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Motion in One Dimensions

141408 A person is running with an uniform velocity towards a flyover. He takes $5 s$ to reach the flyover from a reference point and takes $50 s$ to cross the flyover from the same reference point. If the length of the flyover is $1000 m$ then his velocity in nearly

1

83.1 kmph

2

80.0 kmph

3

75.4 kmph

4

85.2 kmph

Explanation:

B original image
Velocity $= \frac{Displacement}{Time}$
$v = \frac{d}{5}$
$d = 5 v$
$v = \frac{d + 1000}{50}$
$v = \frac{5 v + 1000}{50}$
$50 v = 5 v + 1000$
$v = \frac{1000}{45}$
$v = 22.22 \times \frac{18}{5} kmph$
$v = 80 kmph$

AP EAPCET-12.07.2022

Motion in One Dimensions

141409 Two particles $A$ and $B$ are moving along positive $x$ -and positive $y$ -axes respectively with the same speed $v m s^{- 1}$ . At $t = 0$ , they both pass through the origin. The equation of the straight line joining $A$ and $B$ at time $t$ is

1

y (t) = x (t) + v t

2

y (t) = - x (t) + v t

3

y (t) = + x (t)

4

y (t) = - x (t)

Explanation:

B original image
Both particle are moving with speed $v {ms}^{- 1}$ .
$∵$ Distance $=$ speed $\times$ time
Then, at time $t$ sec
Distance traveled by $A$ on $X$ axis $= vt$
Distance traveled by B on $Y$ axis $= vt$
$∵$ Equation of straight line passing through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$y - 0 = \frac{vt - 0}{0 - vt} (x - vt)$
$y = - x + vt$
or
$y (t) = - x (t) + v t$

AP EAPCET-12.07.2022

Motion in One Dimensions

141410 A truck carrying sand having a total mass of $1600 kg$ is moving with a uniform velocity of 15 ${ms}^{- 1}$ . Sand is dropped into it at a rate of $24 kg$ $min^{- 1}$ . The force needed to keep the truck moving with uniform velocity is

1

6 N

2

1.6 N

3

40 N

4

4 N

Explanation:

A Given that -
Total mass $= 1600 kg$
Velocity $= 15 m / \sec$
Dropped sand permit $= 24 kg / \min = \frac{24}{60} kg / \sec = \frac{2}{5}$
$∴ f = \frac{dp}{dt}$
$\Rightarrow f = \frac{d (mv)}{dt} \Rightarrow f = \frac{v \cdot dm}{dt}$
$= 15 \times \frac{2}{5} kg \frac{m}{\sec^{2}}$
$= 3 \times 2 N = 6 N$

AP EAPCET-12.07.2022

Motion in One Dimensions

141411 A passenger saw a bus standing 10 meters ahead of him. The bus started moving with a constant acceleration $5 {ms}^{- 2}$ away from him. To catch the bus the passenger ran at constant speed towards the bus. The minimum speed of the passenger to catch the bus is

1

10 \sqrt{2} {ms}^{- 1}

2

5 \sqrt{2} {ms}^{- 1}

3

10 {ms}^{- 1}

4

50 {ms}^{- 1}

Explanation:

C Given,
Acceleration of bus (a) $= 5 {ms}^{- 2}$
Initial speed of bus $(u) = 0 m / s$
Let the person catch the bus at time $t$ sec with speed $v$ $m / \sec$ .
$∵ s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} \times 5 \times t^{2} \Rightarrow s = \frac{5}{2} t^{2}$
As the person was at $10 m$ distance from bus.
So, Distance covered by person $= 10 + s$
$= 10 + \frac{5}{2} t^{2}$
$∵$ Distance $=$ speed $\times$ time
$10 + \frac{5}{2} t^{2} = v t$
$5 t^{2} - 2 v t + 20 = 0$
For real root of quadratic equation
$b^{2} - 4 a c \geq 0$
i.e. $(- 2 v)^{2} - 4 \times 5 \times 20 \geq 0$
$4 v^{2} \geq 400$
$v^{2} \geq 100$
$v \geq 10$
So, $V_{min} = 10 m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141408 A person is running with an uniform velocity towards a flyover. He takes $5 s$ to reach the flyover from a reference point and takes $50 s$ to cross the flyover from the same reference point. If the length of the flyover is $1000 m$ then his velocity in nearly

1

83.1 kmph

2

80.0 kmph

3

75.4 kmph

4

85.2 kmph

Explanation:

B original image
Velocity $= \frac{Displacement}{Time}$
$v = \frac{d}{5}$
$d = 5 v$
$v = \frac{d + 1000}{50}$
$v = \frac{5 v + 1000}{50}$
$50 v = 5 v + 1000$
$v = \frac{1000}{45}$
$v = 22.22 \times \frac{18}{5} kmph$
$v = 80 kmph$

AP EAPCET-12.07.2022

Motion in One Dimensions

141409 Two particles $A$ and $B$ are moving along positive $x$ -and positive $y$ -axes respectively with the same speed $v m s^{- 1}$ . At $t = 0$ , they both pass through the origin. The equation of the straight line joining $A$ and $B$ at time $t$ is

1

y (t) = x (t) + v t

2

y (t) = - x (t) + v t

3

y (t) = + x (t)

4

y (t) = - x (t)

Explanation:

B original image
Both particle are moving with speed $v {ms}^{- 1}$ .
$∵$ Distance $=$ speed $\times$ time
Then, at time $t$ sec
Distance traveled by $A$ on $X$ axis $= vt$
Distance traveled by B on $Y$ axis $= vt$
$∵$ Equation of straight line passing through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$y - 0 = \frac{vt - 0}{0 - vt} (x - vt)$
$y = - x + vt$
or
$y (t) = - x (t) + v t$

AP EAPCET-12.07.2022

Motion in One Dimensions

141410 A truck carrying sand having a total mass of $1600 kg$ is moving with a uniform velocity of 15 ${ms}^{- 1}$ . Sand is dropped into it at a rate of $24 kg$ $min^{- 1}$ . The force needed to keep the truck moving with uniform velocity is

1

6 N

2

1.6 N

3

40 N

4

4 N

Explanation:

A Given that -
Total mass $= 1600 kg$
Velocity $= 15 m / \sec$
Dropped sand permit $= 24 kg / \min = \frac{24}{60} kg / \sec = \frac{2}{5}$
$∴ f = \frac{dp}{dt}$
$\Rightarrow f = \frac{d (mv)}{dt} \Rightarrow f = \frac{v \cdot dm}{dt}$
$= 15 \times \frac{2}{5} kg \frac{m}{\sec^{2}}$
$= 3 \times 2 N = 6 N$

AP EAPCET-12.07.2022

Motion in One Dimensions

141411 A passenger saw a bus standing 10 meters ahead of him. The bus started moving with a constant acceleration $5 {ms}^{- 2}$ away from him. To catch the bus the passenger ran at constant speed towards the bus. The minimum speed of the passenger to catch the bus is

1

10 \sqrt{2} {ms}^{- 1}

2

5 \sqrt{2} {ms}^{- 1}

3

10 {ms}^{- 1}

4

50 {ms}^{- 1}

Explanation:

C Given,
Acceleration of bus (a) $= 5 {ms}^{- 2}$
Initial speed of bus $(u) = 0 m / s$
Let the person catch the bus at time $t$ sec with speed $v$ $m / \sec$ .
$∵ s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} \times 5 \times t^{2} \Rightarrow s = \frac{5}{2} t^{2}$
As the person was at $10 m$ distance from bus.
So, Distance covered by person $= 10 + s$
$= 10 + \frac{5}{2} t^{2}$
$∵$ Distance $=$ speed $\times$ time
$10 + \frac{5}{2} t^{2} = v t$
$5 t^{2} - 2 v t + 20 = 0$
For real root of quadratic equation
$b^{2} - 4 a c \geq 0$
i.e. $(- 2 v)^{2} - 4 \times 5 \times 20 \geq 0$
$4 v^{2} \geq 400$
$v^{2} \geq 100$
$v \geq 10$
So, $V_{min} = 10 m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141408 A person is running with an uniform velocity towards a flyover. He takes $5 s$ to reach the flyover from a reference point and takes $50 s$ to cross the flyover from the same reference point. If the length of the flyover is $1000 m$ then his velocity in nearly

1

83.1 kmph

2

80.0 kmph

3

75.4 kmph

4

85.2 kmph

Explanation:

B original image
Velocity $= \frac{Displacement}{Time}$
$v = \frac{d}{5}$
$d = 5 v$
$v = \frac{d + 1000}{50}$
$v = \frac{5 v + 1000}{50}$
$50 v = 5 v + 1000$
$v = \frac{1000}{45}$
$v = 22.22 \times \frac{18}{5} kmph$
$v = 80 kmph$

AP EAPCET-12.07.2022

Motion in One Dimensions

141409 Two particles $A$ and $B$ are moving along positive $x$ -and positive $y$ -axes respectively with the same speed $v m s^{- 1}$ . At $t = 0$ , they both pass through the origin. The equation of the straight line joining $A$ and $B$ at time $t$ is

1

y (t) = x (t) + v t

2

y (t) = - x (t) + v t

3

y (t) = + x (t)

4

y (t) = - x (t)

Explanation:

B original image
Both particle are moving with speed $v {ms}^{- 1}$ .
$∵$ Distance $=$ speed $\times$ time
Then, at time $t$ sec
Distance traveled by $A$ on $X$ axis $= vt$
Distance traveled by B on $Y$ axis $= vt$
$∵$ Equation of straight line passing through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$y - 0 = \frac{vt - 0}{0 - vt} (x - vt)$
$y = - x + vt$
or
$y (t) = - x (t) + v t$

AP EAPCET-12.07.2022

Motion in One Dimensions

141410 A truck carrying sand having a total mass of $1600 kg$ is moving with a uniform velocity of 15 ${ms}^{- 1}$ . Sand is dropped into it at a rate of $24 kg$ $min^{- 1}$ . The force needed to keep the truck moving with uniform velocity is

1

6 N

2

1.6 N

3

40 N

4

4 N

Explanation:

A Given that -
Total mass $= 1600 kg$
Velocity $= 15 m / \sec$
Dropped sand permit $= 24 kg / \min = \frac{24}{60} kg / \sec = \frac{2}{5}$
$∴ f = \frac{dp}{dt}$
$\Rightarrow f = \frac{d (mv)}{dt} \Rightarrow f = \frac{v \cdot dm}{dt}$
$= 15 \times \frac{2}{5} kg \frac{m}{\sec^{2}}$
$= 3 \times 2 N = 6 N$

AP EAPCET-12.07.2022

Motion in One Dimensions

141411 A passenger saw a bus standing 10 meters ahead of him. The bus started moving with a constant acceleration $5 {ms}^{- 2}$ away from him. To catch the bus the passenger ran at constant speed towards the bus. The minimum speed of the passenger to catch the bus is

1

10 \sqrt{2} {ms}^{- 1}

2

5 \sqrt{2} {ms}^{- 1}

3

10 {ms}^{- 1}

4

50 {ms}^{- 1}

Explanation:

C Given,
Acceleration of bus (a) $= 5 {ms}^{- 2}$
Initial speed of bus $(u) = 0 m / s$
Let the person catch the bus at time $t$ sec with speed $v$ $m / \sec$ .
$∵ s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} \times 5 \times t^{2} \Rightarrow s = \frac{5}{2} t^{2}$
As the person was at $10 m$ distance from bus.
So, Distance covered by person $= 10 + s$
$= 10 + \frac{5}{2} t^{2}$
$∵$ Distance $=$ speed $\times$ time
$10 + \frac{5}{2} t^{2} = v t$
$5 t^{2} - 2 v t + 20 = 0$
For real root of quadratic equation
$b^{2} - 4 a c \geq 0$
i.e. $(- 2 v)^{2} - 4 \times 5 \times 20 \geq 0$
$4 v^{2} \geq 400$
$v^{2} \geq 100$
$v \geq 10$
So, $V_{min} = 10 m / \sec$

AP EAMCET-11.07.2022

Motion in One Dimensions

141408 A person is running with an uniform velocity towards a flyover. He takes $5 s$ to reach the flyover from a reference point and takes $50 s$ to cross the flyover from the same reference point. If the length of the flyover is $1000 m$ then his velocity in nearly

1

83.1 kmph

2

80.0 kmph

3

75.4 kmph

4

85.2 kmph

Explanation:

B original image
Velocity $= \frac{Displacement}{Time}$
$v = \frac{d}{5}$
$d = 5 v$
$v = \frac{d + 1000}{50}$
$v = \frac{5 v + 1000}{50}$
$50 v = 5 v + 1000$
$v = \frac{1000}{45}$
$v = 22.22 \times \frac{18}{5} kmph$
$v = 80 kmph$

AP EAPCET-12.07.2022

Motion in One Dimensions

141409 Two particles $A$ and $B$ are moving along positive $x$ -and positive $y$ -axes respectively with the same speed $v m s^{- 1}$ . At $t = 0$ , they both pass through the origin. The equation of the straight line joining $A$ and $B$ at time $t$ is

1

y (t) = x (t) + v t

2

y (t) = - x (t) + v t

3

y (t) = + x (t)

4

y (t) = - x (t)

Explanation:

B original image
Both particle are moving with speed $v {ms}^{- 1}$ .
$∵$ Distance $=$ speed $\times$ time
Then, at time $t$ sec
Distance traveled by $A$ on $X$ axis $= vt$
Distance traveled by B on $Y$ axis $= vt$
$∵$ Equation of straight line passing through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$y - 0 = \frac{vt - 0}{0 - vt} (x - vt)$
$y = - x + vt$
or
$y (t) = - x (t) + v t$

AP EAPCET-12.07.2022

Motion in One Dimensions

141410 A truck carrying sand having a total mass of $1600 kg$ is moving with a uniform velocity of 15 ${ms}^{- 1}$ . Sand is dropped into it at a rate of $24 kg$ $min^{- 1}$ . The force needed to keep the truck moving with uniform velocity is

1

6 N

2

1.6 N

3

40 N

4

4 N

Explanation:

A Given that -
Total mass $= 1600 kg$
Velocity $= 15 m / \sec$
Dropped sand permit $= 24 kg / \min = \frac{24}{60} kg / \sec = \frac{2}{5}$
$∴ f = \frac{dp}{dt}$
$\Rightarrow f = \frac{d (mv)}{dt} \Rightarrow f = \frac{v \cdot dm}{dt}$
$= 15 \times \frac{2}{5} kg \frac{m}{\sec^{2}}$
$= 3 \times 2 N = 6 N$

AP EAPCET-12.07.2022

Motion in One Dimensions

141411 A passenger saw a bus standing 10 meters ahead of him. The bus started moving with a constant acceleration $5 {ms}^{- 2}$ away from him. To catch the bus the passenger ran at constant speed towards the bus. The minimum speed of the passenger to catch the bus is

1

10 \sqrt{2} {ms}^{- 1}

2

5 \sqrt{2} {ms}^{- 1}

3

10 {ms}^{- 1}

4

50 {ms}^{- 1}

Explanation:

C Given,
Acceleration of bus (a) $= 5 {ms}^{- 2}$
Initial speed of bus $(u) = 0 m / s$
Let the person catch the bus at time $t$ sec with speed $v$ $m / \sec$ .
$∵ s = ut + \frac{1}{2} {at}^{2}$
$s = 0 \times t + \frac{1}{2} \times 5 \times t^{2} \Rightarrow s = \frac{5}{2} t^{2}$
As the person was at $10 m$ distance from bus.
So, Distance covered by person $= 10 + s$
$= 10 + \frac{5}{2} t^{2}$
$∵$ Distance $=$ speed $\times$ time
$10 + \frac{5}{2} t^{2} = v t$
$5 t^{2} - 2 v t + 20 = 0$
For real root of quadratic equation
$b^{2} - 4 a c \geq 0$
i.e. $(- 2 v)^{2} - 4 \times 5 \times 20 \geq 0$
$4 v^{2} \geq 400$
$v^{2} \geq 100$
$v \geq 10$
So, $V_{min} = 10 m / \sec$

AP EAMCET-11.07.2022

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