QBManager

Motion in One Dimensions

141286 A particle moves along $Y$ -axis in such a way that its $y$ -coordinate varies with time $t$ according to the relation $y = 3 + 5 t + 7 t^{2}$ . The initial velocity and acceleration of the particle are respectively:

1

14 {ms}^{- 1}, - 5 {ms}^{- 2}

2

19 {ms}^{- 1}, - 9 {ms}^{- 2}

3

- 14 {ms}^{- 1}, - 5 {ms}^{- 2}

4

5 {ms}^{- 1}, 14 {ms}^{- 2}

Explanation:

D Given that,
$y = 3 + 5 t + 7 t^{2}$
$∴ v = \frac{dy}{dt} = \frac{d}{dt} (3 + 5 t + 7 t^{2})$
$v = 5 + 14 t$
Its initial velocity at $t = 0$
Then, $v = 5 m / \sec$
$∴ a = \frac{dv}{dt} = \frac{d}{dt} (5 + 14 t)$
$a = 0 + 14$
$a = 14$
So, acceleration (a) $= 14 m / \sec^{2}$
The initial velocity and acceleration of particle are 5 $m / s$ and $14 m / s^{2}$ respectively

JCECE-2005

Motion in One Dimensions

141287 An aeroplane moves $400 m$ towards the north, $300 m$ towards west and then $1200 m$ vertically upwards, then its displacement from the initial position is

1

1600 m

2

1800 m

3

1500 m

4

1300 m

Explanation:

D Here, CD is perpendicular to the plane of paper. Required distance $= BD$
original image
${CB}^{2} = {BA}^{2} + {AC}^{2}$
${CB}^{2} = 400^{2} + 300^{2}$
$CB = \sqrt{400^{2} + 300^{2}} = 500 m$
${BD}^{2} = {CB}^{2} + {CD}^{2}$
${BD}^{2} = 500^{2} + 1200^{2}$
$BD = \sqrt{500^{2} + 1200^{2}} = 1300 m$

AIIMS-1998

Motion in One Dimensions

141289 An object accelerates from rest to a velocity 27. $5 m / s$ in 10s, then the distance covered after next $10 s$ is :

1

550 m

2

137.5 m

3

412.5 m

4

175 m

Explanation:

C Given,
$u = 0, v = 27.5 m / s, t = 10 \sec, a = ?$
From $v = u +$ at
$∴ a = \frac{v - u}{t} = \frac{27.5 - 0}{10}$
$= \frac{27.5}{10} = 2.75 m / s$
In the first $10 \sec$ , distance travelled
From $S = u t + \frac{1}{2} a t^{2}$
$S_{1} = 0 \times 10 + \frac{1}{2} \times 2.75 \times (10)^{2} = 137.5 m$
In the first $20 \sec$ , distance travelled
$S_{2} = 0 \times 20 + \frac{1}{2} \times 2.75 \times (20)^{2} = 550 m$
Hence, distance travelled in next $10 s$ .
$= 550 - 137.5$
$= 412.5 m$

BCECE-2004

Motion in One Dimensions

141290 A body of mass $m$ thrown horizontally with velocity $v$ , from the top of tower of height $h$ , touches the level of ground at a distance of 250 $m$ from the foot of the tower. A body of mass 2
$m$ , thrown horizontally with velocity $v / 2$ , from the top of a tower of height $4 h$ will touch the level ground at a distance in metre from the foot of the tower is

1

150 m

2

200 m

3

250 m

4

2500 m

Explanation:

C Given data,
Distance from the foot of the tower, $= 250 m$
As we know,
$t = \sqrt{\frac{2 h}{g}}$
$∴$ Distance, $d = vt = v \sqrt{\frac{2 h}{g}} = 250 m$
According to given question,
When, $v = \frac{v}{2}, h = 4 h$
Then, distance, $x = \frac{v}{2} \sqrt{\frac{2 (4 h)}{g}} = v \sqrt{\frac{2 h}{g}}$
$∴ x = 250 m$

EAMCET-1991

Motion in One Dimensions

141291 A particle starts moving from rest under uniform acceleration. It travels a distance $x$ in the first two seconds and a distance of $y$ in the next two seconds. If $y = n x$ , then $n =$

1 1

2 3

3 2

4 4

Explanation:

B Given that,
original image
According to question,
Distance covered in $2 \sec, x = \frac{1}{2} \cdot {at}^{2} = \frac{1}{2} \cdot a \times (2)^{2} = 2 a$
Similarly,
Distance covered in next $4 \sec, y = \frac{1}{2} a (4)^{2} - \frac{1}{2} a (2)^{2}$
$y = 6 a$
From question
$y = nx$
$6 a = n \times 2 a$
$n = 3$

EAMCET-1993

Motion in One Dimensions

141286 A particle moves along $Y$ -axis in such a way that its $y$ -coordinate varies with time $t$ according to the relation $y = 3 + 5 t + 7 t^{2}$ . The initial velocity and acceleration of the particle are respectively:

1

14 {ms}^{- 1}, - 5 {ms}^{- 2}

2

19 {ms}^{- 1}, - 9 {ms}^{- 2}

3

- 14 {ms}^{- 1}, - 5 {ms}^{- 2}

4

5 {ms}^{- 1}, 14 {ms}^{- 2}

Explanation:

D Given that,
$y = 3 + 5 t + 7 t^{2}$
$∴ v = \frac{dy}{dt} = \frac{d}{dt} (3 + 5 t + 7 t^{2})$
$v = 5 + 14 t$
Its initial velocity at $t = 0$
Then, $v = 5 m / \sec$
$∴ a = \frac{dv}{dt} = \frac{d}{dt} (5 + 14 t)$
$a = 0 + 14$
$a = 14$
So, acceleration (a) $= 14 m / \sec^{2}$
The initial velocity and acceleration of particle are 5 $m / s$ and $14 m / s^{2}$ respectively

JCECE-2005

Motion in One Dimensions

141287 An aeroplane moves $400 m$ towards the north, $300 m$ towards west and then $1200 m$ vertically upwards, then its displacement from the initial position is

1

1600 m

2

1800 m

3

1500 m

4

1300 m

Explanation:

D Here, CD is perpendicular to the plane of paper. Required distance $= BD$
original image
${CB}^{2} = {BA}^{2} + {AC}^{2}$
${CB}^{2} = 400^{2} + 300^{2}$
$CB = \sqrt{400^{2} + 300^{2}} = 500 m$
${BD}^{2} = {CB}^{2} + {CD}^{2}$
${BD}^{2} = 500^{2} + 1200^{2}$
$BD = \sqrt{500^{2} + 1200^{2}} = 1300 m$

AIIMS-1998

Motion in One Dimensions

141289 An object accelerates from rest to a velocity 27. $5 m / s$ in 10s, then the distance covered after next $10 s$ is :

1

550 m

2

137.5 m

3

412.5 m

4

175 m

Explanation:

C Given,
$u = 0, v = 27.5 m / s, t = 10 \sec, a = ?$
From $v = u +$ at
$∴ a = \frac{v - u}{t} = \frac{27.5 - 0}{10}$
$= \frac{27.5}{10} = 2.75 m / s$
In the first $10 \sec$ , distance travelled
From $S = u t + \frac{1}{2} a t^{2}$
$S_{1} = 0 \times 10 + \frac{1}{2} \times 2.75 \times (10)^{2} = 137.5 m$
In the first $20 \sec$ , distance travelled
$S_{2} = 0 \times 20 + \frac{1}{2} \times 2.75 \times (20)^{2} = 550 m$
Hence, distance travelled in next $10 s$ .
$= 550 - 137.5$
$= 412.5 m$

BCECE-2004

Motion in One Dimensions

141290 A body of mass $m$ thrown horizontally with velocity $v$ , from the top of tower of height $h$ , touches the level of ground at a distance of 250 $m$ from the foot of the tower. A body of mass 2
$m$ , thrown horizontally with velocity $v / 2$ , from the top of a tower of height $4 h$ will touch the level ground at a distance in metre from the foot of the tower is

1

150 m

2

200 m

3

250 m

4

2500 m

Explanation:

C Given data,
Distance from the foot of the tower, $= 250 m$
As we know,
$t = \sqrt{\frac{2 h}{g}}$
$∴$ Distance, $d = vt = v \sqrt{\frac{2 h}{g}} = 250 m$
According to given question,
When, $v = \frac{v}{2}, h = 4 h$
Then, distance, $x = \frac{v}{2} \sqrt{\frac{2 (4 h)}{g}} = v \sqrt{\frac{2 h}{g}}$
$∴ x = 250 m$

EAMCET-1991

Motion in One Dimensions

141291 A particle starts moving from rest under uniform acceleration. It travels a distance $x$ in the first two seconds and a distance of $y$ in the next two seconds. If $y = n x$ , then $n =$

1 1

2 3

3 2

4 4

Explanation:

B Given that,
original image
According to question,
Distance covered in $2 \sec, x = \frac{1}{2} \cdot {at}^{2} = \frac{1}{2} \cdot a \times (2)^{2} = 2 a$
Similarly,
Distance covered in next $4 \sec, y = \frac{1}{2} a (4)^{2} - \frac{1}{2} a (2)^{2}$
$y = 6 a$
From question
$y = nx$
$6 a = n \times 2 a$
$n = 3$

EAMCET-1993

Motion in One Dimensions

141286 A particle moves along $Y$ -axis in such a way that its $y$ -coordinate varies with time $t$ according to the relation $y = 3 + 5 t + 7 t^{2}$ . The initial velocity and acceleration of the particle are respectively:

1

14 {ms}^{- 1}, - 5 {ms}^{- 2}

2

19 {ms}^{- 1}, - 9 {ms}^{- 2}

3

- 14 {ms}^{- 1}, - 5 {ms}^{- 2}

4

5 {ms}^{- 1}, 14 {ms}^{- 2}

Explanation:

D Given that,
$y = 3 + 5 t + 7 t^{2}$
$∴ v = \frac{dy}{dt} = \frac{d}{dt} (3 + 5 t + 7 t^{2})$
$v = 5 + 14 t$
Its initial velocity at $t = 0$
Then, $v = 5 m / \sec$
$∴ a = \frac{dv}{dt} = \frac{d}{dt} (5 + 14 t)$
$a = 0 + 14$
$a = 14$
So, acceleration (a) $= 14 m / \sec^{2}$
The initial velocity and acceleration of particle are 5 $m / s$ and $14 m / s^{2}$ respectively

JCECE-2005

Motion in One Dimensions

141287 An aeroplane moves $400 m$ towards the north, $300 m$ towards west and then $1200 m$ vertically upwards, then its displacement from the initial position is

1

1600 m

2

1800 m

3

1500 m

4

1300 m

Explanation:

D Here, CD is perpendicular to the plane of paper. Required distance $= BD$
original image
${CB}^{2} = {BA}^{2} + {AC}^{2}$
${CB}^{2} = 400^{2} + 300^{2}$
$CB = \sqrt{400^{2} + 300^{2}} = 500 m$
${BD}^{2} = {CB}^{2} + {CD}^{2}$
${BD}^{2} = 500^{2} + 1200^{2}$
$BD = \sqrt{500^{2} + 1200^{2}} = 1300 m$

AIIMS-1998

Motion in One Dimensions

141289 An object accelerates from rest to a velocity 27. $5 m / s$ in 10s, then the distance covered after next $10 s$ is :

1

550 m

2

137.5 m

3

412.5 m

4

175 m

Explanation:

C Given,
$u = 0, v = 27.5 m / s, t = 10 \sec, a = ?$
From $v = u +$ at
$∴ a = \frac{v - u}{t} = \frac{27.5 - 0}{10}$
$= \frac{27.5}{10} = 2.75 m / s$
In the first $10 \sec$ , distance travelled
From $S = u t + \frac{1}{2} a t^{2}$
$S_{1} = 0 \times 10 + \frac{1}{2} \times 2.75 \times (10)^{2} = 137.5 m$
In the first $20 \sec$ , distance travelled
$S_{2} = 0 \times 20 + \frac{1}{2} \times 2.75 \times (20)^{2} = 550 m$
Hence, distance travelled in next $10 s$ .
$= 550 - 137.5$
$= 412.5 m$

BCECE-2004

Motion in One Dimensions

141290 A body of mass $m$ thrown horizontally with velocity $v$ , from the top of tower of height $h$ , touches the level of ground at a distance of 250 $m$ from the foot of the tower. A body of mass 2
$m$ , thrown horizontally with velocity $v / 2$ , from the top of a tower of height $4 h$ will touch the level ground at a distance in metre from the foot of the tower is

1

150 m

2

200 m

3

250 m

4

2500 m

Explanation:

C Given data,
Distance from the foot of the tower, $= 250 m$
As we know,
$t = \sqrt{\frac{2 h}{g}}$
$∴$ Distance, $d = vt = v \sqrt{\frac{2 h}{g}} = 250 m$
According to given question,
When, $v = \frac{v}{2}, h = 4 h$
Then, distance, $x = \frac{v}{2} \sqrt{\frac{2 (4 h)}{g}} = v \sqrt{\frac{2 h}{g}}$
$∴ x = 250 m$

EAMCET-1991

Motion in One Dimensions

141291 A particle starts moving from rest under uniform acceleration. It travels a distance $x$ in the first two seconds and a distance of $y$ in the next two seconds. If $y = n x$ , then $n =$

1 1

2 3

3 2

4 4

Explanation:

B Given that,
original image
According to question,
Distance covered in $2 \sec, x = \frac{1}{2} \cdot {at}^{2} = \frac{1}{2} \cdot a \times (2)^{2} = 2 a$
Similarly,
Distance covered in next $4 \sec, y = \frac{1}{2} a (4)^{2} - \frac{1}{2} a (2)^{2}$
$y = 6 a$
From question
$y = nx$
$6 a = n \times 2 a$
$n = 3$

EAMCET-1993

Motion in One Dimensions

141286 A particle moves along $Y$ -axis in such a way that its $y$ -coordinate varies with time $t$ according to the relation $y = 3 + 5 t + 7 t^{2}$ . The initial velocity and acceleration of the particle are respectively:

1

14 {ms}^{- 1}, - 5 {ms}^{- 2}

2

19 {ms}^{- 1}, - 9 {ms}^{- 2}

3

- 14 {ms}^{- 1}, - 5 {ms}^{- 2}

4

5 {ms}^{- 1}, 14 {ms}^{- 2}

Explanation:

D Given that,
$y = 3 + 5 t + 7 t^{2}$
$∴ v = \frac{dy}{dt} = \frac{d}{dt} (3 + 5 t + 7 t^{2})$
$v = 5 + 14 t$
Its initial velocity at $t = 0$
Then, $v = 5 m / \sec$
$∴ a = \frac{dv}{dt} = \frac{d}{dt} (5 + 14 t)$
$a = 0 + 14$
$a = 14$
So, acceleration (a) $= 14 m / \sec^{2}$
The initial velocity and acceleration of particle are 5 $m / s$ and $14 m / s^{2}$ respectively

JCECE-2005

Motion in One Dimensions

141287 An aeroplane moves $400 m$ towards the north, $300 m$ towards west and then $1200 m$ vertically upwards, then its displacement from the initial position is

1

1600 m

2

1800 m

3

1500 m

4

1300 m

Explanation:

D Here, CD is perpendicular to the plane of paper. Required distance $= BD$
original image
${CB}^{2} = {BA}^{2} + {AC}^{2}$
${CB}^{2} = 400^{2} + 300^{2}$
$CB = \sqrt{400^{2} + 300^{2}} = 500 m$
${BD}^{2} = {CB}^{2} + {CD}^{2}$
${BD}^{2} = 500^{2} + 1200^{2}$
$BD = \sqrt{500^{2} + 1200^{2}} = 1300 m$

AIIMS-1998

Motion in One Dimensions

141289 An object accelerates from rest to a velocity 27. $5 m / s$ in 10s, then the distance covered after next $10 s$ is :

1

550 m

2

137.5 m

3

412.5 m

4

175 m

Explanation:

C Given,
$u = 0, v = 27.5 m / s, t = 10 \sec, a = ?$
From $v = u +$ at
$∴ a = \frac{v - u}{t} = \frac{27.5 - 0}{10}$
$= \frac{27.5}{10} = 2.75 m / s$
In the first $10 \sec$ , distance travelled
From $S = u t + \frac{1}{2} a t^{2}$
$S_{1} = 0 \times 10 + \frac{1}{2} \times 2.75 \times (10)^{2} = 137.5 m$
In the first $20 \sec$ , distance travelled
$S_{2} = 0 \times 20 + \frac{1}{2} \times 2.75 \times (20)^{2} = 550 m$
Hence, distance travelled in next $10 s$ .
$= 550 - 137.5$
$= 412.5 m$

BCECE-2004

Motion in One Dimensions

141290 A body of mass $m$ thrown horizontally with velocity $v$ , from the top of tower of height $h$ , touches the level of ground at a distance of 250 $m$ from the foot of the tower. A body of mass 2
$m$ , thrown horizontally with velocity $v / 2$ , from the top of a tower of height $4 h$ will touch the level ground at a distance in metre from the foot of the tower is

1

150 m

2

200 m

3

250 m

4

2500 m

Explanation:

C Given data,
Distance from the foot of the tower, $= 250 m$
As we know,
$t = \sqrt{\frac{2 h}{g}}$
$∴$ Distance, $d = vt = v \sqrt{\frac{2 h}{g}} = 250 m$
According to given question,
When, $v = \frac{v}{2}, h = 4 h$
Then, distance, $x = \frac{v}{2} \sqrt{\frac{2 (4 h)}{g}} = v \sqrt{\frac{2 h}{g}}$
$∴ x = 250 m$

EAMCET-1991

Motion in One Dimensions

141291 A particle starts moving from rest under uniform acceleration. It travels a distance $x$ in the first two seconds and a distance of $y$ in the next two seconds. If $y = n x$ , then $n =$

1 1

2 3

3 2

4 4

Explanation:

B Given that,
original image
According to question,
Distance covered in $2 \sec, x = \frac{1}{2} \cdot {at}^{2} = \frac{1}{2} \cdot a \times (2)^{2} = 2 a$
Similarly,
Distance covered in next $4 \sec, y = \frac{1}{2} a (4)^{2} - \frac{1}{2} a (2)^{2}$
$y = 6 a$
From question
$y = nx$
$6 a = n \times 2 a$
$n = 3$

EAMCET-1993

Motion in One Dimensions

141286 A particle moves along $Y$ -axis in such a way that its $y$ -coordinate varies with time $t$ according to the relation $y = 3 + 5 t + 7 t^{2}$ . The initial velocity and acceleration of the particle are respectively:

1

14 {ms}^{- 1}, - 5 {ms}^{- 2}

2

19 {ms}^{- 1}, - 9 {ms}^{- 2}

3

- 14 {ms}^{- 1}, - 5 {ms}^{- 2}

4

5 {ms}^{- 1}, 14 {ms}^{- 2}

Explanation:

D Given that,
$y = 3 + 5 t + 7 t^{2}$
$∴ v = \frac{dy}{dt} = \frac{d}{dt} (3 + 5 t + 7 t^{2})$
$v = 5 + 14 t$
Its initial velocity at $t = 0$
Then, $v = 5 m / \sec$
$∴ a = \frac{dv}{dt} = \frac{d}{dt} (5 + 14 t)$
$a = 0 + 14$
$a = 14$
So, acceleration (a) $= 14 m / \sec^{2}$
The initial velocity and acceleration of particle are 5 $m / s$ and $14 m / s^{2}$ respectively

JCECE-2005

Motion in One Dimensions

141287 An aeroplane moves $400 m$ towards the north, $300 m$ towards west and then $1200 m$ vertically upwards, then its displacement from the initial position is

1

1600 m

2

1800 m

3

1500 m

4

1300 m

Explanation:

D Here, CD is perpendicular to the plane of paper. Required distance $= BD$
original image
${CB}^{2} = {BA}^{2} + {AC}^{2}$
${CB}^{2} = 400^{2} + 300^{2}$
$CB = \sqrt{400^{2} + 300^{2}} = 500 m$
${BD}^{2} = {CB}^{2} + {CD}^{2}$
${BD}^{2} = 500^{2} + 1200^{2}$
$BD = \sqrt{500^{2} + 1200^{2}} = 1300 m$

AIIMS-1998

Motion in One Dimensions

141289 An object accelerates from rest to a velocity 27. $5 m / s$ in 10s, then the distance covered after next $10 s$ is :

1

550 m

2

137.5 m

3

412.5 m

4

175 m

Explanation:

C Given,
$u = 0, v = 27.5 m / s, t = 10 \sec, a = ?$
From $v = u +$ at
$∴ a = \frac{v - u}{t} = \frac{27.5 - 0}{10}$
$= \frac{27.5}{10} = 2.75 m / s$
In the first $10 \sec$ , distance travelled
From $S = u t + \frac{1}{2} a t^{2}$
$S_{1} = 0 \times 10 + \frac{1}{2} \times 2.75 \times (10)^{2} = 137.5 m$
In the first $20 \sec$ , distance travelled
$S_{2} = 0 \times 20 + \frac{1}{2} \times 2.75 \times (20)^{2} = 550 m$
Hence, distance travelled in next $10 s$ .
$= 550 - 137.5$
$= 412.5 m$

BCECE-2004

Motion in One Dimensions

141290 A body of mass $m$ thrown horizontally with velocity $v$ , from the top of tower of height $h$ , touches the level of ground at a distance of 250 $m$ from the foot of the tower. A body of mass 2
$m$ , thrown horizontally with velocity $v / 2$ , from the top of a tower of height $4 h$ will touch the level ground at a distance in metre from the foot of the tower is

1

150 m

2

200 m

3

250 m

4

2500 m

Explanation:

C Given data,
Distance from the foot of the tower, $= 250 m$
As we know,
$t = \sqrt{\frac{2 h}{g}}$
$∴$ Distance, $d = vt = v \sqrt{\frac{2 h}{g}} = 250 m$
According to given question,
When, $v = \frac{v}{2}, h = 4 h$
Then, distance, $x = \frac{v}{2} \sqrt{\frac{2 (4 h)}{g}} = v \sqrt{\frac{2 h}{g}}$
$∴ x = 250 m$

EAMCET-1991

Motion in One Dimensions

141291 A particle starts moving from rest under uniform acceleration. It travels a distance $x$ in the first two seconds and a distance of $y$ in the next two seconds. If $y = n x$ , then $n =$

1 1

2 3

3 2

4 4

Explanation:

B Given that,
original image
According to question,
Distance covered in $2 \sec, x = \frac{1}{2} \cdot {at}^{2} = \frac{1}{2} \cdot a \times (2)^{2} = 2 a$
Similarly,
Distance covered in next $4 \sec, y = \frac{1}{2} a (4)^{2} - \frac{1}{2} a (2)^{2}$
$y = 6 a$
From question
$y = nx$
$6 a = n \times 2 a$
$n = 3$

EAMCET-1993