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PHXI04:MOTION IN A PLANE

362085 A boat is sent across a river with a velocity of 8 $k m$ / $h$ . If the resultant velocity of boat is 10 $k m$ / $h$ , then velocity of river is

1

10 k m / h

2

8 k m / h

3

6 k m / h

4

4 k m / h

Explanation:

Given $A B =$ velocity of boat $= 8 k m / h$
$A C =$ Resultant velocity of boat $= 10 k m / h$
$∴ B C =$ Velocity of river $= \sqrt{A C^{2} - A B^{2}}$
$= \sqrt{{(10)}^{2} - {(8)}^{2}} = 6 k m / h$
supporting img

PHXI04:MOTION IN A PLANE

362086 A swimmer can swim in still water with a speed of $\sqrt{5} m / s .$ While crosses a river his average speed is 3 $m$ / $s$ . If he cross the river in the shortest possible time, what is the speed of flow of water?

1

2 m / s

2

4 m / s

3

6 m / s

4

8 m / s

Explanation:

Let $v_{r}$ is the velocity of river flow.
$3 = \sqrt{{(\sqrt{5})}^{2} + v_{r}^{2}} \Rightarrow v_{r} = 2 m / s$
supporting img

PHXI04:MOTION IN A PLANE

362087 A person swims in a river aiming to reach exactly on the opposite point on the bank of a river. His speed of swimming is 0.5 $m$ / $s$ at an angle with the direction of $120^{\circ}$ flow of water. The speed of water is

1

0.5 m / s

2

1.0 m / s

3

0.43 m / s

4

0.25 m / s

Explanation:

Here, $v = 0.5 m / \sec . u = ?$
So $\sin θ = \frac{v_{r}}{v} \Rightarrow \frac{v_{r}}{0.5} = \frac{1}{2}$ or $u = 0.25 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

362088 A river is flowing at the rate of 6 $k m$ / $h r$ a swimmer swims across the river with a velocity of 9 $k m$ / $h r$ w.r.t water. The resultant velocity of the man is ( in $k m$ / $h$ )

1

\sqrt{117}

2

\sqrt{340}

3

\sqrt{17}

4

3 \sqrt{40}

Explanation:

Assume that river flow is along $+ x$ -axis and the swimmer swims along $+ y$ -axis w.r.to water
Given that
${\vec{v}}_{r} = 6 \hat{i}$
${\vec{v}}_{m r} = 9 \hat{j}$ ( perpendicular to river flow)
${\vec{v}}_{m r} = {\vec{v}}_{m} - {\vec{v}}_{r}$
${\vec{v}}_{m} = ({\vec{v}}_{r} + {\vec{v}}_{m r})$
${\vec{v}}_{m} = 6 \hat{i} + 9 \hat{j}$
$v_{m} = \sqrt{117} k m / h .$

PHXI04:MOTION IN A PLANE

362085 A boat is sent across a river with a velocity of 8 $k m$ / $h$ . If the resultant velocity of boat is 10 $k m$ / $h$ , then velocity of river is

1

10 k m / h

2

8 k m / h

3

6 k m / h

4

4 k m / h

Explanation:

Given $A B =$ velocity of boat $= 8 k m / h$
$A C =$ Resultant velocity of boat $= 10 k m / h$
$∴ B C =$ Velocity of river $= \sqrt{A C^{2} - A B^{2}}$
$= \sqrt{{(10)}^{2} - {(8)}^{2}} = 6 k m / h$
supporting img

PHXI04:MOTION IN A PLANE

362086 A swimmer can swim in still water with a speed of $\sqrt{5} m / s .$ While crosses a river his average speed is 3 $m$ / $s$ . If he cross the river in the shortest possible time, what is the speed of flow of water?

1

2 m / s

2

4 m / s

3

6 m / s

4

8 m / s

Explanation:

Let $v_{r}$ is the velocity of river flow.
$3 = \sqrt{{(\sqrt{5})}^{2} + v_{r}^{2}} \Rightarrow v_{r} = 2 m / s$
supporting img

PHXI04:MOTION IN A PLANE

362087 A person swims in a river aiming to reach exactly on the opposite point on the bank of a river. His speed of swimming is 0.5 $m$ / $s$ at an angle with the direction of $120^{\circ}$ flow of water. The speed of water is

1

0.5 m / s

2

1.0 m / s

3

0.43 m / s

4

0.25 m / s

Explanation:

Here, $v = 0.5 m / \sec . u = ?$
So $\sin θ = \frac{v_{r}}{v} \Rightarrow \frac{v_{r}}{0.5} = \frac{1}{2}$ or $u = 0.25 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

362088 A river is flowing at the rate of 6 $k m$ / $h r$ a swimmer swims across the river with a velocity of 9 $k m$ / $h r$ w.r.t water. The resultant velocity of the man is ( in $k m$ / $h$ )

1

\sqrt{117}

2

\sqrt{340}

3

\sqrt{17}

4

3 \sqrt{40}

Explanation:

Assume that river flow is along $+ x$ -axis and the swimmer swims along $+ y$ -axis w.r.to water
Given that
${\vec{v}}_{r} = 6 \hat{i}$
${\vec{v}}_{m r} = 9 \hat{j}$ ( perpendicular to river flow)
${\vec{v}}_{m r} = {\vec{v}}_{m} - {\vec{v}}_{r}$
${\vec{v}}_{m} = ({\vec{v}}_{r} + {\vec{v}}_{m r})$
${\vec{v}}_{m} = 6 \hat{i} + 9 \hat{j}$
$v_{m} = \sqrt{117} k m / h .$

PHXI04:MOTION IN A PLANE

362085 A boat is sent across a river with a velocity of 8 $k m$ / $h$ . If the resultant velocity of boat is 10 $k m$ / $h$ , then velocity of river is

1

10 k m / h

2

8 k m / h

3

6 k m / h

4

4 k m / h

Explanation:

Given $A B =$ velocity of boat $= 8 k m / h$
$A C =$ Resultant velocity of boat $= 10 k m / h$
$∴ B C =$ Velocity of river $= \sqrt{A C^{2} - A B^{2}}$
$= \sqrt{{(10)}^{2} - {(8)}^{2}} = 6 k m / h$
supporting img

PHXI04:MOTION IN A PLANE

362086 A swimmer can swim in still water with a speed of $\sqrt{5} m / s .$ While crosses a river his average speed is 3 $m$ / $s$ . If he cross the river in the shortest possible time, what is the speed of flow of water?

1

2 m / s

2

4 m / s

3

6 m / s

4

8 m / s

Explanation:

Let $v_{r}$ is the velocity of river flow.
$3 = \sqrt{{(\sqrt{5})}^{2} + v_{r}^{2}} \Rightarrow v_{r} = 2 m / s$
supporting img

PHXI04:MOTION IN A PLANE

362087 A person swims in a river aiming to reach exactly on the opposite point on the bank of a river. His speed of swimming is 0.5 $m$ / $s$ at an angle with the direction of $120^{\circ}$ flow of water. The speed of water is

1

0.5 m / s

2

1.0 m / s

3

0.43 m / s

4

0.25 m / s

Explanation:

Here, $v = 0.5 m / \sec . u = ?$
So $\sin θ = \frac{v_{r}}{v} \Rightarrow \frac{v_{r}}{0.5} = \frac{1}{2}$ or $u = 0.25 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

362088 A river is flowing at the rate of 6 $k m$ / $h r$ a swimmer swims across the river with a velocity of 9 $k m$ / $h r$ w.r.t water. The resultant velocity of the man is ( in $k m$ / $h$ )

1

\sqrt{117}

2

\sqrt{340}

3

\sqrt{17}

4

3 \sqrt{40}

Explanation:

Assume that river flow is along $+ x$ -axis and the swimmer swims along $+ y$ -axis w.r.to water
Given that
${\vec{v}}_{r} = 6 \hat{i}$
${\vec{v}}_{m r} = 9 \hat{j}$ ( perpendicular to river flow)
${\vec{v}}_{m r} = {\vec{v}}_{m} - {\vec{v}}_{r}$
${\vec{v}}_{m} = ({\vec{v}}_{r} + {\vec{v}}_{m r})$
${\vec{v}}_{m} = 6 \hat{i} + 9 \hat{j}$
$v_{m} = \sqrt{117} k m / h .$

PHXI04:MOTION IN A PLANE

362085 A boat is sent across a river with a velocity of 8 $k m$ / $h$ . If the resultant velocity of boat is 10 $k m$ / $h$ , then velocity of river is

1

10 k m / h

2

8 k m / h

3

6 k m / h

4

4 k m / h

Explanation:

Given $A B =$ velocity of boat $= 8 k m / h$
$A C =$ Resultant velocity of boat $= 10 k m / h$
$∴ B C =$ Velocity of river $= \sqrt{A C^{2} - A B^{2}}$
$= \sqrt{{(10)}^{2} - {(8)}^{2}} = 6 k m / h$
supporting img

PHXI04:MOTION IN A PLANE

362086 A swimmer can swim in still water with a speed of $\sqrt{5} m / s .$ While crosses a river his average speed is 3 $m$ / $s$ . If he cross the river in the shortest possible time, what is the speed of flow of water?

1

2 m / s

2

4 m / s

3

6 m / s

4

8 m / s

Explanation:

Let $v_{r}$ is the velocity of river flow.
$3 = \sqrt{{(\sqrt{5})}^{2} + v_{r}^{2}} \Rightarrow v_{r} = 2 m / s$
supporting img

PHXI04:MOTION IN A PLANE

362087 A person swims in a river aiming to reach exactly on the opposite point on the bank of a river. His speed of swimming is 0.5 $m$ / $s$ at an angle with the direction of $120^{\circ}$ flow of water. The speed of water is

1

0.5 m / s

2

1.0 m / s

3

0.43 m / s

4

0.25 m / s

Explanation:

Here, $v = 0.5 m / \sec . u = ?$
So $\sin θ = \frac{v_{r}}{v} \Rightarrow \frac{v_{r}}{0.5} = \frac{1}{2}$ or $u = 0.25 m s^{- 1}$
supporting img

PHXI04:MOTION IN A PLANE

362088 A river is flowing at the rate of 6 $k m$ / $h r$ a swimmer swims across the river with a velocity of 9 $k m$ / $h r$ w.r.t water. The resultant velocity of the man is ( in $k m$ / $h$ )

1

\sqrt{117}

2

\sqrt{340}

3

\sqrt{17}

4

3 \sqrt{40}

Explanation:

Assume that river flow is along $+ x$ -axis and the swimmer swims along $+ y$ -axis w.r.to water
Given that
${\vec{v}}_{r} = 6 \hat{i}$
${\vec{v}}_{m r} = 9 \hat{j}$ ( perpendicular to river flow)
${\vec{v}}_{m r} = {\vec{v}}_{m} - {\vec{v}}_{r}$
${\vec{v}}_{m} = ({\vec{v}}_{r} + {\vec{v}}_{m r})$
${\vec{v}}_{m} = 6 \hat{i} + 9 \hat{j}$
$v_{m} = \sqrt{117} k m / h .$