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Motion in One Dimensions

141657 A particle is moving on a circular path with a constant speed $v$ . It's change of velocity as it moves from $A$ to $B$ in the figure is-
original image

1

2 v \sin \frac{θ}{2}

2

v \sin θ

3

\frac{v \sin 2 θ}{2}

4

2 v \sin θ

Explanation:

A According to the question
original image
$| Δ \vec{v} | = \sqrt{v^{2} + v^{2} - 2 v^{2} \cos θ}$
$= \sqrt{2 v^{2} (1 - \cos θ)}$
$= \sqrt{4 v^{2} \sin^{2} θ / 2}$
$| Δ \vec{v} | = 2 v \sin \frac{θ}{2}$

AP EAMCET-07.07.2022

Motion in One Dimensions

141658 Particle $A$ (which is located at the origin at time $t = 0$ ) is moving along the $x$ - axis with a constant velocity $1 {ms}^{- 1}$ . Another particle $B$ is moving along the $y$ -axis according to, $y = c^{3}$ ; $c = 1 {ms}^{- 1}$ . Then the velocity of $A$ relative to $B$ at $t = 1$ is

1

\sqrt{10} {ms}^{- 1}

2

10 {ms}^{- 1}

3

\sqrt{3} {ms}^{- 1}

4

3 {ms}^{- 1}

Explanation:

A Given,
$v_{x} = 1 m / \sec$
$y = {ct}^{3}$
$\frac{dy}{dt} = v = 3 {ct}^{2}$ [ Differentiating w.r.t time]
$c = 1$
$t = 1$
$v_{y} = 3 \times 1 \times 1^{2}$
$v_{y} = 3 m / \sec$
Hence, the velocity of A relative to $B$
$Δ v = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$= \sqrt{1^{2} + 3^{2}}$
$Δ v = \sqrt{10} m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141659 The variation of induced emf $(F)$ with time ( $t$ ) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as
original image

1

2

3

4

CG PET-2021

Motion in One Dimensions

141660 Two cars $A$ and $B$ initially at rest are moving in same direction with accelerations $a_{1}$ and $a_{2}$ respectively. After a certain time, they achieve velocities $v_{1}$ and $v_{2}$ respectively and separated are by a distance of $50 m$ . If $(a_{1} - a_{2}) = 4 m / s^{2}$ , then the quantity $(v_{1} - v_{2})$ will be

1

24 m / s

2

20 m / s

3

40 m / s

4

12 m / s

Explanation:

B Two car A and B initially at rest are moving in same direction with acceleration $a_{1}$ and $a_{2}$
After a certain time, their velocities $v_{1}$ and $v_{2}$
respectively and separated by distance $(s) = 50 m$ .
Given, $(a_{1} - a_{2}) = 4 m / s^{2}$
Find, $v_{1} - v_{2} =$ ?
By eq., $s = ut + \frac{1}{2} {at}^{2}$
$u = {$ initially at rest $} = 0$
$s_{1} = \frac{1}{2} a_{1} t^{2}$
$s_{2} = \frac{1}{2} a_{2} t^{2}$
$∵ s_{1} - s_{2} = \frac{1}{2} t^{2} (a_{1} - a_{2}) ∵ s_{1} - s_{2} = 50$
$50 = \frac{1}{2} t^{2} (a_{1} - a_{2}) \Rightarrow 100 = t^{2} (4)$
$t^{2} = 25$
$t = 5$
and $V_{1} = u + a_{1} t = a_{1} t$
$V_{2} = u + a_{2} t = a_{2} t$
$V_{1} - V_{2} = (a_{1} - a_{2}) t$
$= 4 \times 5$
$V_{1} - V_{2} = 20 m / s$

TS EAMCET 05.08.2021

Motion in One Dimensions

141657 A particle is moving on a circular path with a constant speed $v$ . It's change of velocity as it moves from $A$ to $B$ in the figure is-
original image

1

2 v \sin \frac{θ}{2}

2

v \sin θ

3

\frac{v \sin 2 θ}{2}

4

2 v \sin θ

Explanation:

A According to the question
original image
$| Δ \vec{v} | = \sqrt{v^{2} + v^{2} - 2 v^{2} \cos θ}$
$= \sqrt{2 v^{2} (1 - \cos θ)}$
$= \sqrt{4 v^{2} \sin^{2} θ / 2}$
$| Δ \vec{v} | = 2 v \sin \frac{θ}{2}$

AP EAMCET-07.07.2022

Motion in One Dimensions

141658 Particle $A$ (which is located at the origin at time $t = 0$ ) is moving along the $x$ - axis with a constant velocity $1 {ms}^{- 1}$ . Another particle $B$ is moving along the $y$ -axis according to, $y = c^{3}$ ; $c = 1 {ms}^{- 1}$ . Then the velocity of $A$ relative to $B$ at $t = 1$ is

1

\sqrt{10} {ms}^{- 1}

2

10 {ms}^{- 1}

3

\sqrt{3} {ms}^{- 1}

4

3 {ms}^{- 1}

Explanation:

A Given,
$v_{x} = 1 m / \sec$
$y = {ct}^{3}$
$\frac{dy}{dt} = v = 3 {ct}^{2}$ [ Differentiating w.r.t time]
$c = 1$
$t = 1$
$v_{y} = 3 \times 1 \times 1^{2}$
$v_{y} = 3 m / \sec$
Hence, the velocity of A relative to $B$
$Δ v = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$= \sqrt{1^{2} + 3^{2}}$
$Δ v = \sqrt{10} m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141659 The variation of induced emf $(F)$ with time ( $t$ ) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as
original image

1

2

3

4

CG PET-2021

Motion in One Dimensions

141660 Two cars $A$ and $B$ initially at rest are moving in same direction with accelerations $a_{1}$ and $a_{2}$ respectively. After a certain time, they achieve velocities $v_{1}$ and $v_{2}$ respectively and separated are by a distance of $50 m$ . If $(a_{1} - a_{2}) = 4 m / s^{2}$ , then the quantity $(v_{1} - v_{2})$ will be

1

24 m / s

2

20 m / s

3

40 m / s

4

12 m / s

Explanation:

B Two car A and B initially at rest are moving in same direction with acceleration $a_{1}$ and $a_{2}$
After a certain time, their velocities $v_{1}$ and $v_{2}$
respectively and separated by distance $(s) = 50 m$ .
Given, $(a_{1} - a_{2}) = 4 m / s^{2}$
Find, $v_{1} - v_{2} =$ ?
By eq., $s = ut + \frac{1}{2} {at}^{2}$
$u = {$ initially at rest $} = 0$
$s_{1} = \frac{1}{2} a_{1} t^{2}$
$s_{2} = \frac{1}{2} a_{2} t^{2}$
$∵ s_{1} - s_{2} = \frac{1}{2} t^{2} (a_{1} - a_{2}) ∵ s_{1} - s_{2} = 50$
$50 = \frac{1}{2} t^{2} (a_{1} - a_{2}) \Rightarrow 100 = t^{2} (4)$
$t^{2} = 25$
$t = 5$
and $V_{1} = u + a_{1} t = a_{1} t$
$V_{2} = u + a_{2} t = a_{2} t$
$V_{1} - V_{2} = (a_{1} - a_{2}) t$
$= 4 \times 5$
$V_{1} - V_{2} = 20 m / s$

TS EAMCET 05.08.2021

Motion in One Dimensions

141657 A particle is moving on a circular path with a constant speed $v$ . It's change of velocity as it moves from $A$ to $B$ in the figure is-
original image

1

2 v \sin \frac{θ}{2}

2

v \sin θ

3

\frac{v \sin 2 θ}{2}

4

2 v \sin θ

Explanation:

A According to the question
original image
$| Δ \vec{v} | = \sqrt{v^{2} + v^{2} - 2 v^{2} \cos θ}$
$= \sqrt{2 v^{2} (1 - \cos θ)}$
$= \sqrt{4 v^{2} \sin^{2} θ / 2}$
$| Δ \vec{v} | = 2 v \sin \frac{θ}{2}$

AP EAMCET-07.07.2022

Motion in One Dimensions

141658 Particle $A$ (which is located at the origin at time $t = 0$ ) is moving along the $x$ - axis with a constant velocity $1 {ms}^{- 1}$ . Another particle $B$ is moving along the $y$ -axis according to, $y = c^{3}$ ; $c = 1 {ms}^{- 1}$ . Then the velocity of $A$ relative to $B$ at $t = 1$ is

1

\sqrt{10} {ms}^{- 1}

2

10 {ms}^{- 1}

3

\sqrt{3} {ms}^{- 1}

4

3 {ms}^{- 1}

Explanation:

A Given,
$v_{x} = 1 m / \sec$
$y = {ct}^{3}$
$\frac{dy}{dt} = v = 3 {ct}^{2}$ [ Differentiating w.r.t time]
$c = 1$
$t = 1$
$v_{y} = 3 \times 1 \times 1^{2}$
$v_{y} = 3 m / \sec$
Hence, the velocity of A relative to $B$
$Δ v = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$= \sqrt{1^{2} + 3^{2}}$
$Δ v = \sqrt{10} m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141659 The variation of induced emf $(F)$ with time ( $t$ ) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as
original image

1

2

3

4

CG PET-2021

Motion in One Dimensions

141660 Two cars $A$ and $B$ initially at rest are moving in same direction with accelerations $a_{1}$ and $a_{2}$ respectively. After a certain time, they achieve velocities $v_{1}$ and $v_{2}$ respectively and separated are by a distance of $50 m$ . If $(a_{1} - a_{2}) = 4 m / s^{2}$ , then the quantity $(v_{1} - v_{2})$ will be

1

24 m / s

2

20 m / s

3

40 m / s

4

12 m / s

Explanation:

B Two car A and B initially at rest are moving in same direction with acceleration $a_{1}$ and $a_{2}$
After a certain time, their velocities $v_{1}$ and $v_{2}$
respectively and separated by distance $(s) = 50 m$ .
Given, $(a_{1} - a_{2}) = 4 m / s^{2}$
Find, $v_{1} - v_{2} =$ ?
By eq., $s = ut + \frac{1}{2} {at}^{2}$
$u = {$ initially at rest $} = 0$
$s_{1} = \frac{1}{2} a_{1} t^{2}$
$s_{2} = \frac{1}{2} a_{2} t^{2}$
$∵ s_{1} - s_{2} = \frac{1}{2} t^{2} (a_{1} - a_{2}) ∵ s_{1} - s_{2} = 50$
$50 = \frac{1}{2} t^{2} (a_{1} - a_{2}) \Rightarrow 100 = t^{2} (4)$
$t^{2} = 25$
$t = 5$
and $V_{1} = u + a_{1} t = a_{1} t$
$V_{2} = u + a_{2} t = a_{2} t$
$V_{1} - V_{2} = (a_{1} - a_{2}) t$
$= 4 \times 5$
$V_{1} - V_{2} = 20 m / s$

TS EAMCET 05.08.2021

Motion in One Dimensions

141657 A particle is moving on a circular path with a constant speed $v$ . It's change of velocity as it moves from $A$ to $B$ in the figure is-
original image

1

2 v \sin \frac{θ}{2}

2

v \sin θ

3

\frac{v \sin 2 θ}{2}

4

2 v \sin θ

Explanation:

A According to the question
original image
$| Δ \vec{v} | = \sqrt{v^{2} + v^{2} - 2 v^{2} \cos θ}$
$= \sqrt{2 v^{2} (1 - \cos θ)}$
$= \sqrt{4 v^{2} \sin^{2} θ / 2}$
$| Δ \vec{v} | = 2 v \sin \frac{θ}{2}$

AP EAMCET-07.07.2022

Motion in One Dimensions

141658 Particle $A$ (which is located at the origin at time $t = 0$ ) is moving along the $x$ - axis with a constant velocity $1 {ms}^{- 1}$ . Another particle $B$ is moving along the $y$ -axis according to, $y = c^{3}$ ; $c = 1 {ms}^{- 1}$ . Then the velocity of $A$ relative to $B$ at $t = 1$ is

1

\sqrt{10} {ms}^{- 1}

2

10 {ms}^{- 1}

3

\sqrt{3} {ms}^{- 1}

4

3 {ms}^{- 1}

Explanation:

A Given,
$v_{x} = 1 m / \sec$
$y = {ct}^{3}$
$\frac{dy}{dt} = v = 3 {ct}^{2}$ [ Differentiating w.r.t time]
$c = 1$
$t = 1$
$v_{y} = 3 \times 1 \times 1^{2}$
$v_{y} = 3 m / \sec$
Hence, the velocity of A relative to $B$
$Δ v = \sqrt{v_{x}^{2} + v_{y}^{2}}$
$= \sqrt{1^{2} + 3^{2}}$
$Δ v = \sqrt{10} m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141659 The variation of induced emf $(F)$ with time ( $t$ ) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as
original image

1

2

3

4

CG PET-2021

Motion in One Dimensions

141660 Two cars $A$ and $B$ initially at rest are moving in same direction with accelerations $a_{1}$ and $a_{2}$ respectively. After a certain time, they achieve velocities $v_{1}$ and $v_{2}$ respectively and separated are by a distance of $50 m$ . If $(a_{1} - a_{2}) = 4 m / s^{2}$ , then the quantity $(v_{1} - v_{2})$ will be

1

24 m / s

2

20 m / s

3

40 m / s

4

12 m / s

Explanation:

B Two car A and B initially at rest are moving in same direction with acceleration $a_{1}$ and $a_{2}$
After a certain time, their velocities $v_{1}$ and $v_{2}$
respectively and separated by distance $(s) = 50 m$ .
Given, $(a_{1} - a_{2}) = 4 m / s^{2}$
Find, $v_{1} - v_{2} =$ ?
By eq., $s = ut + \frac{1}{2} {at}^{2}$
$u = {$ initially at rest $} = 0$
$s_{1} = \frac{1}{2} a_{1} t^{2}$
$s_{2} = \frac{1}{2} a_{2} t^{2}$
$∵ s_{1} - s_{2} = \frac{1}{2} t^{2} (a_{1} - a_{2}) ∵ s_{1} - s_{2} = 50$
$50 = \frac{1}{2} t^{2} (a_{1} - a_{2}) \Rightarrow 100 = t^{2} (4)$
$t^{2} = 25$
$t = 5$
and $V_{1} = u + a_{1} t = a_{1} t$
$V_{2} = u + a_{2} t = a_{2} t$
$V_{1} - V_{2} = (a_{1} - a_{2}) t$
$= 4 \times 5$
$V_{1} - V_{2} = 20 m / s$

TS EAMCET 05.08.2021