QBManager

Motion in Plane

143737 A swimmer wants to cross a $200 m$ wide river which is flowing at a speed of $2 m / s$ . The velocity of the swimmer with respect to the river is $1 m / s$ . How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1

200 m

2

400 m

3

600 m

4

800 m

Explanation:

B Given,
velocity of $Man (v_{man}) = 1 m / s$
velocity of river $(v_{river}) = 2 m / s$
width of river $(w) = 200 m$
We know that,
$∵ \frac{W}{v_{man}} = \frac{d}{v_{river}}$
$d = \frac{v_{river} \times W}{v_{man}} \Rightarrow d = \frac{2 \times 200}{1}$
$d = 400 m$

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time $t = 0$ , a force $F = α t$ , where $t$ is time in seconds, applied to a body of mass $1 kg$ , resting on a smooth horizontal plane. If the direction of the force makes an angle of $45^{\circ}$ with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1

\frac{100}{α} m / s

2

\frac{50 \sqrt{2}}{α} m / s

3

\frac{50 α}{\sqrt{2}} m / s

4

\frac{50}{α} m / s

Explanation:

B Given, $t = 0$ , A Force $F = α t$ is applied to small body of mass $m$ resting on a smooth horizontal plane where $α$ is constant.

The normal reaction will be zero where the body will leave the ground surface.
$\Rightarrow N = mg - α t_{0} \sin θ$
$\Rightarrow mg - α t_{0} \sin θ = 0$
$t_{0} = \frac{mg}{α \sin θ}$
$t_{0} =$ time where the body leaves the ground. The force component on horizontal direction is
$\Rightarrow {ma}_{x} = α t \cos θ$
$∵ a_{x} = \frac{{dv}_{x}}{dt}$
$∴ m \times \frac{{dv}_{x}}{dt} = α t \cos θ$
$m \cdot d v_{x} = α \cos θ \cdot t d t$
Integrating both side,
$m \int_{0}^{v} d v_{x} = α \cos θ \int_{0}^{t_{0}} t d t$
$m . v = α \cos θ \cdot (\frac{t_{0}^{2}}{2})$
$\Rightarrow v = \frac{α \cos θ \cdot t_{0}^{2}}{2 m}$
Put $t_{0} = \frac{mg}{α \sin θ}$ in equation (iii)
$v = \frac{α \cos θ}{2 m} \cdot {(\frac{mg}{α \sin θ})}^{2}$
$v = \frac{α \cos θ}{2 m} \cdot \frac{m^{2} g^{2}}{α^{2} \sin^{2} θ}$
$v = \frac{α \cos θ \times m^{2} g^{2}}{α^{2} \sin^{2} θ 2 m}$
$v = \frac{{mg}^{2} \cos θ}{2 α \sin^{2} θ}$
$v = \frac{{mg}^{2}}{2 α \sin θ \cdot \tan θ}$
Given, $m = 1 kg, θ = 45^{\circ}$
$v = \frac{1 \times (10)^{2}}{2 α \cdot \sin 45^{\circ} \tan 45^{\circ}} {∵ g = 10 m / s^{2}}$
$v = \frac{100}{2 α \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow v = \frac{50 \sqrt{2}}{α} m / s$

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass $m_{1}$ slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass $m_{2}$ and can also slide frictionless on the table the velocity of the particle when it exits the track at $B$ is

1

\sqrt{gR}

2

\sqrt{g R (1 + \frac{m_{1}}{m_{2}})}

3

\sqrt{2 gR (m_{1} + m_{2})}

4

\sqrt{\frac{2 g {gm}_{2}}{(m_{1} + m_{2})}}

Explanation:

D From question
We have
$m_{1} v = m_{2} u$
$u = \frac{m_{1} v}{m_{2}}$
By conservation of Energy
$∴ \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} u^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} {(\frac{m_{1} v}{m_{2}})}^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} \times \frac{m_{1}^{2} v^{2}}{m_{2}^{2}} = m_{1} g R$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} \frac{m_{1}^{2} v^{2}}{m_{2}} = m_{1} g R$
$\frac{1}{2} v^{2} + \frac{1}{2} \frac{m_{1}}{m_{2}} v^{2} = g R$
$\frac{1}{2} v^{2} (1 + \frac{m_{1}}{m_{2}}) = g R$
$\frac{1}{2} v^{2} (\frac{m_{2} + m_{1}}{m_{2}}) = g R$
$v^{2} = \frac{2 g R m_{2}}{(m_{1} + m_{2})}$
$v = \sqrt{\frac{2 g R m_{2}}{(m_{1} + m_{2})}}$

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of $50 km / h$ . If the speed remains unchanged after turning through $90^{\circ}$ , the increase in the velocity of bus in the turning process is

1

70.7 km / h

along South-West direction

2 zero

3

50 km / h

along West

4

70.7 km / h

along North-West direction

Explanation:

A Let ${\vec{v}}_{1} = 50 \hat{j} kmph$
and ${\vec{v}}_{2} = - 50 \hat{i} kmph$
Let North $= Y$ axis
West $= X$ -axis (when the car turns left)
Change in velocity is
${\vec{v}}_{2} - {\vec{v}}_{1} = (- 50 \hat{i} - 50 \hat{j}) kmph$
$∴$ Magnitude of change in velocity is $=$
$| {\vec{V}}_{2} - {\vec{V}}_{1} | = \sqrt{50^{2} + 50^{2}}$
$= 50 \sqrt{2} = 50 \times 1.414$
$= 70.7 km / h$ along south - west direction

AIPMT 1989

Motion in Plane

143737 A swimmer wants to cross a $200 m$ wide river which is flowing at a speed of $2 m / s$ . The velocity of the swimmer with respect to the river is $1 m / s$ . How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1

200 m

2

400 m

3

600 m

4

800 m

Explanation:

B Given,
velocity of $Man (v_{man}) = 1 m / s$
velocity of river $(v_{river}) = 2 m / s$
width of river $(w) = 200 m$
We know that,
$∵ \frac{W}{v_{man}} = \frac{d}{v_{river}}$
$d = \frac{v_{river} \times W}{v_{man}} \Rightarrow d = \frac{2 \times 200}{1}$
$d = 400 m$

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time $t = 0$ , a force $F = α t$ , where $t$ is time in seconds, applied to a body of mass $1 kg$ , resting on a smooth horizontal plane. If the direction of the force makes an angle of $45^{\circ}$ with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1

\frac{100}{α} m / s

2

\frac{50 \sqrt{2}}{α} m / s

3

\frac{50 α}{\sqrt{2}} m / s

4

\frac{50}{α} m / s

Explanation:

B Given, $t = 0$ , A Force $F = α t$ is applied to small body of mass $m$ resting on a smooth horizontal plane where $α$ is constant.

The normal reaction will be zero where the body will leave the ground surface.
$\Rightarrow N = mg - α t_{0} \sin θ$
$\Rightarrow mg - α t_{0} \sin θ = 0$
$t_{0} = \frac{mg}{α \sin θ}$
$t_{0} =$ time where the body leaves the ground. The force component on horizontal direction is
$\Rightarrow {ma}_{x} = α t \cos θ$
$∵ a_{x} = \frac{{dv}_{x}}{dt}$
$∴ m \times \frac{{dv}_{x}}{dt} = α t \cos θ$
$m \cdot d v_{x} = α \cos θ \cdot t d t$
Integrating both side,
$m \int_{0}^{v} d v_{x} = α \cos θ \int_{0}^{t_{0}} t d t$
$m . v = α \cos θ \cdot (\frac{t_{0}^{2}}{2})$
$\Rightarrow v = \frac{α \cos θ \cdot t_{0}^{2}}{2 m}$
Put $t_{0} = \frac{mg}{α \sin θ}$ in equation (iii)
$v = \frac{α \cos θ}{2 m} \cdot {(\frac{mg}{α \sin θ})}^{2}$
$v = \frac{α \cos θ}{2 m} \cdot \frac{m^{2} g^{2}}{α^{2} \sin^{2} θ}$
$v = \frac{α \cos θ \times m^{2} g^{2}}{α^{2} \sin^{2} θ 2 m}$
$v = \frac{{mg}^{2} \cos θ}{2 α \sin^{2} θ}$
$v = \frac{{mg}^{2}}{2 α \sin θ \cdot \tan θ}$
Given, $m = 1 kg, θ = 45^{\circ}$
$v = \frac{1 \times (10)^{2}}{2 α \cdot \sin 45^{\circ} \tan 45^{\circ}} {∵ g = 10 m / s^{2}}$
$v = \frac{100}{2 α \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow v = \frac{50 \sqrt{2}}{α} m / s$

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass $m_{1}$ slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass $m_{2}$ and can also slide frictionless on the table the velocity of the particle when it exits the track at $B$ is

1

\sqrt{gR}

2

\sqrt{g R (1 + \frac{m_{1}}{m_{2}})}

3

\sqrt{2 gR (m_{1} + m_{2})}

4

\sqrt{\frac{2 g {gm}_{2}}{(m_{1} + m_{2})}}

Explanation:

D From question
We have
$m_{1} v = m_{2} u$
$u = \frac{m_{1} v}{m_{2}}$
By conservation of Energy
$∴ \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} u^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} {(\frac{m_{1} v}{m_{2}})}^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} \times \frac{m_{1}^{2} v^{2}}{m_{2}^{2}} = m_{1} g R$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} \frac{m_{1}^{2} v^{2}}{m_{2}} = m_{1} g R$
$\frac{1}{2} v^{2} + \frac{1}{2} \frac{m_{1}}{m_{2}} v^{2} = g R$
$\frac{1}{2} v^{2} (1 + \frac{m_{1}}{m_{2}}) = g R$
$\frac{1}{2} v^{2} (\frac{m_{2} + m_{1}}{m_{2}}) = g R$
$v^{2} = \frac{2 g R m_{2}}{(m_{1} + m_{2})}$
$v = \sqrt{\frac{2 g R m_{2}}{(m_{1} + m_{2})}}$

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of $50 km / h$ . If the speed remains unchanged after turning through $90^{\circ}$ , the increase in the velocity of bus in the turning process is

1

70.7 km / h

along South-West direction

2 zero

3

50 km / h

along West

4

70.7 km / h

along North-West direction

Explanation:

A Let ${\vec{v}}_{1} = 50 \hat{j} kmph$
and ${\vec{v}}_{2} = - 50 \hat{i} kmph$
Let North $= Y$ axis
West $= X$ -axis (when the car turns left)
Change in velocity is
${\vec{v}}_{2} - {\vec{v}}_{1} = (- 50 \hat{i} - 50 \hat{j}) kmph$
$∴$ Magnitude of change in velocity is $=$
$| {\vec{V}}_{2} - {\vec{V}}_{1} | = \sqrt{50^{2} + 50^{2}}$
$= 50 \sqrt{2} = 50 \times 1.414$
$= 70.7 km / h$ along south - west direction

AIPMT 1989

Motion in Plane

143737 A swimmer wants to cross a $200 m$ wide river which is flowing at a speed of $2 m / s$ . The velocity of the swimmer with respect to the river is $1 m / s$ . How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1

200 m

2

400 m

3

600 m

4

800 m

Explanation:

B Given,
velocity of $Man (v_{man}) = 1 m / s$
velocity of river $(v_{river}) = 2 m / s$
width of river $(w) = 200 m$
We know that,
$∵ \frac{W}{v_{man}} = \frac{d}{v_{river}}$
$d = \frac{v_{river} \times W}{v_{man}} \Rightarrow d = \frac{2 \times 200}{1}$
$d = 400 m$

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time $t = 0$ , a force $F = α t$ , where $t$ is time in seconds, applied to a body of mass $1 kg$ , resting on a smooth horizontal plane. If the direction of the force makes an angle of $45^{\circ}$ with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1

\frac{100}{α} m / s

2

\frac{50 \sqrt{2}}{α} m / s

3

\frac{50 α}{\sqrt{2}} m / s

4

\frac{50}{α} m / s

Explanation:

B Given, $t = 0$ , A Force $F = α t$ is applied to small body of mass $m$ resting on a smooth horizontal plane where $α$ is constant.

The normal reaction will be zero where the body will leave the ground surface.
$\Rightarrow N = mg - α t_{0} \sin θ$
$\Rightarrow mg - α t_{0} \sin θ = 0$
$t_{0} = \frac{mg}{α \sin θ}$
$t_{0} =$ time where the body leaves the ground. The force component on horizontal direction is
$\Rightarrow {ma}_{x} = α t \cos θ$
$∵ a_{x} = \frac{{dv}_{x}}{dt}$
$∴ m \times \frac{{dv}_{x}}{dt} = α t \cos θ$
$m \cdot d v_{x} = α \cos θ \cdot t d t$
Integrating both side,
$m \int_{0}^{v} d v_{x} = α \cos θ \int_{0}^{t_{0}} t d t$
$m . v = α \cos θ \cdot (\frac{t_{0}^{2}}{2})$
$\Rightarrow v = \frac{α \cos θ \cdot t_{0}^{2}}{2 m}$
Put $t_{0} = \frac{mg}{α \sin θ}$ in equation (iii)
$v = \frac{α \cos θ}{2 m} \cdot {(\frac{mg}{α \sin θ})}^{2}$
$v = \frac{α \cos θ}{2 m} \cdot \frac{m^{2} g^{2}}{α^{2} \sin^{2} θ}$
$v = \frac{α \cos θ \times m^{2} g^{2}}{α^{2} \sin^{2} θ 2 m}$
$v = \frac{{mg}^{2} \cos θ}{2 α \sin^{2} θ}$
$v = \frac{{mg}^{2}}{2 α \sin θ \cdot \tan θ}$
Given, $m = 1 kg, θ = 45^{\circ}$
$v = \frac{1 \times (10)^{2}}{2 α \cdot \sin 45^{\circ} \tan 45^{\circ}} {∵ g = 10 m / s^{2}}$
$v = \frac{100}{2 α \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow v = \frac{50 \sqrt{2}}{α} m / s$

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass $m_{1}$ slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass $m_{2}$ and can also slide frictionless on the table the velocity of the particle when it exits the track at $B$ is

1

\sqrt{gR}

2

\sqrt{g R (1 + \frac{m_{1}}{m_{2}})}

3

\sqrt{2 gR (m_{1} + m_{2})}

4

\sqrt{\frac{2 g {gm}_{2}}{(m_{1} + m_{2})}}

Explanation:

D From question
We have
$m_{1} v = m_{2} u$
$u = \frac{m_{1} v}{m_{2}}$
By conservation of Energy
$∴ \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} u^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} {(\frac{m_{1} v}{m_{2}})}^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} \times \frac{m_{1}^{2} v^{2}}{m_{2}^{2}} = m_{1} g R$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} \frac{m_{1}^{2} v^{2}}{m_{2}} = m_{1} g R$
$\frac{1}{2} v^{2} + \frac{1}{2} \frac{m_{1}}{m_{2}} v^{2} = g R$
$\frac{1}{2} v^{2} (1 + \frac{m_{1}}{m_{2}}) = g R$
$\frac{1}{2} v^{2} (\frac{m_{2} + m_{1}}{m_{2}}) = g R$
$v^{2} = \frac{2 g R m_{2}}{(m_{1} + m_{2})}$
$v = \sqrt{\frac{2 g R m_{2}}{(m_{1} + m_{2})}}$

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of $50 km / h$ . If the speed remains unchanged after turning through $90^{\circ}$ , the increase in the velocity of bus in the turning process is

1

70.7 km / h

along South-West direction

2 zero

3

50 km / h

along West

4

70.7 km / h

along North-West direction

Explanation:

A Let ${\vec{v}}_{1} = 50 \hat{j} kmph$
and ${\vec{v}}_{2} = - 50 \hat{i} kmph$
Let North $= Y$ axis
West $= X$ -axis (when the car turns left)
Change in velocity is
${\vec{v}}_{2} - {\vec{v}}_{1} = (- 50 \hat{i} - 50 \hat{j}) kmph$
$∴$ Magnitude of change in velocity is $=$
$| {\vec{V}}_{2} - {\vec{V}}_{1} | = \sqrt{50^{2} + 50^{2}}$
$= 50 \sqrt{2} = 50 \times 1.414$
$= 70.7 km / h$ along south - west direction

AIPMT 1989

Motion in Plane

143737 A swimmer wants to cross a $200 m$ wide river which is flowing at a speed of $2 m / s$ . The velocity of the swimmer with respect to the river is $1 m / s$ . How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1

200 m

2

400 m

3

600 m

4

800 m

Explanation:

B Given,
velocity of $Man (v_{man}) = 1 m / s$
velocity of river $(v_{river}) = 2 m / s$
width of river $(w) = 200 m$
We know that,
$∵ \frac{W}{v_{man}} = \frac{d}{v_{river}}$
$d = \frac{v_{river} \times W}{v_{man}} \Rightarrow d = \frac{2 \times 200}{1}$
$d = 400 m$

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time $t = 0$ , a force $F = α t$ , where $t$ is time in seconds, applied to a body of mass $1 kg$ , resting on a smooth horizontal plane. If the direction of the force makes an angle of $45^{\circ}$ with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1

\frac{100}{α} m / s

2

\frac{50 \sqrt{2}}{α} m / s

3

\frac{50 α}{\sqrt{2}} m / s

4

\frac{50}{α} m / s

Explanation:

B Given, $t = 0$ , A Force $F = α t$ is applied to small body of mass $m$ resting on a smooth horizontal plane where $α$ is constant.

The normal reaction will be zero where the body will leave the ground surface.
$\Rightarrow N = mg - α t_{0} \sin θ$
$\Rightarrow mg - α t_{0} \sin θ = 0$
$t_{0} = \frac{mg}{α \sin θ}$
$t_{0} =$ time where the body leaves the ground. The force component on horizontal direction is
$\Rightarrow {ma}_{x} = α t \cos θ$
$∵ a_{x} = \frac{{dv}_{x}}{dt}$
$∴ m \times \frac{{dv}_{x}}{dt} = α t \cos θ$
$m \cdot d v_{x} = α \cos θ \cdot t d t$
Integrating both side,
$m \int_{0}^{v} d v_{x} = α \cos θ \int_{0}^{t_{0}} t d t$
$m . v = α \cos θ \cdot (\frac{t_{0}^{2}}{2})$
$\Rightarrow v = \frac{α \cos θ \cdot t_{0}^{2}}{2 m}$
Put $t_{0} = \frac{mg}{α \sin θ}$ in equation (iii)
$v = \frac{α \cos θ}{2 m} \cdot {(\frac{mg}{α \sin θ})}^{2}$
$v = \frac{α \cos θ}{2 m} \cdot \frac{m^{2} g^{2}}{α^{2} \sin^{2} θ}$
$v = \frac{α \cos θ \times m^{2} g^{2}}{α^{2} \sin^{2} θ 2 m}$
$v = \frac{{mg}^{2} \cos θ}{2 α \sin^{2} θ}$
$v = \frac{{mg}^{2}}{2 α \sin θ \cdot \tan θ}$
Given, $m = 1 kg, θ = 45^{\circ}$
$v = \frac{1 \times (10)^{2}}{2 α \cdot \sin 45^{\circ} \tan 45^{\circ}} {∵ g = 10 m / s^{2}}$
$v = \frac{100}{2 α \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow v = \frac{50 \sqrt{2}}{α} m / s$

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass $m_{1}$ slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass $m_{2}$ and can also slide frictionless on the table the velocity of the particle when it exits the track at $B$ is

1

\sqrt{gR}

2

\sqrt{g R (1 + \frac{m_{1}}{m_{2}})}

3

\sqrt{2 gR (m_{1} + m_{2})}

4

\sqrt{\frac{2 g {gm}_{2}}{(m_{1} + m_{2})}}

Explanation:

D From question
We have
$m_{1} v = m_{2} u$
$u = \frac{m_{1} v}{m_{2}}$
By conservation of Energy
$∴ \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} u^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} {(\frac{m_{1} v}{m_{2}})}^{2} = m_{1} gR$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} \times \frac{m_{1}^{2} v^{2}}{m_{2}^{2}} = m_{1} g R$
$\frac{1}{2} m_{1} v^{2} + \frac{1}{2} \frac{m_{1}^{2} v^{2}}{m_{2}} = m_{1} g R$
$\frac{1}{2} v^{2} + \frac{1}{2} \frac{m_{1}}{m_{2}} v^{2} = g R$
$\frac{1}{2} v^{2} (1 + \frac{m_{1}}{m_{2}}) = g R$
$\frac{1}{2} v^{2} (\frac{m_{2} + m_{1}}{m_{2}}) = g R$
$v^{2} = \frac{2 g R m_{2}}{(m_{1} + m_{2})}$
$v = \sqrt{\frac{2 g R m_{2}}{(m_{1} + m_{2})}}$

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of $50 km / h$ . If the speed remains unchanged after turning through $90^{\circ}$ , the increase in the velocity of bus in the turning process is

1

70.7 km / h

along South-West direction

2 zero

3

50 km / h

along West

4

70.7 km / h

along North-West direction

Explanation:

A Let ${\vec{v}}_{1} = 50 \hat{j} kmph$
and ${\vec{v}}_{2} = - 50 \hat{i} kmph$
Let North $= Y$ axis
West $= X$ -axis (when the car turns left)
Change in velocity is
${\vec{v}}_{2} - {\vec{v}}_{1} = (- 50 \hat{i} - 50 \hat{j}) kmph$
$∴$ Magnitude of change in velocity is $=$
$| {\vec{V}}_{2} - {\vec{V}}_{1} | = \sqrt{50^{2} + 50^{2}}$
$= 50 \sqrt{2} = 50 \times 1.414$
$= 70.7 km / h$ along south - west direction

AIPMT 1989

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Question Bank Series