QBManager

Motion in One Dimensions

141429 To reach point-B from point-A, a person travels $500 m$ North, $400 m$ East and $200 m$ south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person

1

\frac{5}{12} m \cdot s^{- 1}

2

\frac{5}{20} m \cdot s^{- 1}

3

\frac{10}{3} m \cdot s^{- 1}

4

\frac{12}{5} m \cdot s^{- 1}

Explanation:

A original image
Distance $= AB + BC + CD$
$= (500 + 400 + 200)$
$= 1100 m$
Displacement $= AD = \sqrt{(AB - CD)^{2} + {BC}^{2}}$
$= \sqrt{(500 - 200)^{2} + 400^{2}}$
$= 500 m$
Average speed $= \frac{Total distance}{Total time}$
$\frac{1100}{1200} = \frac{11}{12} m / s$
$Average Velocity = \frac{A D}{t}$
$= \frac{500}{1200}$
Average velocity $= \frac{5}{12} {ms}^{- 1}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141430 A particle is at $x = 0$ when $t = 0$ . It moves among $x$ -axis with a velocity given by $v = 5 \sqrt{x}$ . Find the acceleration of this particle

1

8.5 m . s^{- 2}

2

12.5 m \cdot s^{- 2}

3

10.5 m \cdot s^{- 2}

4

11.5 m . s^{- 2}

Explanation:

B We know that
Acceleration $=$ derivative of speed with respect to time Therefore,
$a = \frac{d v}{d t}$
$a = \frac{d v}{d x} \times \frac{d x}{d t}$
$a = d (5 \sqrt{x}) / d x \times \frac{d x}{d t}$
$a = \frac{5}{2 \sqrt{x}} \times v (v = 5 \sqrt{x})$
$a = (\frac{5}{2 \sqrt{x}}) \times 5 \sqrt{x}$
$a = \frac{5}{2} \times 5 = \frac{25}{2} = 12.5 m / s^{2}$
Therefore, acceleration is $12.5 m / s^{2}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141432 The acceleration (a) of an object varies as a function of its velocity $(v)$ as $a = λ \sqrt{v}$ where $λ$ is a constant. If at $t = 0, v = 0$ , then the velocity as a function of time $(t)$ is given as

1

\frac{3}{4} λ^{2} t^{3}

2

\frac{1}{4} λ t

3

\frac{1}{4} λ^{2} t^{2}

4

\frac{1}{4} λ^{2} t

Explanation:

C We know that,
$a = \frac{dv}{dt} {a = λ \sqrt{v} - given}$
$dv = a \cdot dt$
Integration both side
$\int_{0}^{v} \frac{dv}{\sqrt{v}} = λ \int_{t = 0}^{t} dt$
$[2 \sqrt{v}]_{0}^{v} = λ [t]_{0}^{t}$
$2 \sqrt{v} = λ [t - 0]$
Squaring both side
$4 v = λ^{2} t^{2}$
$v = \frac{λ^{2} t^{2}}{4}$

UPSEE 2020

Motion in One Dimensions

141433 Find the relation between ' $x$ ' and ' $y$ ', when a body starting from rest, moves with uniform acceleration and travels a distance $x$ in the first $n$ seconds and $y$ in the next $n$ seconds:

1

3 x = y

2

x = 3 y

3

x = y

4

x = 2 y

Explanation:

A From given question.
The distance covered in $2 \sec, x = \frac{1}{2} a (2)^{2} = 2 a$
and the distance covered in next $2 \sec$ ,
$y = \frac{1}{2} \cdot a (4)^{2} - \frac{1}{2} a \cdot (2)^{2}$
$y = 6 a$
So, $\frac{x}{y} = \frac{2 a}{6 a} = \frac{1}{3}$
$3 x = y \Rightarrow y = 3 x$

AP EAMCET-25.09.2020

Motion in One Dimensions

141429 To reach point-B from point-A, a person travels $500 m$ North, $400 m$ East and $200 m$ south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person

1

\frac{5}{12} m \cdot s^{- 1}

2

\frac{5}{20} m \cdot s^{- 1}

3

\frac{10}{3} m \cdot s^{- 1}

4

\frac{12}{5} m \cdot s^{- 1}

Explanation:

A original image
Distance $= AB + BC + CD$
$= (500 + 400 + 200)$
$= 1100 m$
Displacement $= AD = \sqrt{(AB - CD)^{2} + {BC}^{2}}$
$= \sqrt{(500 - 200)^{2} + 400^{2}}$
$= 500 m$
Average speed $= \frac{Total distance}{Total time}$
$\frac{1100}{1200} = \frac{11}{12} m / s$
$Average Velocity = \frac{A D}{t}$
$= \frac{500}{1200}$
Average velocity $= \frac{5}{12} {ms}^{- 1}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141430 A particle is at $x = 0$ when $t = 0$ . It moves among $x$ -axis with a velocity given by $v = 5 \sqrt{x}$ . Find the acceleration of this particle

1

8.5 m . s^{- 2}

2

12.5 m \cdot s^{- 2}

3

10.5 m \cdot s^{- 2}

4

11.5 m . s^{- 2}

Explanation:

B We know that
Acceleration $=$ derivative of speed with respect to time Therefore,
$a = \frac{d v}{d t}$
$a = \frac{d v}{d x} \times \frac{d x}{d t}$
$a = d (5 \sqrt{x}) / d x \times \frac{d x}{d t}$
$a = \frac{5}{2 \sqrt{x}} \times v (v = 5 \sqrt{x})$
$a = (\frac{5}{2 \sqrt{x}}) \times 5 \sqrt{x}$
$a = \frac{5}{2} \times 5 = \frac{25}{2} = 12.5 m / s^{2}$
Therefore, acceleration is $12.5 m / s^{2}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141432 The acceleration (a) of an object varies as a function of its velocity $(v)$ as $a = λ \sqrt{v}$ where $λ$ is a constant. If at $t = 0, v = 0$ , then the velocity as a function of time $(t)$ is given as

1

\frac{3}{4} λ^{2} t^{3}

2

\frac{1}{4} λ t

3

\frac{1}{4} λ^{2} t^{2}

4

\frac{1}{4} λ^{2} t

Explanation:

C We know that,
$a = \frac{dv}{dt} {a = λ \sqrt{v} - given}$
$dv = a \cdot dt$
Integration both side
$\int_{0}^{v} \frac{dv}{\sqrt{v}} = λ \int_{t = 0}^{t} dt$
$[2 \sqrt{v}]_{0}^{v} = λ [t]_{0}^{t}$
$2 \sqrt{v} = λ [t - 0]$
Squaring both side
$4 v = λ^{2} t^{2}$
$v = \frac{λ^{2} t^{2}}{4}$

UPSEE 2020

Motion in One Dimensions

141433 Find the relation between ' $x$ ' and ' $y$ ', when a body starting from rest, moves with uniform acceleration and travels a distance $x$ in the first $n$ seconds and $y$ in the next $n$ seconds:

1

3 x = y

2

x = 3 y

3

x = y

4

x = 2 y

Explanation:

A From given question.
The distance covered in $2 \sec, x = \frac{1}{2} a (2)^{2} = 2 a$
and the distance covered in next $2 \sec$ ,
$y = \frac{1}{2} \cdot a (4)^{2} - \frac{1}{2} a \cdot (2)^{2}$
$y = 6 a$
So, $\frac{x}{y} = \frac{2 a}{6 a} = \frac{1}{3}$
$3 x = y \Rightarrow y = 3 x$

AP EAMCET-25.09.2020

Motion in One Dimensions

141429 To reach point-B from point-A, a person travels $500 m$ North, $400 m$ East and $200 m$ south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person

1

\frac{5}{12} m \cdot s^{- 1}

2

\frac{5}{20} m \cdot s^{- 1}

3

\frac{10}{3} m \cdot s^{- 1}

4

\frac{12}{5} m \cdot s^{- 1}

Explanation:

A original image
Distance $= AB + BC + CD$
$= (500 + 400 + 200)$
$= 1100 m$
Displacement $= AD = \sqrt{(AB - CD)^{2} + {BC}^{2}}$
$= \sqrt{(500 - 200)^{2} + 400^{2}}$
$= 500 m$
Average speed $= \frac{Total distance}{Total time}$
$\frac{1100}{1200} = \frac{11}{12} m / s$
$Average Velocity = \frac{A D}{t}$
$= \frac{500}{1200}$
Average velocity $= \frac{5}{12} {ms}^{- 1}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141430 A particle is at $x = 0$ when $t = 0$ . It moves among $x$ -axis with a velocity given by $v = 5 \sqrt{x}$ . Find the acceleration of this particle

1

8.5 m . s^{- 2}

2

12.5 m \cdot s^{- 2}

3

10.5 m \cdot s^{- 2}

4

11.5 m . s^{- 2}

Explanation:

B We know that
Acceleration $=$ derivative of speed with respect to time Therefore,
$a = \frac{d v}{d t}$
$a = \frac{d v}{d x} \times \frac{d x}{d t}$
$a = d (5 \sqrt{x}) / d x \times \frac{d x}{d t}$
$a = \frac{5}{2 \sqrt{x}} \times v (v = 5 \sqrt{x})$
$a = (\frac{5}{2 \sqrt{x}}) \times 5 \sqrt{x}$
$a = \frac{5}{2} \times 5 = \frac{25}{2} = 12.5 m / s^{2}$
Therefore, acceleration is $12.5 m / s^{2}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141432 The acceleration (a) of an object varies as a function of its velocity $(v)$ as $a = λ \sqrt{v}$ where $λ$ is a constant. If at $t = 0, v = 0$ , then the velocity as a function of time $(t)$ is given as

1

\frac{3}{4} λ^{2} t^{3}

2

\frac{1}{4} λ t

3

\frac{1}{4} λ^{2} t^{2}

4

\frac{1}{4} λ^{2} t

Explanation:

C We know that,
$a = \frac{dv}{dt} {a = λ \sqrt{v} - given}$
$dv = a \cdot dt$
Integration both side
$\int_{0}^{v} \frac{dv}{\sqrt{v}} = λ \int_{t = 0}^{t} dt$
$[2 \sqrt{v}]_{0}^{v} = λ [t]_{0}^{t}$
$2 \sqrt{v} = λ [t - 0]$
Squaring both side
$4 v = λ^{2} t^{2}$
$v = \frac{λ^{2} t^{2}}{4}$

UPSEE 2020

Motion in One Dimensions

141433 Find the relation between ' $x$ ' and ' $y$ ', when a body starting from rest, moves with uniform acceleration and travels a distance $x$ in the first $n$ seconds and $y$ in the next $n$ seconds:

1

3 x = y

2

x = 3 y

3

x = y

4

x = 2 y

Explanation:

A From given question.
The distance covered in $2 \sec, x = \frac{1}{2} a (2)^{2} = 2 a$
and the distance covered in next $2 \sec$ ,
$y = \frac{1}{2} \cdot a (4)^{2} - \frac{1}{2} a \cdot (2)^{2}$
$y = 6 a$
So, $\frac{x}{y} = \frac{2 a}{6 a} = \frac{1}{3}$
$3 x = y \Rightarrow y = 3 x$

AP EAMCET-25.09.2020

Motion in One Dimensions

141429 To reach point-B from point-A, a person travels $500 m$ North, $400 m$ East and $200 m$ south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person

1

\frac{5}{12} m \cdot s^{- 1}

2

\frac{5}{20} m \cdot s^{- 1}

3

\frac{10}{3} m \cdot s^{- 1}

4

\frac{12}{5} m \cdot s^{- 1}

Explanation:

A original image
Distance $= AB + BC + CD$
$= (500 + 400 + 200)$
$= 1100 m$
Displacement $= AD = \sqrt{(AB - CD)^{2} + {BC}^{2}}$
$= \sqrt{(500 - 200)^{2} + 400^{2}}$
$= 500 m$
Average speed $= \frac{Total distance}{Total time}$
$\frac{1100}{1200} = \frac{11}{12} m / s$
$Average Velocity = \frac{A D}{t}$
$= \frac{500}{1200}$
Average velocity $= \frac{5}{12} {ms}^{- 1}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141430 A particle is at $x = 0$ when $t = 0$ . It moves among $x$ -axis with a velocity given by $v = 5 \sqrt{x}$ . Find the acceleration of this particle

1

8.5 m . s^{- 2}

2

12.5 m \cdot s^{- 2}

3

10.5 m \cdot s^{- 2}

4

11.5 m . s^{- 2}

Explanation:

B We know that
Acceleration $=$ derivative of speed with respect to time Therefore,
$a = \frac{d v}{d t}$
$a = \frac{d v}{d x} \times \frac{d x}{d t}$
$a = d (5 \sqrt{x}) / d x \times \frac{d x}{d t}$
$a = \frac{5}{2 \sqrt{x}} \times v (v = 5 \sqrt{x})$
$a = (\frac{5}{2 \sqrt{x}}) \times 5 \sqrt{x}$
$a = \frac{5}{2} \times 5 = \frac{25}{2} = 12.5 m / s^{2}$
Therefore, acceleration is $12.5 m / s^{2}$

AP EAMCET-05.10.2021

Motion in One Dimensions

141432 The acceleration (a) of an object varies as a function of its velocity $(v)$ as $a = λ \sqrt{v}$ where $λ$ is a constant. If at $t = 0, v = 0$ , then the velocity as a function of time $(t)$ is given as

1

\frac{3}{4} λ^{2} t^{3}

2

\frac{1}{4} λ t

3

\frac{1}{4} λ^{2} t^{2}

4

\frac{1}{4} λ^{2} t

Explanation:

C We know that,
$a = \frac{dv}{dt} {a = λ \sqrt{v} - given}$
$dv = a \cdot dt$
Integration both side
$\int_{0}^{v} \frac{dv}{\sqrt{v}} = λ \int_{t = 0}^{t} dt$
$[2 \sqrt{v}]_{0}^{v} = λ [t]_{0}^{t}$
$2 \sqrt{v} = λ [t - 0]$
Squaring both side
$4 v = λ^{2} t^{2}$
$v = \frac{λ^{2} t^{2}}{4}$

UPSEE 2020

Motion in One Dimensions

141433 Find the relation between ' $x$ ' and ' $y$ ', when a body starting from rest, moves with uniform acceleration and travels a distance $x$ in the first $n$ seconds and $y$ in the next $n$ seconds:

1

3 x = y

2

x = 3 y

3

x = y

4

x = 2 y

Explanation:

A From given question.
The distance covered in $2 \sec, x = \frac{1}{2} a (2)^{2} = 2 a$
and the distance covered in next $2 \sec$ ,
$y = \frac{1}{2} \cdot a (4)^{2} - \frac{1}{2} a \cdot (2)^{2}$
$y = 6 a$
So, $\frac{x}{y} = \frac{2 a}{6 a} = \frac{1}{3}$
$3 x = y \Rightarrow y = 3 x$

AP EAMCET-25.09.2020

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