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Motion in One Dimensions

141344 The position of a particle moving along $x$ -axis is given by $x = 3 t - 4 t^{2} + t^{3}$ , Where $x$ is in meter and $t$ in seconds. The average velocity of the particle in the time interval from $t = 2$ seconds to $t = 4$ seconds is

1

7 m / s

2

1 m / s

3

13 m / s

4 None of these

Explanation:

A Position of the particle is given by
$x (t) = 3 t - 4 t^{2} + t^{3}$
Average velocity of the particle is
$v_{avg} = \frac{x (t_{2}) - x (t_{1})}{t_{2} - t_{1}} t_{1} = 2, t_{2} = 4$
$v_{avg} = \frac{x (4) - x (2)}{4 - 2}$
Now, $x (4) = 3 (4) - 4 (4)^{2} + 4^{3} = 12 m$
$x (2) = 3 (2) - 4 (2)^{2} + 2^{3} = - 2 m$
$∴ v_{avg} = \frac{12 - (- 2)}{2} = \frac{14}{2} = 7 m / s$

SRM JEE - 2012

Motion in One Dimensions

141346 A particle starts from rest. Its acceleration (a) versus time ( $t$ ) is as shown in the figure. The maximum speed of the particle will be
original image

1

110 m / s

2

45 m / s

3

900 m / s

4

90 m / s

Explanation:

B Area under the acceleration time graph will give maximum speed of the particle.
$v_{max} = \frac{1}{2} \times base \times height$
$v_{max} = \frac{1}{2} \times 10 \times 9 = 45 m / s$

SRM JEE - 2013

Motion in One Dimensions

141347 A car is moving along a straight road with a uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocity $30 km / h$ and $40 km / h$ respectively. The velocity of the car midway between $P$ and $Q$ is

1

33.3 km / h

2

20 \sqrt{2} km / h

3

25 \sqrt{2} km / h

4

0.35 km / h

Explanation:

C By using the equation of motion $b / w$ point $P$ and $M$ ,
original image
$v^{2} = u^{2} + 2 as$
$\Rightarrow v^{2} = (30)^{2} + 2$ as
$\Rightarrow v^{2} - (30)^{2} = 2 as$
$\Rightarrow v^{2} - 900 = 2 a \frac{x}{2}$
$\Rightarrow v^{2} - 900 = ax$
$a = \frac{v^{2} - 900}{x}$
Using equation of motion between point $M$ and $Q$ ,
$(40)^{2} = v^{2} + 2 a \frac{x}{2}$
$1600 = v^{2} + a x$
$a = \frac{1600 - v^{2}}{x}$
From equation (i) and (ii)
$\frac{v^{2} - 900}{x} = \frac{1600 - v^{2}}{x}$
$2 v^{2} = 2500$
$v^{2} = \frac{2500}{2}$
$v = \sqrt{\frac{2500}{2}}$
$v = \frac{50}{\sqrt{2}}$
Multiplying by $\sqrt{2}$
$v = \frac{50 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
$v = \frac{50 \sqrt{2}}{2}$
$v = 25 \sqrt{2} km / h$

AIPMT 1988

Motion in One Dimensions

141343 The velocity time graphs of two bodies $A$ and $B$ are shown in figure. The ratio of their acceleration is
original image

1

1 : \sqrt{3}

2

1 : 3

3

\sqrt{3} : 1

4

\sqrt{3} : \sqrt{2}

Explanation:

A Slope of the velocity time graph gives acceleration (a)
$∴ \frac{a_{A}}{a_{B}} = \frac{\tan 45^{\circ}}{\tan 60^{\circ}} = \frac{1}{\sqrt{3}}$

SRM JEE - 2012

Motion in One Dimensions

141344 The position of a particle moving along $x$ -axis is given by $x = 3 t - 4 t^{2} + t^{3}$ , Where $x$ is in meter and $t$ in seconds. The average velocity of the particle in the time interval from $t = 2$ seconds to $t = 4$ seconds is

1

7 m / s

2

1 m / s

3

13 m / s

4 None of these

Explanation:

A Position of the particle is given by
$x (t) = 3 t - 4 t^{2} + t^{3}$
Average velocity of the particle is
$v_{avg} = \frac{x (t_{2}) - x (t_{1})}{t_{2} - t_{1}} t_{1} = 2, t_{2} = 4$
$v_{avg} = \frac{x (4) - x (2)}{4 - 2}$
Now, $x (4) = 3 (4) - 4 (4)^{2} + 4^{3} = 12 m$
$x (2) = 3 (2) - 4 (2)^{2} + 2^{3} = - 2 m$
$∴ v_{avg} = \frac{12 - (- 2)}{2} = \frac{14}{2} = 7 m / s$

SRM JEE - 2012

Motion in One Dimensions

141346 A particle starts from rest. Its acceleration (a) versus time ( $t$ ) is as shown in the figure. The maximum speed of the particle will be
original image

1

110 m / s

2

45 m / s

3

900 m / s

4

90 m / s

Explanation:

B Area under the acceleration time graph will give maximum speed of the particle.
$v_{max} = \frac{1}{2} \times base \times height$
$v_{max} = \frac{1}{2} \times 10 \times 9 = 45 m / s$

SRM JEE - 2013

Motion in One Dimensions

141347 A car is moving along a straight road with a uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocity $30 km / h$ and $40 km / h$ respectively. The velocity of the car midway between $P$ and $Q$ is

1

33.3 km / h

2

20 \sqrt{2} km / h

3

25 \sqrt{2} km / h

4

0.35 km / h

Explanation:

C By using the equation of motion $b / w$ point $P$ and $M$ ,
original image
$v^{2} = u^{2} + 2 as$
$\Rightarrow v^{2} = (30)^{2} + 2$ as
$\Rightarrow v^{2} - (30)^{2} = 2 as$
$\Rightarrow v^{2} - 900 = 2 a \frac{x}{2}$
$\Rightarrow v^{2} - 900 = ax$
$a = \frac{v^{2} - 900}{x}$
Using equation of motion between point $M$ and $Q$ ,
$(40)^{2} = v^{2} + 2 a \frac{x}{2}$
$1600 = v^{2} + a x$
$a = \frac{1600 - v^{2}}{x}$
From equation (i) and (ii)
$\frac{v^{2} - 900}{x} = \frac{1600 - v^{2}}{x}$
$2 v^{2} = 2500$
$v^{2} = \frac{2500}{2}$
$v = \sqrt{\frac{2500}{2}}$
$v = \frac{50}{\sqrt{2}}$
Multiplying by $\sqrt{2}$
$v = \frac{50 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
$v = \frac{50 \sqrt{2}}{2}$
$v = 25 \sqrt{2} km / h$

AIPMT 1988

Motion in One Dimensions

141343 The velocity time graphs of two bodies $A$ and $B$ are shown in figure. The ratio of their acceleration is
original image

1

1 : \sqrt{3}

2

1 : 3

3

\sqrt{3} : 1

4

\sqrt{3} : \sqrt{2}

Explanation:

A Slope of the velocity time graph gives acceleration (a)
$∴ \frac{a_{A}}{a_{B}} = \frac{\tan 45^{\circ}}{\tan 60^{\circ}} = \frac{1}{\sqrt{3}}$

SRM JEE - 2012

Motion in One Dimensions

141344 The position of a particle moving along $x$ -axis is given by $x = 3 t - 4 t^{2} + t^{3}$ , Where $x$ is in meter and $t$ in seconds. The average velocity of the particle in the time interval from $t = 2$ seconds to $t = 4$ seconds is

1

7 m / s

2

1 m / s

3

13 m / s

4 None of these

Explanation:

A Position of the particle is given by
$x (t) = 3 t - 4 t^{2} + t^{3}$
Average velocity of the particle is
$v_{avg} = \frac{x (t_{2}) - x (t_{1})}{t_{2} - t_{1}} t_{1} = 2, t_{2} = 4$
$v_{avg} = \frac{x (4) - x (2)}{4 - 2}$
Now, $x (4) = 3 (4) - 4 (4)^{2} + 4^{3} = 12 m$
$x (2) = 3 (2) - 4 (2)^{2} + 2^{3} = - 2 m$
$∴ v_{avg} = \frac{12 - (- 2)}{2} = \frac{14}{2} = 7 m / s$

SRM JEE - 2012

Motion in One Dimensions

141346 A particle starts from rest. Its acceleration (a) versus time ( $t$ ) is as shown in the figure. The maximum speed of the particle will be
original image

1

110 m / s

2

45 m / s

3

900 m / s

4

90 m / s

Explanation:

B Area under the acceleration time graph will give maximum speed of the particle.
$v_{max} = \frac{1}{2} \times base \times height$
$v_{max} = \frac{1}{2} \times 10 \times 9 = 45 m / s$

SRM JEE - 2013

Motion in One Dimensions

141347 A car is moving along a straight road with a uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocity $30 km / h$ and $40 km / h$ respectively. The velocity of the car midway between $P$ and $Q$ is

1

33.3 km / h

2

20 \sqrt{2} km / h

3

25 \sqrt{2} km / h

4

0.35 km / h

Explanation:

C By using the equation of motion $b / w$ point $P$ and $M$ ,
original image
$v^{2} = u^{2} + 2 as$
$\Rightarrow v^{2} = (30)^{2} + 2$ as
$\Rightarrow v^{2} - (30)^{2} = 2 as$
$\Rightarrow v^{2} - 900 = 2 a \frac{x}{2}$
$\Rightarrow v^{2} - 900 = ax$
$a = \frac{v^{2} - 900}{x}$
Using equation of motion between point $M$ and $Q$ ,
$(40)^{2} = v^{2} + 2 a \frac{x}{2}$
$1600 = v^{2} + a x$
$a = \frac{1600 - v^{2}}{x}$
From equation (i) and (ii)
$\frac{v^{2} - 900}{x} = \frac{1600 - v^{2}}{x}$
$2 v^{2} = 2500$
$v^{2} = \frac{2500}{2}$
$v = \sqrt{\frac{2500}{2}}$
$v = \frac{50}{\sqrt{2}}$
Multiplying by $\sqrt{2}$
$v = \frac{50 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
$v = \frac{50 \sqrt{2}}{2}$
$v = 25 \sqrt{2} km / h$

AIPMT 1988

Motion in One Dimensions

141343 The velocity time graphs of two bodies $A$ and $B$ are shown in figure. The ratio of their acceleration is
original image

1

1 : \sqrt{3}

2

1 : 3

3

\sqrt{3} : 1

4

\sqrt{3} : \sqrt{2}

Explanation:

A Slope of the velocity time graph gives acceleration (a)
$∴ \frac{a_{A}}{a_{B}} = \frac{\tan 45^{\circ}}{\tan 60^{\circ}} = \frac{1}{\sqrt{3}}$

SRM JEE - 2012

Motion in One Dimensions

141344 The position of a particle moving along $x$ -axis is given by $x = 3 t - 4 t^{2} + t^{3}$ , Where $x$ is in meter and $t$ in seconds. The average velocity of the particle in the time interval from $t = 2$ seconds to $t = 4$ seconds is

1

7 m / s

2

1 m / s

3

13 m / s

4 None of these

Explanation:

A Position of the particle is given by
$x (t) = 3 t - 4 t^{2} + t^{3}$
Average velocity of the particle is
$v_{avg} = \frac{x (t_{2}) - x (t_{1})}{t_{2} - t_{1}} t_{1} = 2, t_{2} = 4$
$v_{avg} = \frac{x (4) - x (2)}{4 - 2}$
Now, $x (4) = 3 (4) - 4 (4)^{2} + 4^{3} = 12 m$
$x (2) = 3 (2) - 4 (2)^{2} + 2^{3} = - 2 m$
$∴ v_{avg} = \frac{12 - (- 2)}{2} = \frac{14}{2} = 7 m / s$

SRM JEE - 2012

Motion in One Dimensions

141346 A particle starts from rest. Its acceleration (a) versus time ( $t$ ) is as shown in the figure. The maximum speed of the particle will be
original image

1

110 m / s

2

45 m / s

3

900 m / s

4

90 m / s

Explanation:

B Area under the acceleration time graph will give maximum speed of the particle.
$v_{max} = \frac{1}{2} \times base \times height$
$v_{max} = \frac{1}{2} \times 10 \times 9 = 45 m / s$

SRM JEE - 2013

Motion in One Dimensions

141347 A car is moving along a straight road with a uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocity $30 km / h$ and $40 km / h$ respectively. The velocity of the car midway between $P$ and $Q$ is

1

33.3 km / h

2

20 \sqrt{2} km / h

3

25 \sqrt{2} km / h

4

0.35 km / h

Explanation:

C By using the equation of motion $b / w$ point $P$ and $M$ ,
original image
$v^{2} = u^{2} + 2 as$
$\Rightarrow v^{2} = (30)^{2} + 2$ as
$\Rightarrow v^{2} - (30)^{2} = 2 as$
$\Rightarrow v^{2} - 900 = 2 a \frac{x}{2}$
$\Rightarrow v^{2} - 900 = ax$
$a = \frac{v^{2} - 900}{x}$
Using equation of motion between point $M$ and $Q$ ,
$(40)^{2} = v^{2} + 2 a \frac{x}{2}$
$1600 = v^{2} + a x$
$a = \frac{1600 - v^{2}}{x}$
From equation (i) and (ii)
$\frac{v^{2} - 900}{x} = \frac{1600 - v^{2}}{x}$
$2 v^{2} = 2500$
$v^{2} = \frac{2500}{2}$
$v = \sqrt{\frac{2500}{2}}$
$v = \frac{50}{\sqrt{2}}$
Multiplying by $\sqrt{2}$
$v = \frac{50 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
$v = \frac{50 \sqrt{2}}{2}$
$v = 25 \sqrt{2} km / h$

AIPMT 1988

Motion in One Dimensions

141343 The velocity time graphs of two bodies $A$ and $B$ are shown in figure. The ratio of their acceleration is
original image

1

1 : \sqrt{3}

2

1 : 3

3

\sqrt{3} : 1

4

\sqrt{3} : \sqrt{2}

Explanation:

A Slope of the velocity time graph gives acceleration (a)
$∴ \frac{a_{A}}{a_{B}} = \frac{\tan 45^{\circ}}{\tan 60^{\circ}} = \frac{1}{\sqrt{3}}$

SRM JEE - 2012