QBManager

Motion in Plane

143704 A block of mass $1 kg$ starts from rest at $x = 0$ and moves along the $X$ -axis under the action of a force $F = k t$ , where $t$ is time and $k = 1 N^{- 1}$ . The distance the block will travel in 6 seconds is

1

36 m

2

72 m

3

108 m

4

18 m

Explanation:

A Given,
$F = kt, m = 1 kg, k = 1 {Ns}^{- 1}, t = 6 \sec$
$F = ma$
$∴ ma = kt$
$1 \times a = kt$
$a = kt$
$\frac{dv}{dt} = kt {∵ a = \frac{dv}{dt}}$
$dv = ktdt$
Integrating both side,
$\int dv = \int ktdt$
$v = k \frac{t^{2}}{2}$
$\frac{dx}{dt} = k \frac{t^{2}}{2} {∵ v = \frac{dx}{dt}}$
$dx = k \frac{t^{2}}{2} \cdot dt$
Again Integrating both side,
$\begin{array}{r} \int dx = \int k \frac{t^{2}}{2} \cdot dt = \frac{k}{2} \frac{t^{3}}{3} = \frac{{kt}^{3}}{6} \\ t = 6 \\ At = k \frac{t^{3}}{6} = 1 \times \frac{6^{3}}{6} = 36 m \end{array}$

WB JEE 2017

Motion in Plane

143705 A cyclist is moving along a curvature of radius ' $r$ ' at a constant speed ' $v$ '. The angle made by the cyclist with the vertical plane is

1

\tan^{- 1} (\frac{v^{2}}{rg})

2

\tan^{- 1} (\frac{rg}{v^{2}})

3

\tan^{- 1} (\frac{v^{2} r}{g})

4

\tan^{- 1} (\frac{r}{v^{2} g})

Explanation:

A From Newton's second law,

$∴$ Vertical component
$mg = R \cos θ$
Horizontal component
Centripetal Force,
$F_{c} = R \sin θ$
$\frac{{mv}^{2}}{r} = R \sin θ$
Equation (ii) divided by eq $^{n}$ (i)
$\frac{\frac{{mv}^{2}}{r}}{mg} = \frac{R \sin θ}{R \cos θ}$
$\tan θ = \frac{v^{2}}{rg}$
$θ = \tan^{- 1} (\frac{v^{2}}{rg})$

WB JEE-2007

Motion in Plane

143706 A boat crosses a river with a velocity of $8 km / h$ . If the resulting velocity of boat is $10 km / h$ , then the velocity of river water is

1

4 km / h

2

6 km / h

3

8 km / h

4

10 km / h

Explanation:

B Resulting velocity of boat $(u_{net}) = 10 km / hr$
velocity of boat $(u_{b}) = 8 km / hr$
We know that,

${\vec{u}}_{net} = {\vec{u}}_{b} + {\vec{u}}_{r}$
$| {\vec{u}}_{net} | = \sqrt{u_{b}^{2} + u_{r}^{2}}$
$10 = \sqrt{(8)^{2} + u_{r}^{2}}$
$100 = 64 + u_{r}^{2}$
$u_{r} = 6 km / h$

UP CPMT-2001

Motion in Plane

143707 An aircraft is flying at a height of ' $H$ ' above the ground and at a speed of ' $V$ '. The maximum angle subtended at a ground observation point by the aircraft after time $T$ is

1

\tan^{- 1} (\frac{VT}{H})

2

\tan^{- 1} (\frac{VT}{2 H})

3

2 \tan^{- 1} (\frac{2 VT}{H})

4

2 \tan^{- 1} (\frac{VT}{2 H})

Explanation:

D Given,
Aircraft is flying at a height $= H$
Speed of aircraft $= V$
Time $= T$ .

Let A, B , and C denotes the position of aircraft as it flies above the observation point $O$ and the angle subtend here is $θ$ from end to end.
Using triangle geometry in $△ AOB$
$\tan \frac{θ}{2} = \frac{AB}{BO}$
$\tan \frac{θ}{2} = \frac{VT}{2 H}$
$\frac{θ}{2} = \tan^{- 1} (\frac{VT}{2 H})$
$θ = 2 \tan^{- 1} (\frac{VT}{2 H})$

TS EAMCET 19.07.2022

Motion in Plane

143709 Two boys are standing at the ends $A$ and $B$ of a ground, where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ with velocity $v_{1}$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$ , where $t$ is

1

\frac{a}{\sqrt{v^{2} + v_{1}^{2}}}

2

\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}

3

\frac{a}{v - v_{1}}

4

\frac{a}{v + v_{1}}

Explanation:

B Let two boys meet at point $C$ after time ' $t$ ' from the starting,

$AC = vt and BC = v_{1} t$
We know that,
$(A C)^{2} = (A B)^{2} + (B C)^{2}$
$v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

JIPMER-2013

Motion in Plane

143704 A block of mass $1 kg$ starts from rest at $x = 0$ and moves along the $X$ -axis under the action of a force $F = k t$ , where $t$ is time and $k = 1 N^{- 1}$ . The distance the block will travel in 6 seconds is

1

36 m

2

72 m

3

108 m

4

18 m

Explanation:

A Given,
$F = kt, m = 1 kg, k = 1 {Ns}^{- 1}, t = 6 \sec$
$F = ma$
$∴ ma = kt$
$1 \times a = kt$
$a = kt$
$\frac{dv}{dt} = kt {∵ a = \frac{dv}{dt}}$
$dv = ktdt$
Integrating both side,
$\int dv = \int ktdt$
$v = k \frac{t^{2}}{2}$
$\frac{dx}{dt} = k \frac{t^{2}}{2} {∵ v = \frac{dx}{dt}}$
$dx = k \frac{t^{2}}{2} \cdot dt$
Again Integrating both side,
$\begin{array}{r} \int dx = \int k \frac{t^{2}}{2} \cdot dt = \frac{k}{2} \frac{t^{3}}{3} = \frac{{kt}^{3}}{6} \\ t = 6 \\ At = k \frac{t^{3}}{6} = 1 \times \frac{6^{3}}{6} = 36 m \end{array}$

WB JEE 2017

Motion in Plane

143705 A cyclist is moving along a curvature of radius ' $r$ ' at a constant speed ' $v$ '. The angle made by the cyclist with the vertical plane is

1

\tan^{- 1} (\frac{v^{2}}{rg})

2

\tan^{- 1} (\frac{rg}{v^{2}})

3

\tan^{- 1} (\frac{v^{2} r}{g})

4

\tan^{- 1} (\frac{r}{v^{2} g})

Explanation:

A From Newton's second law,

$∴$ Vertical component
$mg = R \cos θ$
Horizontal component
Centripetal Force,
$F_{c} = R \sin θ$
$\frac{{mv}^{2}}{r} = R \sin θ$
Equation (ii) divided by eq $^{n}$ (i)
$\frac{\frac{{mv}^{2}}{r}}{mg} = \frac{R \sin θ}{R \cos θ}$
$\tan θ = \frac{v^{2}}{rg}$
$θ = \tan^{- 1} (\frac{v^{2}}{rg})$

WB JEE-2007

Motion in Plane

143706 A boat crosses a river with a velocity of $8 km / h$ . If the resulting velocity of boat is $10 km / h$ , then the velocity of river water is

1

4 km / h

2

6 km / h

3

8 km / h

4

10 km / h

Explanation:

B Resulting velocity of boat $(u_{net}) = 10 km / hr$
velocity of boat $(u_{b}) = 8 km / hr$
We know that,

${\vec{u}}_{net} = {\vec{u}}_{b} + {\vec{u}}_{r}$
$| {\vec{u}}_{net} | = \sqrt{u_{b}^{2} + u_{r}^{2}}$
$10 = \sqrt{(8)^{2} + u_{r}^{2}}$
$100 = 64 + u_{r}^{2}$
$u_{r} = 6 km / h$

UP CPMT-2001

Motion in Plane

143707 An aircraft is flying at a height of ' $H$ ' above the ground and at a speed of ' $V$ '. The maximum angle subtended at a ground observation point by the aircraft after time $T$ is

1

\tan^{- 1} (\frac{VT}{H})

2

\tan^{- 1} (\frac{VT}{2 H})

3

2 \tan^{- 1} (\frac{2 VT}{H})

4

2 \tan^{- 1} (\frac{VT}{2 H})

Explanation:

D Given,
Aircraft is flying at a height $= H$
Speed of aircraft $= V$
Time $= T$ .

Let A, B , and C denotes the position of aircraft as it flies above the observation point $O$ and the angle subtend here is $θ$ from end to end.
Using triangle geometry in $△ AOB$
$\tan \frac{θ}{2} = \frac{AB}{BO}$
$\tan \frac{θ}{2} = \frac{VT}{2 H}$
$\frac{θ}{2} = \tan^{- 1} (\frac{VT}{2 H})$
$θ = 2 \tan^{- 1} (\frac{VT}{2 H})$

TS EAMCET 19.07.2022

Motion in Plane

143709 Two boys are standing at the ends $A$ and $B$ of a ground, where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ with velocity $v_{1}$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$ , where $t$ is

1

\frac{a}{\sqrt{v^{2} + v_{1}^{2}}}

2

\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}

3

\frac{a}{v - v_{1}}

4

\frac{a}{v + v_{1}}

Explanation:

B Let two boys meet at point $C$ after time ' $t$ ' from the starting,

$AC = vt and BC = v_{1} t$
We know that,
$(A C)^{2} = (A B)^{2} + (B C)^{2}$
$v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

JIPMER-2013

Motion in Plane

143704 A block of mass $1 kg$ starts from rest at $x = 0$ and moves along the $X$ -axis under the action of a force $F = k t$ , where $t$ is time and $k = 1 N^{- 1}$ . The distance the block will travel in 6 seconds is

1

36 m

2

72 m

3

108 m

4

18 m

Explanation:

A Given,
$F = kt, m = 1 kg, k = 1 {Ns}^{- 1}, t = 6 \sec$
$F = ma$
$∴ ma = kt$
$1 \times a = kt$
$a = kt$
$\frac{dv}{dt} = kt {∵ a = \frac{dv}{dt}}$
$dv = ktdt$
Integrating both side,
$\int dv = \int ktdt$
$v = k \frac{t^{2}}{2}$
$\frac{dx}{dt} = k \frac{t^{2}}{2} {∵ v = \frac{dx}{dt}}$
$dx = k \frac{t^{2}}{2} \cdot dt$
Again Integrating both side,
$\begin{array}{r} \int dx = \int k \frac{t^{2}}{2} \cdot dt = \frac{k}{2} \frac{t^{3}}{3} = \frac{{kt}^{3}}{6} \\ t = 6 \\ At = k \frac{t^{3}}{6} = 1 \times \frac{6^{3}}{6} = 36 m \end{array}$

WB JEE 2017

Motion in Plane

143705 A cyclist is moving along a curvature of radius ' $r$ ' at a constant speed ' $v$ '. The angle made by the cyclist with the vertical plane is

1

\tan^{- 1} (\frac{v^{2}}{rg})

2

\tan^{- 1} (\frac{rg}{v^{2}})

3

\tan^{- 1} (\frac{v^{2} r}{g})

4

\tan^{- 1} (\frac{r}{v^{2} g})

Explanation:

A From Newton's second law,

$∴$ Vertical component
$mg = R \cos θ$
Horizontal component
Centripetal Force,
$F_{c} = R \sin θ$
$\frac{{mv}^{2}}{r} = R \sin θ$
Equation (ii) divided by eq $^{n}$ (i)
$\frac{\frac{{mv}^{2}}{r}}{mg} = \frac{R \sin θ}{R \cos θ}$
$\tan θ = \frac{v^{2}}{rg}$
$θ = \tan^{- 1} (\frac{v^{2}}{rg})$

WB JEE-2007

Motion in Plane

143706 A boat crosses a river with a velocity of $8 km / h$ . If the resulting velocity of boat is $10 km / h$ , then the velocity of river water is

1

4 km / h

2

6 km / h

3

8 km / h

4

10 km / h

Explanation:

B Resulting velocity of boat $(u_{net}) = 10 km / hr$
velocity of boat $(u_{b}) = 8 km / hr$
We know that,

${\vec{u}}_{net} = {\vec{u}}_{b} + {\vec{u}}_{r}$
$| {\vec{u}}_{net} | = \sqrt{u_{b}^{2} + u_{r}^{2}}$
$10 = \sqrt{(8)^{2} + u_{r}^{2}}$
$100 = 64 + u_{r}^{2}$
$u_{r} = 6 km / h$

UP CPMT-2001

Motion in Plane

143707 An aircraft is flying at a height of ' $H$ ' above the ground and at a speed of ' $V$ '. The maximum angle subtended at a ground observation point by the aircraft after time $T$ is

1

\tan^{- 1} (\frac{VT}{H})

2

\tan^{- 1} (\frac{VT}{2 H})

3

2 \tan^{- 1} (\frac{2 VT}{H})

4

2 \tan^{- 1} (\frac{VT}{2 H})

Explanation:

D Given,
Aircraft is flying at a height $= H$
Speed of aircraft $= V$
Time $= T$ .

Let A, B , and C denotes the position of aircraft as it flies above the observation point $O$ and the angle subtend here is $θ$ from end to end.
Using triangle geometry in $△ AOB$
$\tan \frac{θ}{2} = \frac{AB}{BO}$
$\tan \frac{θ}{2} = \frac{VT}{2 H}$
$\frac{θ}{2} = \tan^{- 1} (\frac{VT}{2 H})$
$θ = 2 \tan^{- 1} (\frac{VT}{2 H})$

TS EAMCET 19.07.2022

Motion in Plane

143709 Two boys are standing at the ends $A$ and $B$ of a ground, where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ with velocity $v_{1}$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$ , where $t$ is

1

\frac{a}{\sqrt{v^{2} + v_{1}^{2}}}

2

\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}

3

\frac{a}{v - v_{1}}

4

\frac{a}{v + v_{1}}

Explanation:

B Let two boys meet at point $C$ after time ' $t$ ' from the starting,

$AC = vt and BC = v_{1} t$
We know that,
$(A C)^{2} = (A B)^{2} + (B C)^{2}$
$v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

JIPMER-2013

Motion in Plane

143704 A block of mass $1 kg$ starts from rest at $x = 0$ and moves along the $X$ -axis under the action of a force $F = k t$ , where $t$ is time and $k = 1 N^{- 1}$ . The distance the block will travel in 6 seconds is

1

36 m

2

72 m

3

108 m

4

18 m

Explanation:

A Given,
$F = kt, m = 1 kg, k = 1 {Ns}^{- 1}, t = 6 \sec$
$F = ma$
$∴ ma = kt$
$1 \times a = kt$
$a = kt$
$\frac{dv}{dt} = kt {∵ a = \frac{dv}{dt}}$
$dv = ktdt$
Integrating both side,
$\int dv = \int ktdt$
$v = k \frac{t^{2}}{2}$
$\frac{dx}{dt} = k \frac{t^{2}}{2} {∵ v = \frac{dx}{dt}}$
$dx = k \frac{t^{2}}{2} \cdot dt$
Again Integrating both side,
$\begin{array}{r} \int dx = \int k \frac{t^{2}}{2} \cdot dt = \frac{k}{2} \frac{t^{3}}{3} = \frac{{kt}^{3}}{6} \\ t = 6 \\ At = k \frac{t^{3}}{6} = 1 \times \frac{6^{3}}{6} = 36 m \end{array}$

WB JEE 2017

Motion in Plane

143705 A cyclist is moving along a curvature of radius ' $r$ ' at a constant speed ' $v$ '. The angle made by the cyclist with the vertical plane is

1

\tan^{- 1} (\frac{v^{2}}{rg})

2

\tan^{- 1} (\frac{rg}{v^{2}})

3

\tan^{- 1} (\frac{v^{2} r}{g})

4

\tan^{- 1} (\frac{r}{v^{2} g})

Explanation:

A From Newton's second law,

$∴$ Vertical component
$mg = R \cos θ$
Horizontal component
Centripetal Force,
$F_{c} = R \sin θ$
$\frac{{mv}^{2}}{r} = R \sin θ$
Equation (ii) divided by eq $^{n}$ (i)
$\frac{\frac{{mv}^{2}}{r}}{mg} = \frac{R \sin θ}{R \cos θ}$
$\tan θ = \frac{v^{2}}{rg}$
$θ = \tan^{- 1} (\frac{v^{2}}{rg})$

WB JEE-2007

Motion in Plane

143706 A boat crosses a river with a velocity of $8 km / h$ . If the resulting velocity of boat is $10 km / h$ , then the velocity of river water is

1

4 km / h

2

6 km / h

3

8 km / h

4

10 km / h

Explanation:

B Resulting velocity of boat $(u_{net}) = 10 km / hr$
velocity of boat $(u_{b}) = 8 km / hr$
We know that,

${\vec{u}}_{net} = {\vec{u}}_{b} + {\vec{u}}_{r}$
$| {\vec{u}}_{net} | = \sqrt{u_{b}^{2} + u_{r}^{2}}$
$10 = \sqrt{(8)^{2} + u_{r}^{2}}$
$100 = 64 + u_{r}^{2}$
$u_{r} = 6 km / h$

UP CPMT-2001

Motion in Plane

143707 An aircraft is flying at a height of ' $H$ ' above the ground and at a speed of ' $V$ '. The maximum angle subtended at a ground observation point by the aircraft after time $T$ is

1

\tan^{- 1} (\frac{VT}{H})

2

\tan^{- 1} (\frac{VT}{2 H})

3

2 \tan^{- 1} (\frac{2 VT}{H})

4

2 \tan^{- 1} (\frac{VT}{2 H})

Explanation:

D Given,
Aircraft is flying at a height $= H$
Speed of aircraft $= V$
Time $= T$ .

Let A, B , and C denotes the position of aircraft as it flies above the observation point $O$ and the angle subtend here is $θ$ from end to end.
Using triangle geometry in $△ AOB$
$\tan \frac{θ}{2} = \frac{AB}{BO}$
$\tan \frac{θ}{2} = \frac{VT}{2 H}$
$\frac{θ}{2} = \tan^{- 1} (\frac{VT}{2 H})$
$θ = 2 \tan^{- 1} (\frac{VT}{2 H})$

TS EAMCET 19.07.2022

Motion in Plane

143709 Two boys are standing at the ends $A$ and $B$ of a ground, where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ with velocity $v_{1}$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$ , where $t$ is

1

\frac{a}{\sqrt{v^{2} + v_{1}^{2}}}

2

\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}

3

\frac{a}{v - v_{1}}

4

\frac{a}{v + v_{1}}

Explanation:

B Let two boys meet at point $C$ after time ' $t$ ' from the starting,

$AC = vt and BC = v_{1} t$
We know that,
$(A C)^{2} = (A B)^{2} + (B C)^{2}$
$v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

JIPMER-2013

Motion in Plane

143704 A block of mass $1 kg$ starts from rest at $x = 0$ and moves along the $X$ -axis under the action of a force $F = k t$ , where $t$ is time and $k = 1 N^{- 1}$ . The distance the block will travel in 6 seconds is

1

36 m

2

72 m

3

108 m

4

18 m

Explanation:

A Given,
$F = kt, m = 1 kg, k = 1 {Ns}^{- 1}, t = 6 \sec$
$F = ma$
$∴ ma = kt$
$1 \times a = kt$
$a = kt$
$\frac{dv}{dt} = kt {∵ a = \frac{dv}{dt}}$
$dv = ktdt$
Integrating both side,
$\int dv = \int ktdt$
$v = k \frac{t^{2}}{2}$
$\frac{dx}{dt} = k \frac{t^{2}}{2} {∵ v = \frac{dx}{dt}}$
$dx = k \frac{t^{2}}{2} \cdot dt$
Again Integrating both side,
$\begin{array}{r} \int dx = \int k \frac{t^{2}}{2} \cdot dt = \frac{k}{2} \frac{t^{3}}{3} = \frac{{kt}^{3}}{6} \\ t = 6 \\ At = k \frac{t^{3}}{6} = 1 \times \frac{6^{3}}{6} = 36 m \end{array}$

WB JEE 2017

Motion in Plane

143705 A cyclist is moving along a curvature of radius ' $r$ ' at a constant speed ' $v$ '. The angle made by the cyclist with the vertical plane is

1

\tan^{- 1} (\frac{v^{2}}{rg})

2

\tan^{- 1} (\frac{rg}{v^{2}})

3

\tan^{- 1} (\frac{v^{2} r}{g})

4

\tan^{- 1} (\frac{r}{v^{2} g})

Explanation:

A From Newton's second law,

$∴$ Vertical component
$mg = R \cos θ$
Horizontal component
Centripetal Force,
$F_{c} = R \sin θ$
$\frac{{mv}^{2}}{r} = R \sin θ$
Equation (ii) divided by eq $^{n}$ (i)
$\frac{\frac{{mv}^{2}}{r}}{mg} = \frac{R \sin θ}{R \cos θ}$
$\tan θ = \frac{v^{2}}{rg}$
$θ = \tan^{- 1} (\frac{v^{2}}{rg})$

WB JEE-2007

Motion in Plane

143706 A boat crosses a river with a velocity of $8 km / h$ . If the resulting velocity of boat is $10 km / h$ , then the velocity of river water is

1

4 km / h

2

6 km / h

3

8 km / h

4

10 km / h

Explanation:

B Resulting velocity of boat $(u_{net}) = 10 km / hr$
velocity of boat $(u_{b}) = 8 km / hr$
We know that,

${\vec{u}}_{net} = {\vec{u}}_{b} + {\vec{u}}_{r}$
$| {\vec{u}}_{net} | = \sqrt{u_{b}^{2} + u_{r}^{2}}$
$10 = \sqrt{(8)^{2} + u_{r}^{2}}$
$100 = 64 + u_{r}^{2}$
$u_{r} = 6 km / h$

UP CPMT-2001

Motion in Plane

143707 An aircraft is flying at a height of ' $H$ ' above the ground and at a speed of ' $V$ '. The maximum angle subtended at a ground observation point by the aircraft after time $T$ is

1

\tan^{- 1} (\frac{VT}{H})

2

\tan^{- 1} (\frac{VT}{2 H})

3

2 \tan^{- 1} (\frac{2 VT}{H})

4

2 \tan^{- 1} (\frac{VT}{2 H})

Explanation:

D Given,
Aircraft is flying at a height $= H$
Speed of aircraft $= V$
Time $= T$ .

Let A, B , and C denotes the position of aircraft as it flies above the observation point $O$ and the angle subtend here is $θ$ from end to end.
Using triangle geometry in $△ AOB$
$\tan \frac{θ}{2} = \frac{AB}{BO}$
$\tan \frac{θ}{2} = \frac{VT}{2 H}$
$\frac{θ}{2} = \tan^{- 1} (\frac{VT}{2 H})$
$θ = 2 \tan^{- 1} (\frac{VT}{2 H})$

TS EAMCET 19.07.2022

Motion in Plane

143709 Two boys are standing at the ends $A$ and $B$ of a ground, where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ with velocity $v_{1}$ The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$ , where $t$ is

1

\frac{a}{\sqrt{v^{2} + v_{1}^{2}}}

2

\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}

3

\frac{a}{v - v_{1}}

4

\frac{a}{v + v_{1}}

Explanation:

B Let two boys meet at point $C$ after time ' $t$ ' from the starting,

$AC = vt and BC = v_{1} t$
We know that,
$(A C)^{2} = (A B)^{2} + (B C)^{2}$
$v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

JIPMER-2013

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: