 m

2

16 m

3

27 m

4

36 m

Explanation:

A Given,
Variation of velocity with $t$ is given as $v = kt$
$\Rightarrow \frac{dx}{dt} = kt$
$\Rightarrow dx = ktdt$
Distance covered in first 3 seconds will be
$\int_{0}^{x} dx = k \int_{0}^{3} tdt$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9 (k = 2 m / s^{2})$
$∴ x = 9 m$

SRMJEE - 2013

Motion in One Dimensions

141770 A body moving with uniform acceleration 6 $m / s^{2}$ starts from rest. The distance covered by it in $4^{th}$ second will be

1

21 m

2

35 m

3

53 m

4

12 m

Explanation:

A Given, initial velocity $(u) = 0$
Acceleration of the body, $a = 6 m / s^{2}$
Velocity acquired in first three seconds of the motion $v = u + at = 6 \times 3 m / s = 18 m / s$
This will be the initial velocity for the last one second of the motion
$∴$ The distance covered in $4^{th}$ second
$s = 18 \times 1 + \frac{1}{2} \times 6 \times 1^{2} = 18 + 3 = 21 m$

SRMJEE - 2014

Motion in One Dimensions

141772 A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 {ms}^{- 1}$ to $20 {ms}^{- 1}$ while passing through a distance $135 m$ in $t$ sec. The value of $t$ is

1 10

2 1.8

3 12

4 9

Explanation:

D Given, $u = 10 m / s, v = 20 m / s, s = 135 m$
We know that, $v = u +$ at
$20 = 10 + a t$
$a = \frac{10}{t}$
Now, $s = ut + \frac{1}{2} {at}^{2}$
$135 = 10 \times t + \frac{1}{2} \times \frac{10}{t} (t)^{2}$
$135 = 10 t + 5 t \Rightarrow 135 = 15 t$
$t = 9 \sec$

AIPMT 2008

Motion in One Dimensions

141768 Distance travelled by a body falling from rest in the first, second and third second is in the ratio of

1

1 : 2 : 3

2

1 : 3 : 5

3

1 : 4 : 9

4 None of the above

Explanation:

B Given, initial velocity $(u) = 0$
Distance travelled in first second
$s_{1} = \frac{1}{2} {at}^{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 1 = 5 m$
$S_{2} = u + \frac{1}{2} g (2 t - 1)$
$= 0 + \frac{1}{2} \times 10 (2 \times 2 - 1)$
$S_{2} = 15 m \dots \dots .$ . (ii)
$S_{3} = u + \frac{1}{2} g (2 t - 1)$
$S_{3} = 25 m$
from eqn (i), (ii) and (iii)
Now, $s_{1} : s_{2} : s_{3} = 1 : 3 : 5$

SRMJEE - 2013

Motion in One Dimensions

141769 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first 3 seconds will be :

1

9 m

2

16 m

3

27 m

4

36 m

Explanation:

A Given,
Variation of velocity with $t$ is given as $v = kt$
$\Rightarrow \frac{dx}{dt} = kt$
$\Rightarrow dx = ktdt$
Distance covered in first 3 seconds will be
$\int_{0}^{x} dx = k \int_{0}^{3} tdt$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9 (k = 2 m / s^{2})$
$∴ x = 9 m$

SRMJEE - 2013

Motion in One Dimensions

141770 A body moving with uniform acceleration 6 $m / s^{2}$ starts from rest. The distance covered by it in $4^{th}$ second will be

1

21 m

2

35 m

3

53 m

4

12 m

Explanation:

A Given, initial velocity $(u) = 0$
Acceleration of the body, $a = 6 m / s^{2}$
Velocity acquired in first three seconds of the motion $v = u + at = 6 \times 3 m / s = 18 m / s$
This will be the initial velocity for the last one second of the motion
$∴$ The distance covered in $4^{th}$ second
$s = 18 \times 1 + \frac{1}{2} \times 6 \times 1^{2} = 18 + 3 = 21 m$

SRMJEE - 2014

Motion in One Dimensions

141772 A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 {ms}^{- 1}$ to $20 {ms}^{- 1}$ while passing through a distance $135 m$ in $t$ sec. The value of $t$ is

1 10

2 1.8

3 12

4 9

Explanation:

D Given, $u = 10 m / s, v = 20 m / s, s = 135 m$
We know that, $v = u +$ at
$20 = 10 + a t$
$a = \frac{10}{t}$
Now, $s = ut + \frac{1}{2} {at}^{2}$
$135 = 10 \times t + \frac{1}{2} \times \frac{10}{t} (t)^{2}$
$135 = 10 t + 5 t \Rightarrow 135 = 15 t$
$t = 9 \sec$

AIPMT 2008

Question Bank Series

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