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Motion in One Dimensions

141750 A man runs at a speed of $4 m / s$ to overtake a standing bus. When he is $6 m$ behind the door at $t = 0$ , the bus moves forward and continuous with a constant acceleration of $1.2 m / s^{2}$ . The man reaches the door in time $t$ . Then,

1

4 t = 6 + 0.6 t^{2}

2

1.2 t^{2} = 4 t

3

4 t^{2} = 1.2 t

4

6 + 4 t = 0.2 t^{2}

Explanation:

A Given that,
Speed of a man $= 4 m / s$
Acceleration of bus $= 1.2 m / s^{2}$
Distance between man and standing bus door $(d) = 6 m$
We know that,
Distance $(s) =$ velocity $\times$ time
$s = vt$
$s = 4 t$
At standing position $u = 0$
Using the second equation of motion,
$s - d = ut + \frac{1}{2} {at}^{2}$
$s = d + ut + \frac{1}{2} {at}^{2}$
$s = 6 + \frac{1}{2} \times 1.2 \times t^{2}$
$(∵ u = 0 and d = 6)$
Putting $s = 4 t$ in equation (i) -
$4 t = 6 + 0.6 t^{2}$

BITSAT-2013

Motion in One Dimensions

141751 A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the $X$ -axis and 8 $m / s^{2}$ along the $Y$ - axis. Its distance from the origin after $4 s$ will be

1

56 m

2

64 m

3

80 m

4

128 m

Explanation:

C Acceleration along $x$ -axis, $a_{x} = 6 m / s^{2}$ Acceleration along $y$ -axis, $a_{y} = 8 m / s^{2}$
original image
Resultant acceleration is given by
$a = \sqrt{a_{x}^{2} + a_{y}^{2}}$
$a = \sqrt{6^{2} + 8^{2}}$
$a = 10 m / s^{2}$
For calculating distance,
Using second equation of motion
$s = ut + \frac{1}{2} {at}^{2}$
$u = 0$ , distance at time $t = 4 \sec$
$s = 0 + \frac{1}{2} \times 10 \times 4^{2}$
$s = 5 \times 16 = 80 m$
$s = 80 m$

CG PET- 2004

Motion in One Dimensions

141752 A bullet comes out of the barrel of gun of length $2 m$ with a speed $80 m / s$ . The average acceleration of the bullet is

1

1.6 m / s^{2}

2

160 m / s^{2}

3

1600 m / s^{2}

4

16 m / s^{2}

Explanation:

C Given that,
Initial velocity of bullet $(u) = 0 m / s$
Final velocity of bullet $(v) = 80 m / s$
Total length of gun $= 2 m$
Average acceleration of bullet $(a) =$ ?
We know that,
Third equation of motion-
$v^{2} = u^{2} + 2 a s$
$∵ u^{2} = 0$
$∴ v^{2} = 2 as$ ,
$a = \frac{v^{2}}{2 s} = \frac{(80)^{2}}{2 \times 2} = \frac{6400}{4}$
$a = 1600 m / s^{2}$

Manipal UGET-2010

Motion in One Dimensions

141753 A ball rolls off the top of stairway with a horizontal velocity of magnitude $1.8 m / s$ . The steps are $0.20 m$ high and $0.20 m$ width. Which step will the ball hit first?

1 First

2 Second

3 Third

4 Fourth

Explanation:

D Given,
Height of the step (h) $= 0.20 m$
Width of step $(w) = 0.20 m$
Initial velocity $(u) = 1.8 m / s$ (In Horizontal direction)
original image
Let the ball strikes the $n^{th}$ step of stair
then vertical distance $= n . h$
From equation of motion -
$nh = ut + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = 0 + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = \frac{1}{2} {gt}^{2}$
Horizontal distance $= nw$
$n . w = ut$
$t = \frac{nw}{u}$
$t = \frac{n \times 0.20}{u}$
Put the value of ' $t$ ' in equation no. (i) $n \times (0.20) = \frac{1}{2} \times (9.8) \times {(\frac{n \times 0.20}{u})}^{2}$
$n \times 0.20 = \frac{4.9 \times n^{2} \times 0.20 \times 0.20}{u^{2}}$
$n = \frac{u^{2}}{4.9 \times 0.20} = \frac{1.8 \times 1.8}{4.9 \times 0.20}$
$n = 3.30 \approx 4$
$n = 4$

Manipal UGET-2010

Motion in One Dimensions

141750 A man runs at a speed of $4 m / s$ to overtake a standing bus. When he is $6 m$ behind the door at $t = 0$ , the bus moves forward and continuous with a constant acceleration of $1.2 m / s^{2}$ . The man reaches the door in time $t$ . Then,

1

4 t = 6 + 0.6 t^{2}

2

1.2 t^{2} = 4 t

3

4 t^{2} = 1.2 t

4

6 + 4 t = 0.2 t^{2}

Explanation:

A Given that,
Speed of a man $= 4 m / s$
Acceleration of bus $= 1.2 m / s^{2}$
Distance between man and standing bus door $(d) = 6 m$
We know that,
Distance $(s) =$ velocity $\times$ time
$s = vt$
$s = 4 t$
At standing position $u = 0$
Using the second equation of motion,
$s - d = ut + \frac{1}{2} {at}^{2}$
$s = d + ut + \frac{1}{2} {at}^{2}$
$s = 6 + \frac{1}{2} \times 1.2 \times t^{2}$
$(∵ u = 0 and d = 6)$
Putting $s = 4 t$ in equation (i) -
$4 t = 6 + 0.6 t^{2}$

BITSAT-2013

Motion in One Dimensions

141751 A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the $X$ -axis and 8 $m / s^{2}$ along the $Y$ - axis. Its distance from the origin after $4 s$ will be

1

56 m

2

64 m

3

80 m

4

128 m

Explanation:

C Acceleration along $x$ -axis, $a_{x} = 6 m / s^{2}$ Acceleration along $y$ -axis, $a_{y} = 8 m / s^{2}$
original image
Resultant acceleration is given by
$a = \sqrt{a_{x}^{2} + a_{y}^{2}}$
$a = \sqrt{6^{2} + 8^{2}}$
$a = 10 m / s^{2}$
For calculating distance,
Using second equation of motion
$s = ut + \frac{1}{2} {at}^{2}$
$u = 0$ , distance at time $t = 4 \sec$
$s = 0 + \frac{1}{2} \times 10 \times 4^{2}$
$s = 5 \times 16 = 80 m$
$s = 80 m$

CG PET- 2004

Motion in One Dimensions

141752 A bullet comes out of the barrel of gun of length $2 m$ with a speed $80 m / s$ . The average acceleration of the bullet is

1

1.6 m / s^{2}

2

160 m / s^{2}

3

1600 m / s^{2}

4

16 m / s^{2}

Explanation:

C Given that,
Initial velocity of bullet $(u) = 0 m / s$
Final velocity of bullet $(v) = 80 m / s$
Total length of gun $= 2 m$
Average acceleration of bullet $(a) =$ ?
We know that,
Third equation of motion-
$v^{2} = u^{2} + 2 a s$
$∵ u^{2} = 0$
$∴ v^{2} = 2 as$ ,
$a = \frac{v^{2}}{2 s} = \frac{(80)^{2}}{2 \times 2} = \frac{6400}{4}$
$a = 1600 m / s^{2}$

Manipal UGET-2010

Motion in One Dimensions

141753 A ball rolls off the top of stairway with a horizontal velocity of magnitude $1.8 m / s$ . The steps are $0.20 m$ high and $0.20 m$ width. Which step will the ball hit first?

1 First

2 Second

3 Third

4 Fourth

Explanation:

D Given,
Height of the step (h) $= 0.20 m$
Width of step $(w) = 0.20 m$
Initial velocity $(u) = 1.8 m / s$ (In Horizontal direction)
original image
Let the ball strikes the $n^{th}$ step of stair
then vertical distance $= n . h$
From equation of motion -
$nh = ut + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = 0 + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = \frac{1}{2} {gt}^{2}$
Horizontal distance $= nw$
$n . w = ut$
$t = \frac{nw}{u}$
$t = \frac{n \times 0.20}{u}$
Put the value of ' $t$ ' in equation no. (i) $n \times (0.20) = \frac{1}{2} \times (9.8) \times {(\frac{n \times 0.20}{u})}^{2}$
$n \times 0.20 = \frac{4.9 \times n^{2} \times 0.20 \times 0.20}{u^{2}}$
$n = \frac{u^{2}}{4.9 \times 0.20} = \frac{1.8 \times 1.8}{4.9 \times 0.20}$
$n = 3.30 \approx 4$
$n = 4$

Manipal UGET-2010

Motion in One Dimensions

141750 A man runs at a speed of $4 m / s$ to overtake a standing bus. When he is $6 m$ behind the door at $t = 0$ , the bus moves forward and continuous with a constant acceleration of $1.2 m / s^{2}$ . The man reaches the door in time $t$ . Then,

1

4 t = 6 + 0.6 t^{2}

2

1.2 t^{2} = 4 t

3

4 t^{2} = 1.2 t

4

6 + 4 t = 0.2 t^{2}

Explanation:

A Given that,
Speed of a man $= 4 m / s$
Acceleration of bus $= 1.2 m / s^{2}$
Distance between man and standing bus door $(d) = 6 m$
We know that,
Distance $(s) =$ velocity $\times$ time
$s = vt$
$s = 4 t$
At standing position $u = 0$
Using the second equation of motion,
$s - d = ut + \frac{1}{2} {at}^{2}$
$s = d + ut + \frac{1}{2} {at}^{2}$
$s = 6 + \frac{1}{2} \times 1.2 \times t^{2}$
$(∵ u = 0 and d = 6)$
Putting $s = 4 t$ in equation (i) -
$4 t = 6 + 0.6 t^{2}$

BITSAT-2013

Motion in One Dimensions

141751 A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the $X$ -axis and 8 $m / s^{2}$ along the $Y$ - axis. Its distance from the origin after $4 s$ will be

1

56 m

2

64 m

3

80 m

4

128 m

Explanation:

C Acceleration along $x$ -axis, $a_{x} = 6 m / s^{2}$ Acceleration along $y$ -axis, $a_{y} = 8 m / s^{2}$
original image
Resultant acceleration is given by
$a = \sqrt{a_{x}^{2} + a_{y}^{2}}$
$a = \sqrt{6^{2} + 8^{2}}$
$a = 10 m / s^{2}$
For calculating distance,
Using second equation of motion
$s = ut + \frac{1}{2} {at}^{2}$
$u = 0$ , distance at time $t = 4 \sec$
$s = 0 + \frac{1}{2} \times 10 \times 4^{2}$
$s = 5 \times 16 = 80 m$
$s = 80 m$

CG PET- 2004

Motion in One Dimensions

141752 A bullet comes out of the barrel of gun of length $2 m$ with a speed $80 m / s$ . The average acceleration of the bullet is

1

1.6 m / s^{2}

2

160 m / s^{2}

3

1600 m / s^{2}

4

16 m / s^{2}

Explanation:

C Given that,
Initial velocity of bullet $(u) = 0 m / s$
Final velocity of bullet $(v) = 80 m / s$
Total length of gun $= 2 m$
Average acceleration of bullet $(a) =$ ?
We know that,
Third equation of motion-
$v^{2} = u^{2} + 2 a s$
$∵ u^{2} = 0$
$∴ v^{2} = 2 as$ ,
$a = \frac{v^{2}}{2 s} = \frac{(80)^{2}}{2 \times 2} = \frac{6400}{4}$
$a = 1600 m / s^{2}$

Manipal UGET-2010

Motion in One Dimensions

141753 A ball rolls off the top of stairway with a horizontal velocity of magnitude $1.8 m / s$ . The steps are $0.20 m$ high and $0.20 m$ width. Which step will the ball hit first?

1 First

2 Second

3 Third

4 Fourth

Explanation:

D Given,
Height of the step (h) $= 0.20 m$
Width of step $(w) = 0.20 m$
Initial velocity $(u) = 1.8 m / s$ (In Horizontal direction)
original image
Let the ball strikes the $n^{th}$ step of stair
then vertical distance $= n . h$
From equation of motion -
$nh = ut + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = 0 + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = \frac{1}{2} {gt}^{2}$
Horizontal distance $= nw$
$n . w = ut$
$t = \frac{nw}{u}$
$t = \frac{n \times 0.20}{u}$
Put the value of ' $t$ ' in equation no. (i) $n \times (0.20) = \frac{1}{2} \times (9.8) \times {(\frac{n \times 0.20}{u})}^{2}$
$n \times 0.20 = \frac{4.9 \times n^{2} \times 0.20 \times 0.20}{u^{2}}$
$n = \frac{u^{2}}{4.9 \times 0.20} = \frac{1.8 \times 1.8}{4.9 \times 0.20}$
$n = 3.30 \approx 4$
$n = 4$

Manipal UGET-2010

Motion in One Dimensions

141750 A man runs at a speed of $4 m / s$ to overtake a standing bus. When he is $6 m$ behind the door at $t = 0$ , the bus moves forward and continuous with a constant acceleration of $1.2 m / s^{2}$ . The man reaches the door in time $t$ . Then,

1

4 t = 6 + 0.6 t^{2}

2

1.2 t^{2} = 4 t

3

4 t^{2} = 1.2 t

4

6 + 4 t = 0.2 t^{2}

Explanation:

A Given that,
Speed of a man $= 4 m / s$
Acceleration of bus $= 1.2 m / s^{2}$
Distance between man and standing bus door $(d) = 6 m$
We know that,
Distance $(s) =$ velocity $\times$ time
$s = vt$
$s = 4 t$
At standing position $u = 0$
Using the second equation of motion,
$s - d = ut + \frac{1}{2} {at}^{2}$
$s = d + ut + \frac{1}{2} {at}^{2}$
$s = 6 + \frac{1}{2} \times 1.2 \times t^{2}$
$(∵ u = 0 and d = 6)$
Putting $s = 4 t$ in equation (i) -
$4 t = 6 + 0.6 t^{2}$

BITSAT-2013

Motion in One Dimensions

141751 A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the $X$ -axis and 8 $m / s^{2}$ along the $Y$ - axis. Its distance from the origin after $4 s$ will be

1

56 m

2

64 m

3

80 m

4

128 m

Explanation:

C Acceleration along $x$ -axis, $a_{x} = 6 m / s^{2}$ Acceleration along $y$ -axis, $a_{y} = 8 m / s^{2}$
original image
Resultant acceleration is given by
$a = \sqrt{a_{x}^{2} + a_{y}^{2}}$
$a = \sqrt{6^{2} + 8^{2}}$
$a = 10 m / s^{2}$
For calculating distance,
Using second equation of motion
$s = ut + \frac{1}{2} {at}^{2}$
$u = 0$ , distance at time $t = 4 \sec$
$s = 0 + \frac{1}{2} \times 10 \times 4^{2}$
$s = 5 \times 16 = 80 m$
$s = 80 m$

CG PET- 2004

Motion in One Dimensions

141752 A bullet comes out of the barrel of gun of length $2 m$ with a speed $80 m / s$ . The average acceleration of the bullet is

1

1.6 m / s^{2}

2

160 m / s^{2}

3

1600 m / s^{2}

4

16 m / s^{2}

Explanation:

C Given that,
Initial velocity of bullet $(u) = 0 m / s$
Final velocity of bullet $(v) = 80 m / s$
Total length of gun $= 2 m$
Average acceleration of bullet $(a) =$ ?
We know that,
Third equation of motion-
$v^{2} = u^{2} + 2 a s$
$∵ u^{2} = 0$
$∴ v^{2} = 2 as$ ,
$a = \frac{v^{2}}{2 s} = \frac{(80)^{2}}{2 \times 2} = \frac{6400}{4}$
$a = 1600 m / s^{2}$

Manipal UGET-2010

Motion in One Dimensions

141753 A ball rolls off the top of stairway with a horizontal velocity of magnitude $1.8 m / s$ . The steps are $0.20 m$ high and $0.20 m$ width. Which step will the ball hit first?

1 First

2 Second

3 Third

4 Fourth

Explanation:

D Given,
Height of the step (h) $= 0.20 m$
Width of step $(w) = 0.20 m$
Initial velocity $(u) = 1.8 m / s$ (In Horizontal direction)
original image
Let the ball strikes the $n^{th}$ step of stair
then vertical distance $= n . h$
From equation of motion -
$nh = ut + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = 0 + \frac{1}{2} {gt}^{2}$
$n \times 0.20 = \frac{1}{2} {gt}^{2}$
Horizontal distance $= nw$
$n . w = ut$
$t = \frac{nw}{u}$
$t = \frac{n \times 0.20}{u}$
Put the value of ' $t$ ' in equation no. (i) $n \times (0.20) = \frac{1}{2} \times (9.8) \times {(\frac{n \times 0.20}{u})}^{2}$
$n \times 0.20 = \frac{4.9 \times n^{2} \times 0.20 \times 0.20}{u^{2}}$
$n = \frac{u^{2}}{4.9 \times 0.20} = \frac{1.8 \times 1.8}{4.9 \times 0.20}$
$n = 3.30 \approx 4$
$n = 4$

Manipal UGET-2010

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