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Motion in One Dimensions

141188 $y = ({P t}^{2} - {Q t}^{3}) m$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

1

\frac{27 P^{3}}{4 Q^{2}}

2

\frac{4 Q^{2}}{27 P^{3}}

3

\frac{4 P^{3}}{27 Q^{2}}

4

\frac{27 Q^{2}}{4 P^{3}}

Explanation:

C In vertical plane the vertical displacement of ball is given as -
$y = ({Pt}^{2} - {Qt}^{3})$
For maximum height,
$\frac{dy}{dt} = 0$
$\frac{d}{dt} ({Pt}^{2} - {Qt}^{3}) = 0$
$(2 Pt - 3 {Qt}^{2}) = 0$
$t (2 P - 3 Qt) = 0$
$2 P - 3 Qt = 0$
$t = \frac{2 P}{3 Q}$
Putting the value of $t$ in equation (i),
$y_{max} = \frac{4 P^{3}}{9 Q^{2}} - \frac{8 P^{3}}{27 Q^{2}}$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}} [3 - 2]$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}}$

AP EAMCET-04.07.2022

Motion in One Dimensions

141189 A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60 m$ , then the magnitude of displacement would be: $($ Given $\cos 135^{\circ} = - 0.7)$
original image

1

42 m

2

47 m

3

19 m

4

40 m

Explanation:

B Distance travelled by Person $= 60 m$ (Arc ACB) Let, $R =$ Radius
original image
Displacement is shortest distance between two point. So, Displacement of person $= AB$ (Straight line) Using consines rule in triangle $BOA$ . $(AB)^{2} = (OA)^{2} + (OB)^{2} - 2 (OA) (OB) \cos (BOA)$ $(A B)^{2} = R^{2} + R^{2} - 2 R \times R \cos 135^{\circ}$
$(A B)^{2} = 2 R^{2} - 2 R^{2} \times (- 0.7) (∵ \cos 135^{\circ} = - 0.7)$
$AB = 1.84 R$
Circumference of circle $= 2 π R$
$R = 60 \times \frac{360}{135 \times 2 π} = \frac{80}{π}$
$= \frac{80}{22} \times 7$
$R = 25.45 m$
Putting value of $R$ in ${eq}^{n}$ (i) we have.
$AB = 1.84 R = 1.84 \times 25.45$
$AB = 46.92 m$
$AB ≅ 47 m$

JEE Main-25.07.2022

Motion in One Dimensions

141191 The relation between time $t$ and distance $x$ for a moving body is given as $t = {m x}^{2} + n x$ , where $m$ and $n$ are constants. The retardation of the motion is (when $v$ stands for velocity)

1

2 {mv}^{3}

2

2 {mnv}^{3}

3

2 {nv}^{3}

4

2 n^{2} v^{3}

Explanation:

A Given,
$t = m x^{2} + nx$
Differentiating the above equation with respect to time, we have
$\frac{dt}{dt} = m \frac{d (x^{2})}{dt} + n \frac{dx}{dt}$
$1 = 2 mx \frac{dx}{dt} + n \frac{dx}{dt} = 2 mxv + nv$
$1 = v (2 mx + n)$
$v = \frac{1}{2 mx + n}$
$a = \frac{dv}{dt} = \frac{- 2 m (dx / dt)}{(2 mx + n)^{2}} = - 2 m \times v^{2} \times v = - 2 {mv}^{3}$
$∴$ The retardation of the motion is $2 {mv}^{3}$ .

JEE Main-25.07.2021

Motion in One Dimensions

141192 The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between $t = 0 s$ and $t = 10$ s will be
original image

1

120 m

2

90 m

3

60 m

4

30 m

Explanation:

B original image
Distance travelled by the particle $=$ Area under the v-t graph
Area of $◻ ABCO +$ area of $△ BCP$
$= 5 \times 12 + \frac{1}{2} \times 12 \times 5$
$= 60 + 30 = 90 m$

Assam CEE-2021

Motion in One Dimensions

141188 $y = ({P t}^{2} - {Q t}^{3}) m$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

1

\frac{27 P^{3}}{4 Q^{2}}

2

\frac{4 Q^{2}}{27 P^{3}}

3

\frac{4 P^{3}}{27 Q^{2}}

4

\frac{27 Q^{2}}{4 P^{3}}

Explanation:

C In vertical plane the vertical displacement of ball is given as -
$y = ({Pt}^{2} - {Qt}^{3})$
For maximum height,
$\frac{dy}{dt} = 0$
$\frac{d}{dt} ({Pt}^{2} - {Qt}^{3}) = 0$
$(2 Pt - 3 {Qt}^{2}) = 0$
$t (2 P - 3 Qt) = 0$
$2 P - 3 Qt = 0$
$t = \frac{2 P}{3 Q}$
Putting the value of $t$ in equation (i),
$y_{max} = \frac{4 P^{3}}{9 Q^{2}} - \frac{8 P^{3}}{27 Q^{2}}$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}} [3 - 2]$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}}$

AP EAMCET-04.07.2022

Motion in One Dimensions

141189 A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60 m$ , then the magnitude of displacement would be: $($ Given $\cos 135^{\circ} = - 0.7)$
original image

1

42 m

2

47 m

3

19 m

4

40 m

Explanation:

B Distance travelled by Person $= 60 m$ (Arc ACB) Let, $R =$ Radius
original image
Displacement is shortest distance between two point. So, Displacement of person $= AB$ (Straight line) Using consines rule in triangle $BOA$ . $(AB)^{2} = (OA)^{2} + (OB)^{2} - 2 (OA) (OB) \cos (BOA)$ $(A B)^{2} = R^{2} + R^{2} - 2 R \times R \cos 135^{\circ}$
$(A B)^{2} = 2 R^{2} - 2 R^{2} \times (- 0.7) (∵ \cos 135^{\circ} = - 0.7)$
$AB = 1.84 R$
Circumference of circle $= 2 π R$
$R = 60 \times \frac{360}{135 \times 2 π} = \frac{80}{π}$
$= \frac{80}{22} \times 7$
$R = 25.45 m$
Putting value of $R$ in ${eq}^{n}$ (i) we have.
$AB = 1.84 R = 1.84 \times 25.45$
$AB = 46.92 m$
$AB ≅ 47 m$

JEE Main-25.07.2022

Motion in One Dimensions

141191 The relation between time $t$ and distance $x$ for a moving body is given as $t = {m x}^{2} + n x$ , where $m$ and $n$ are constants. The retardation of the motion is (when $v$ stands for velocity)

1

2 {mv}^{3}

2

2 {mnv}^{3}

3

2 {nv}^{3}

4

2 n^{2} v^{3}

Explanation:

A Given,
$t = m x^{2} + nx$
Differentiating the above equation with respect to time, we have
$\frac{dt}{dt} = m \frac{d (x^{2})}{dt} + n \frac{dx}{dt}$
$1 = 2 mx \frac{dx}{dt} + n \frac{dx}{dt} = 2 mxv + nv$
$1 = v (2 mx + n)$
$v = \frac{1}{2 mx + n}$
$a = \frac{dv}{dt} = \frac{- 2 m (dx / dt)}{(2 mx + n)^{2}} = - 2 m \times v^{2} \times v = - 2 {mv}^{3}$
$∴$ The retardation of the motion is $2 {mv}^{3}$ .

JEE Main-25.07.2021

Motion in One Dimensions

141192 The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between $t = 0 s$ and $t = 10$ s will be
original image

1

120 m

2

90 m

3

60 m

4

30 m

Explanation:

B original image
Distance travelled by the particle $=$ Area under the v-t graph
Area of $◻ ABCO +$ area of $△ BCP$
$= 5 \times 12 + \frac{1}{2} \times 12 \times 5$
$= 60 + 30 = 90 m$

Assam CEE-2021

Motion in One Dimensions

141188 $y = ({P t}^{2} - {Q t}^{3}) m$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

1

\frac{27 P^{3}}{4 Q^{2}}

2

\frac{4 Q^{2}}{27 P^{3}}

3

\frac{4 P^{3}}{27 Q^{2}}

4

\frac{27 Q^{2}}{4 P^{3}}

Explanation:

C In vertical plane the vertical displacement of ball is given as -
$y = ({Pt}^{2} - {Qt}^{3})$
For maximum height,
$\frac{dy}{dt} = 0$
$\frac{d}{dt} ({Pt}^{2} - {Qt}^{3}) = 0$
$(2 Pt - 3 {Qt}^{2}) = 0$
$t (2 P - 3 Qt) = 0$
$2 P - 3 Qt = 0$
$t = \frac{2 P}{3 Q}$
Putting the value of $t$ in equation (i),
$y_{max} = \frac{4 P^{3}}{9 Q^{2}} - \frac{8 P^{3}}{27 Q^{2}}$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}} [3 - 2]$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}}$

AP EAMCET-04.07.2022

Motion in One Dimensions

141189 A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60 m$ , then the magnitude of displacement would be: $($ Given $\cos 135^{\circ} = - 0.7)$
original image

1

42 m

2

47 m

3

19 m

4

40 m

Explanation:

B Distance travelled by Person $= 60 m$ (Arc ACB) Let, $R =$ Radius
original image
Displacement is shortest distance between two point. So, Displacement of person $= AB$ (Straight line) Using consines rule in triangle $BOA$ . $(AB)^{2} = (OA)^{2} + (OB)^{2} - 2 (OA) (OB) \cos (BOA)$ $(A B)^{2} = R^{2} + R^{2} - 2 R \times R \cos 135^{\circ}$
$(A B)^{2} = 2 R^{2} - 2 R^{2} \times (- 0.7) (∵ \cos 135^{\circ} = - 0.7)$
$AB = 1.84 R$
Circumference of circle $= 2 π R$
$R = 60 \times \frac{360}{135 \times 2 π} = \frac{80}{π}$
$= \frac{80}{22} \times 7$
$R = 25.45 m$
Putting value of $R$ in ${eq}^{n}$ (i) we have.
$AB = 1.84 R = 1.84 \times 25.45$
$AB = 46.92 m$
$AB ≅ 47 m$

JEE Main-25.07.2022

Motion in One Dimensions

141191 The relation between time $t$ and distance $x$ for a moving body is given as $t = {m x}^{2} + n x$ , where $m$ and $n$ are constants. The retardation of the motion is (when $v$ stands for velocity)

1

2 {mv}^{3}

2

2 {mnv}^{3}

3

2 {nv}^{3}

4

2 n^{2} v^{3}

Explanation:

A Given,
$t = m x^{2} + nx$
Differentiating the above equation with respect to time, we have
$\frac{dt}{dt} = m \frac{d (x^{2})}{dt} + n \frac{dx}{dt}$
$1 = 2 mx \frac{dx}{dt} + n \frac{dx}{dt} = 2 mxv + nv$
$1 = v (2 mx + n)$
$v = \frac{1}{2 mx + n}$
$a = \frac{dv}{dt} = \frac{- 2 m (dx / dt)}{(2 mx + n)^{2}} = - 2 m \times v^{2} \times v = - 2 {mv}^{3}$
$∴$ The retardation of the motion is $2 {mv}^{3}$ .

JEE Main-25.07.2021

Motion in One Dimensions

141192 The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between $t = 0 s$ and $t = 10$ s will be
original image

1

120 m

2

90 m

3

60 m

4

30 m

Explanation:

B original image
Distance travelled by the particle $=$ Area under the v-t graph
Area of $◻ ABCO +$ area of $△ BCP$
$= 5 \times 12 + \frac{1}{2} \times 12 \times 5$
$= 60 + 30 = 90 m$

Assam CEE-2021

Motion in One Dimensions

141188 $y = ({P t}^{2} - {Q t}^{3}) m$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

1

\frac{27 P^{3}}{4 Q^{2}}

2

\frac{4 Q^{2}}{27 P^{3}}

3

\frac{4 P^{3}}{27 Q^{2}}

4

\frac{27 Q^{2}}{4 P^{3}}

Explanation:

C In vertical plane the vertical displacement of ball is given as -
$y = ({Pt}^{2} - {Qt}^{3})$
For maximum height,
$\frac{dy}{dt} = 0$
$\frac{d}{dt} ({Pt}^{2} - {Qt}^{3}) = 0$
$(2 Pt - 3 {Qt}^{2}) = 0$
$t (2 P - 3 Qt) = 0$
$2 P - 3 Qt = 0$
$t = \frac{2 P}{3 Q}$
Putting the value of $t$ in equation (i),
$y_{max} = \frac{4 P^{3}}{9 Q^{2}} - \frac{8 P^{3}}{27 Q^{2}}$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}} [3 - 2]$
$y_{max} = \frac{4 P^{3}}{27 Q^{2}}$

AP EAMCET-04.07.2022

Motion in One Dimensions

141189 A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60 m$ , then the magnitude of displacement would be: $($ Given $\cos 135^{\circ} = - 0.7)$
original image

1

42 m

2

47 m

3

19 m

4

40 m

Explanation:

B Distance travelled by Person $= 60 m$ (Arc ACB) Let, $R =$ Radius
original image
Displacement is shortest distance between two point. So, Displacement of person $= AB$ (Straight line) Using consines rule in triangle $BOA$ . $(AB)^{2} = (OA)^{2} + (OB)^{2} - 2 (OA) (OB) \cos (BOA)$ $(A B)^{2} = R^{2} + R^{2} - 2 R \times R \cos 135^{\circ}$
$(A B)^{2} = 2 R^{2} - 2 R^{2} \times (- 0.7) (∵ \cos 135^{\circ} = - 0.7)$
$AB = 1.84 R$
Circumference of circle $= 2 π R$
$R = 60 \times \frac{360}{135 \times 2 π} = \frac{80}{π}$
$= \frac{80}{22} \times 7$
$R = 25.45 m$
Putting value of $R$ in ${eq}^{n}$ (i) we have.
$AB = 1.84 R = 1.84 \times 25.45$
$AB = 46.92 m$
$AB ≅ 47 m$

JEE Main-25.07.2022

Motion in One Dimensions

141191 The relation between time $t$ and distance $x$ for a moving body is given as $t = {m x}^{2} + n x$ , where $m$ and $n$ are constants. The retardation of the motion is (when $v$ stands for velocity)

1

2 {mv}^{3}

2

2 {mnv}^{3}

3

2 {nv}^{3}

4

2 n^{2} v^{3}

Explanation:

A Given,
$t = m x^{2} + nx$
Differentiating the above equation with respect to time, we have
$\frac{dt}{dt} = m \frac{d (x^{2})}{dt} + n \frac{dx}{dt}$
$1 = 2 mx \frac{dx}{dt} + n \frac{dx}{dt} = 2 mxv + nv$
$1 = v (2 mx + n)$
$v = \frac{1}{2 mx + n}$
$a = \frac{dv}{dt} = \frac{- 2 m (dx / dt)}{(2 mx + n)^{2}} = - 2 m \times v^{2} \times v = - 2 {mv}^{3}$
$∴$ The retardation of the motion is $2 {mv}^{3}$ .

JEE Main-25.07.2021

Motion in One Dimensions

141192 The speed-time graph of a particle along a fixed direction is shown below. The distance traversed by the particle between $t = 0 s$ and $t = 10$ s will be
original image

1

120 m

2

90 m

3

60 m

4

30 m

Explanation:

B original image
Distance travelled by the particle $=$ Area under the v-t graph
Area of $◻ ABCO +$ area of $△ BCP$
$= 5 \times 12 + \frac{1}{2} \times 12 \times 5$
$= 60 + 30 = 90 m$

Assam CEE-2021