QBManager

PHXI03:MOTION IN A STRAIGHT LINE

362254 The displacement time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the $X$ -axis. The ratio of their velocities is

1

\sqrt{3} : 2

2

1 : 1

3

1 : 2

4

1 : \sqrt{3}

Explanation:

Slope of displacement-time graph gives velocity.
$∴ \frac{v_{1}}{v_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1}{\sqrt{3}}$
supporting img

PHXI03:MOTION IN A STRAIGHT LINE

362255 The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of $30^{\circ} a n d 45^{\circ}$ respectively with the time axis. If the velocity of $A$ is $v_{A}$ and that of $B$ is $v_{B}$ . The value of $v_{A} / v_{B}$ is

1

1 / 2

2

1 / \sqrt{3}

3

\sqrt{3}

4

1 / 3

Explanation:

$v_{A} = \tan 30^{\circ}$ and $v_{B} = \tan 45^{\circ}$
$∴ \frac{v_{A}}{v_{B}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1 / \sqrt{3}}{\sqrt{1}} = \frac{1}{\sqrt{3}}$

PHXI03:MOTION IN A STRAIGHT LINE

362256 The displacement of a body along $x$ -axis depends on time as $\sqrt{x} = t + 1.$ Then the velocity of body

1 Decreases with time

2 Independent of time

3 Increases with time

4 None of these

Explanation:

$\sqrt{x} = t + 1$
Squaring both sides, we get,
$x = {(t + 1)}^{2} = t^{2} + 2 t + 1$
Differentiating it w.r.t. time $t$ , we get, $\frac{d x}{d t} = 2 t + 2$
velocity, $v = \frac{d x}{d t} = 2 t + 2.$

PHXI03:MOTION IN A STRAIGHT LINE

362257 The displacement $y$ (in metre) of a body varies with time $t$ (in second) as $y = - \frac{2}{3} t^{2} + 16 t + 2$ . How long does the body take to come to rest?

1

8 s

2

10 s

3

12 s

4

16 s

Explanation:

Velocity, $v = \frac{d y}{d t} = - \frac{4}{3} t + 16$
For body to be at rest, $v = 0$
$- \frac{4}{3} t + 16 = 0 \Rightarrow t = 12$

PHXI03:MOTION IN A STRAIGHT LINE

362258 The distance travelled by a particle is related to time $t$ as $x = 4 t^{2}$ The velocity of the particle at $t = 5 s$ is

1

20 m s^{- 1}

2

25 m s^{- 1}

3

8 m s^{- 1}

4

40 m s^{- 1}

Explanation:

Given, $x = 4 t^{2}$
$v = \frac{d x}{d t} = 8 t$ at $t = 5 \sec$
$v = 8 \times 5 = 40 m / s$

JEE - 2023

PHXI03:MOTION IN A STRAIGHT LINE

362254 The displacement time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the $X$ -axis. The ratio of their velocities is

1

\sqrt{3} : 2

2

1 : 1

3

1 : 2

4

1 : \sqrt{3}

Explanation:

Slope of displacement-time graph gives velocity.
$∴ \frac{v_{1}}{v_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1}{\sqrt{3}}$
supporting img

PHXI03:MOTION IN A STRAIGHT LINE

362255 The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of $30^{\circ} a n d 45^{\circ}$ respectively with the time axis. If the velocity of $A$ is $v_{A}$ and that of $B$ is $v_{B}$ . The value of $v_{A} / v_{B}$ is

1

1 / 2

2

1 / \sqrt{3}

3

\sqrt{3}

4

1 / 3

Explanation:

$v_{A} = \tan 30^{\circ}$ and $v_{B} = \tan 45^{\circ}$
$∴ \frac{v_{A}}{v_{B}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1 / \sqrt{3}}{\sqrt{1}} = \frac{1}{\sqrt{3}}$

PHXI03:MOTION IN A STRAIGHT LINE

362256 The displacement of a body along $x$ -axis depends on time as $\sqrt{x} = t + 1.$ Then the velocity of body

1 Decreases with time

2 Independent of time

3 Increases with time

4 None of these

Explanation:

$\sqrt{x} = t + 1$
Squaring both sides, we get,
$x = {(t + 1)}^{2} = t^{2} + 2 t + 1$
Differentiating it w.r.t. time $t$ , we get, $\frac{d x}{d t} = 2 t + 2$
velocity, $v = \frac{d x}{d t} = 2 t + 2.$

PHXI03:MOTION IN A STRAIGHT LINE

362257 The displacement $y$ (in metre) of a body varies with time $t$ (in second) as $y = - \frac{2}{3} t^{2} + 16 t + 2$ . How long does the body take to come to rest?

1

8 s

2

10 s

3

12 s

4

16 s

Explanation:

Velocity, $v = \frac{d y}{d t} = - \frac{4}{3} t + 16$
For body to be at rest, $v = 0$
$- \frac{4}{3} t + 16 = 0 \Rightarrow t = 12$

PHXI03:MOTION IN A STRAIGHT LINE

362258 The distance travelled by a particle is related to time $t$ as $x = 4 t^{2}$ The velocity of the particle at $t = 5 s$ is

1

20 m s^{- 1}

2

25 m s^{- 1}

3

8 m s^{- 1}

4

40 m s^{- 1}

Explanation:

Given, $x = 4 t^{2}$
$v = \frac{d x}{d t} = 8 t$ at $t = 5 \sec$
$v = 8 \times 5 = 40 m / s$

JEE - 2023

PHXI03:MOTION IN A STRAIGHT LINE

362254 The displacement time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the $X$ -axis. The ratio of their velocities is

1

\sqrt{3} : 2

2

1 : 1

3

1 : 2

4

1 : \sqrt{3}

Explanation:

Slope of displacement-time graph gives velocity.
$∴ \frac{v_{1}}{v_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1}{\sqrt{3}}$
supporting img

PHXI03:MOTION IN A STRAIGHT LINE

362255 The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of $30^{\circ} a n d 45^{\circ}$ respectively with the time axis. If the velocity of $A$ is $v_{A}$ and that of $B$ is $v_{B}$ . The value of $v_{A} / v_{B}$ is

1

1 / 2

2

1 / \sqrt{3}

3

\sqrt{3}

4

1 / 3

Explanation:

$v_{A} = \tan 30^{\circ}$ and $v_{B} = \tan 45^{\circ}$
$∴ \frac{v_{A}}{v_{B}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1 / \sqrt{3}}{\sqrt{1}} = \frac{1}{\sqrt{3}}$

PHXI03:MOTION IN A STRAIGHT LINE

362256 The displacement of a body along $x$ -axis depends on time as $\sqrt{x} = t + 1.$ Then the velocity of body

1 Decreases with time

2 Independent of time

3 Increases with time

4 None of these

Explanation:

$\sqrt{x} = t + 1$
Squaring both sides, we get,
$x = {(t + 1)}^{2} = t^{2} + 2 t + 1$
Differentiating it w.r.t. time $t$ , we get, $\frac{d x}{d t} = 2 t + 2$
velocity, $v = \frac{d x}{d t} = 2 t + 2.$

PHXI03:MOTION IN A STRAIGHT LINE

362257 The displacement $y$ (in metre) of a body varies with time $t$ (in second) as $y = - \frac{2}{3} t^{2} + 16 t + 2$ . How long does the body take to come to rest?

1

8 s

2

10 s

3

12 s

4

16 s

Explanation:

Velocity, $v = \frac{d y}{d t} = - \frac{4}{3} t + 16$
For body to be at rest, $v = 0$
$- \frac{4}{3} t + 16 = 0 \Rightarrow t = 12$

PHXI03:MOTION IN A STRAIGHT LINE

362258 The distance travelled by a particle is related to time $t$ as $x = 4 t^{2}$ The velocity of the particle at $t = 5 s$ is

1

20 m s^{- 1}

2

25 m s^{- 1}

3

8 m s^{- 1}

4

40 m s^{- 1}

Explanation:

Given, $x = 4 t^{2}$
$v = \frac{d x}{d t} = 8 t$ at $t = 5 \sec$
$v = 8 \times 5 = 40 m / s$

JEE - 2023

PHXI03:MOTION IN A STRAIGHT LINE

362254 The displacement time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the $X$ -axis. The ratio of their velocities is

1

\sqrt{3} : 2

2

1 : 1

3

1 : 2

4

1 : \sqrt{3}

Explanation:

Slope of displacement-time graph gives velocity.
$∴ \frac{v_{1}}{v_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1}{\sqrt{3}}$
supporting img

PHXI03:MOTION IN A STRAIGHT LINE

362255 The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of $30^{\circ} a n d 45^{\circ}$ respectively with the time axis. If the velocity of $A$ is $v_{A}$ and that of $B$ is $v_{B}$ . The value of $v_{A} / v_{B}$ is

1

1 / 2

2

1 / \sqrt{3}

3

\sqrt{3}

4

1 / 3

Explanation:

$v_{A} = \tan 30^{\circ}$ and $v_{B} = \tan 45^{\circ}$
$∴ \frac{v_{A}}{v_{B}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1 / \sqrt{3}}{\sqrt{1}} = \frac{1}{\sqrt{3}}$

PHXI03:MOTION IN A STRAIGHT LINE

362256 The displacement of a body along $x$ -axis depends on time as $\sqrt{x} = t + 1.$ Then the velocity of body

1 Decreases with time

2 Independent of time

3 Increases with time

4 None of these

Explanation:

$\sqrt{x} = t + 1$
Squaring both sides, we get,
$x = {(t + 1)}^{2} = t^{2} + 2 t + 1$
Differentiating it w.r.t. time $t$ , we get, $\frac{d x}{d t} = 2 t + 2$
velocity, $v = \frac{d x}{d t} = 2 t + 2.$

PHXI03:MOTION IN A STRAIGHT LINE

362257 The displacement $y$ (in metre) of a body varies with time $t$ (in second) as $y = - \frac{2}{3} t^{2} + 16 t + 2$ . How long does the body take to come to rest?

1

8 s

2

10 s

3

12 s

4

16 s

Explanation:

Velocity, $v = \frac{d y}{d t} = - \frac{4}{3} t + 16$
For body to be at rest, $v = 0$
$- \frac{4}{3} t + 16 = 0 \Rightarrow t = 12$

PHXI03:MOTION IN A STRAIGHT LINE

362258 The distance travelled by a particle is related to time $t$ as $x = 4 t^{2}$ The velocity of the particle at $t = 5 s$ is

1

20 m s^{- 1}

2

25 m s^{- 1}

3

8 m s^{- 1}

4

40 m s^{- 1}

Explanation:

Given, $x = 4 t^{2}$
$v = \frac{d x}{d t} = 8 t$ at $t = 5 \sec$
$v = 8 \times 5 = 40 m / s$

JEE - 2023

PHXI03:MOTION IN A STRAIGHT LINE

362254 The displacement time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the $X$ -axis. The ratio of their velocities is

1

\sqrt{3} : 2

2

1 : 1

3

1 : 2

4

1 : \sqrt{3}

Explanation:

Slope of displacement-time graph gives velocity.
$∴ \frac{v_{1}}{v_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1}{\sqrt{3}}$
supporting img

PHXI03:MOTION IN A STRAIGHT LINE

362255 The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of $30^{\circ} a n d 45^{\circ}$ respectively with the time axis. If the velocity of $A$ is $v_{A}$ and that of $B$ is $v_{B}$ . The value of $v_{A} / v_{B}$ is

1

1 / 2

2

1 / \sqrt{3}

3

\sqrt{3}

4

1 / 3

Explanation:

$v_{A} = \tan 30^{\circ}$ and $v_{B} = \tan 45^{\circ}$
$∴ \frac{v_{A}}{v_{B}} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}} = \frac{1 / \sqrt{3}}{\sqrt{1}} = \frac{1}{\sqrt{3}}$

PHXI03:MOTION IN A STRAIGHT LINE

362256 The displacement of a body along $x$ -axis depends on time as $\sqrt{x} = t + 1.$ Then the velocity of body

1 Decreases with time

2 Independent of time

3 Increases with time

4 None of these

Explanation:

$\sqrt{x} = t + 1$
Squaring both sides, we get,
$x = {(t + 1)}^{2} = t^{2} + 2 t + 1$
Differentiating it w.r.t. time $t$ , we get, $\frac{d x}{d t} = 2 t + 2$
velocity, $v = \frac{d x}{d t} = 2 t + 2.$

PHXI03:MOTION IN A STRAIGHT LINE

362257 The displacement $y$ (in metre) of a body varies with time $t$ (in second) as $y = - \frac{2}{3} t^{2} + 16 t + 2$ . How long does the body take to come to rest?

1

8 s

2

10 s

3

12 s

4

16 s

Explanation:

Velocity, $v = \frac{d y}{d t} = - \frac{4}{3} t + 16$
For body to be at rest, $v = 0$
$- \frac{4}{3} t + 16 = 0 \Rightarrow t = 12$

PHXI03:MOTION IN A STRAIGHT LINE

362258 The distance travelled by a particle is related to time $t$ as $x = 4 t^{2}$ The velocity of the particle at $t = 5 s$ is

1

20 m s^{- 1}

2

25 m s^{- 1}

3

8 m s^{- 1}

4

40 m s^{- 1}

Explanation:

Given, $x = 4 t^{2}$
$v = \frac{d x}{d t} = 8 t$ at $t = 5 \sec$
$v = 8 \times 5 = 40 m / s$

JEE - 2023

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: