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Motion in One Dimensions

141376 An object moving with a speed of $6.25 m / s$ , is decelerated at a rate given by $dv / dt = - 2.5 \sqrt{v}$ where $v$ is the instantaneous speed(d) The time taken by the object to come to rest, would be

1

2 s

2

4 s

3

8 s

4

1 s

Explanation:

A Given,
$t = 0, v = 6.25 m / s$
$t = t, v = 0$
$\frac{dv}{dt} = - 2.5 \sqrt{v}$
$\frac{dv}{\sqrt{v}} = - 2.5 dt \Rightarrow \int_{6.25}^{0} (V)^{- 1 / 2} = - 2.5 \int_{0}^{t} dt$
$| 2 \sqrt{v} |_{625}^{0} = - 2.5 | t |_{0}^{t}$
$2 \sqrt{6.25} = 2.5 t \Rightarrow t = \frac{2 \times 2.5}{2.5} = 2 \sec$

AIEEE 2011

Motion in One Dimensions

141377 The equation of motion of a body is $\frac{d v (t)}{d t} = 9 - 3 v (t)$ , where $v (t)$ , where $v (t)$ is the speed (in $m / s$ ) at time $t$ (in second). If the body was at rest at $t = 0$ , then which of the following is correct?

1 The terminal speed is

3 m / s

2 Initial acceleration is

9 m / s^{2}

3

v (t) = 3 (1 - e^{- 3 t})

4 All of these are correct

Explanation:

D Given,
$\frac{d v (t)}{d t} = 9 - 3 v (t)$
$d v = [9 - 3 v (t)] d t$
$\frac{d v}{9 - 3 v (t)} = d t$
Integrating both side of above equation, we get
$\int_{0}^{v} \frac{dv}{9 - 3 v (t)} = \int_{0}^{t} dt$
${[\frac{\ln [9 - 3 v]}{- 3}]}_{0}^{v} = t$
$\ln (9 - 3 v) - \ln 9 = - 3 t$
$\ln [\frac{9 - 3 v}{9}] = - 3 t$
$\ln (1 - \frac{v}{3}) = - 3 t$
$1 - \frac{v}{3} = e^{- 3 t}$
$\frac{v}{3} = 1 - e^{- 3 t}$
$v = 3 (1 - e^{- 3 t})$
(ii) The terminal speed I the constant speed when acceleration is zero.
Thus, $0 = 9 - 3 v$
$v = 3 m / s$
(iii) at $t = 0, v = 0$
$∴$ initial acceleration $a = 9 - 0 = 9 m / s^{2}$

Assam CEE-2016

Motion in One Dimensions

141378 A particle moves in the $x$ -y plane with velocity $v_{x} = 8 t - 2$ and $v_{y} = 2$ . If it passes through the point $x = 14$ and $y = 4$ at $t = 2 s$ , then the equation of the path will be

1

x = y^{2} - y + 2

2

x = 2 y^{2} + 2 y - 3

3

x = 3 y^{2} + 5

4 None of the above

Explanation:

D Given,
$v_{x} = 8 t - 2$
$\frac{dx (t)}{dt} = 8 t - 2$
$x (t) = \frac{8 t^{2}}{2} - 2 t + x (0)$
$x (t) = 4 t^{2} - 2 t + x (0)$
Condition, $x = 14$ at $t = 2$
$x (2) = 4 \times (2)^{2} - 2 \times 2 + x (0)$
$14 = 16 - 4 + x (0)$
$\Rightarrow x (0) = 2$
$∴$ equation (i) becomes, we get
$x (t) = 4 t^{2} - 2 t + 2$
Now,
$v_{y} = 2$
$\frac{d y (t)}{d t} = 2$
$y (t) = 2 t + y (0)$
Condition, $y = 4$ at $t = 2 s$ ,
$y (4) = 2 \times 2 + y (0)$
$4 = 4 + y (0)$
$\Rightarrow y (0) = 0$
$∴$ Equation (iii) becomes, wet get,
$y (t) = 2 t$
$t = y / 2$
$∴$ equation (ii) becomes
$x (y / 2) = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2$
$x (t) = y^{2} - y + 2$

Assam CEE-2014

Motion in One Dimensions

141379 Engine of a vehicle can give it an acceleration of $1 m / s^{2}$ , while the brake of the vehicle can retard it at $8 m / s^{2}$ . Then the minimum time in which the vehicle can complete a journey of $600 m$ will be.

1

80 s

2

60 s

3

40 s

4 None of the above

Explanation:

C Area of the curve gives total distance
original image
$\frac{1}{2} \times v_{max} \times t_{1} + \frac{1}{2} \times v_{max} \times t_{2} = 600$
$v_{max} = {at}_{1}$
$v_{max} = 1 \times t_{1} = t_{1}$
$\frac{1}{2} \times t_{1} \times t_{1} + \frac{1}{2} t_{1} t_{2} = 600$
$\frac{1}{2} \times t_{1}^{2} + \frac{1}{2} t_{1} t_{2} = 600$
$v = v_{max} - β t_{2}$
$0 = v_{max} - 8 t_{2}$
$v_{max} = 8 t_{2}$
$t_{1} = 8 t_{2}$
From equation (i) and (ii)
$\frac{1}{2} {(8 t_{2})}^{2} + \frac{1}{2} \times (8 t_{2}) (t_{2}) = 600$
$64 t_{2}^{2} + 8 t_{2}^{2} = 1200$
$72 t_{2}^{2} = 1200$
$t_{2} = \sqrt{\frac{100}{6}} = 4.08 \sec$
$t_{1} = 8 t_{2}$
$= 8 (4.08)$
$= 32.65$
Total time to complete the journey
$t_{1} + t_{2} = 32.65 + 4.08$
$= 36.73 \sec$

Assam CEE-2014

Motion in One Dimensions

141376 An object moving with a speed of $6.25 m / s$ , is decelerated at a rate given by $dv / dt = - 2.5 \sqrt{v}$ where $v$ is the instantaneous speed(d) The time taken by the object to come to rest, would be

1

2 s

2

4 s

3

8 s

4

1 s

Explanation:

A Given,
$t = 0, v = 6.25 m / s$
$t = t, v = 0$
$\frac{dv}{dt} = - 2.5 \sqrt{v}$
$\frac{dv}{\sqrt{v}} = - 2.5 dt \Rightarrow \int_{6.25}^{0} (V)^{- 1 / 2} = - 2.5 \int_{0}^{t} dt$
$| 2 \sqrt{v} |_{625}^{0} = - 2.5 | t |_{0}^{t}$
$2 \sqrt{6.25} = 2.5 t \Rightarrow t = \frac{2 \times 2.5}{2.5} = 2 \sec$

AIEEE 2011

Motion in One Dimensions

141377 The equation of motion of a body is $\frac{d v (t)}{d t} = 9 - 3 v (t)$ , where $v (t)$ , where $v (t)$ is the speed (in $m / s$ ) at time $t$ (in second). If the body was at rest at $t = 0$ , then which of the following is correct?

1 The terminal speed is

3 m / s

2 Initial acceleration is

9 m / s^{2}

3

v (t) = 3 (1 - e^{- 3 t})

4 All of these are correct

Explanation:

D Given,
$\frac{d v (t)}{d t} = 9 - 3 v (t)$
$d v = [9 - 3 v (t)] d t$
$\frac{d v}{9 - 3 v (t)} = d t$
Integrating both side of above equation, we get
$\int_{0}^{v} \frac{dv}{9 - 3 v (t)} = \int_{0}^{t} dt$
${[\frac{\ln [9 - 3 v]}{- 3}]}_{0}^{v} = t$
$\ln (9 - 3 v) - \ln 9 = - 3 t$
$\ln [\frac{9 - 3 v}{9}] = - 3 t$
$\ln (1 - \frac{v}{3}) = - 3 t$
$1 - \frac{v}{3} = e^{- 3 t}$
$\frac{v}{3} = 1 - e^{- 3 t}$
$v = 3 (1 - e^{- 3 t})$
(ii) The terminal speed I the constant speed when acceleration is zero.
Thus, $0 = 9 - 3 v$
$v = 3 m / s$
(iii) at $t = 0, v = 0$
$∴$ initial acceleration $a = 9 - 0 = 9 m / s^{2}$

Assam CEE-2016

Motion in One Dimensions

141378 A particle moves in the $x$ -y plane with velocity $v_{x} = 8 t - 2$ and $v_{y} = 2$ . If it passes through the point $x = 14$ and $y = 4$ at $t = 2 s$ , then the equation of the path will be

1

x = y^{2} - y + 2

2

x = 2 y^{2} + 2 y - 3

3

x = 3 y^{2} + 5

4 None of the above

Explanation:

D Given,
$v_{x} = 8 t - 2$
$\frac{dx (t)}{dt} = 8 t - 2$
$x (t) = \frac{8 t^{2}}{2} - 2 t + x (0)$
$x (t) = 4 t^{2} - 2 t + x (0)$
Condition, $x = 14$ at $t = 2$
$x (2) = 4 \times (2)^{2} - 2 \times 2 + x (0)$
$14 = 16 - 4 + x (0)$
$\Rightarrow x (0) = 2$
$∴$ equation (i) becomes, we get
$x (t) = 4 t^{2} - 2 t + 2$
Now,
$v_{y} = 2$
$\frac{d y (t)}{d t} = 2$
$y (t) = 2 t + y (0)$
Condition, $y = 4$ at $t = 2 s$ ,
$y (4) = 2 \times 2 + y (0)$
$4 = 4 + y (0)$
$\Rightarrow y (0) = 0$
$∴$ Equation (iii) becomes, wet get,
$y (t) = 2 t$
$t = y / 2$
$∴$ equation (ii) becomes
$x (y / 2) = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2$
$x (t) = y^{2} - y + 2$

Assam CEE-2014

Motion in One Dimensions

141379 Engine of a vehicle can give it an acceleration of $1 m / s^{2}$ , while the brake of the vehicle can retard it at $8 m / s^{2}$ . Then the minimum time in which the vehicle can complete a journey of $600 m$ will be.

1

80 s

2

60 s

3

40 s

4 None of the above

Explanation:

C Area of the curve gives total distance
original image
$\frac{1}{2} \times v_{max} \times t_{1} + \frac{1}{2} \times v_{max} \times t_{2} = 600$
$v_{max} = {at}_{1}$
$v_{max} = 1 \times t_{1} = t_{1}$
$\frac{1}{2} \times t_{1} \times t_{1} + \frac{1}{2} t_{1} t_{2} = 600$
$\frac{1}{2} \times t_{1}^{2} + \frac{1}{2} t_{1} t_{2} = 600$
$v = v_{max} - β t_{2}$
$0 = v_{max} - 8 t_{2}$
$v_{max} = 8 t_{2}$
$t_{1} = 8 t_{2}$
From equation (i) and (ii)
$\frac{1}{2} {(8 t_{2})}^{2} + \frac{1}{2} \times (8 t_{2}) (t_{2}) = 600$
$64 t_{2}^{2} + 8 t_{2}^{2} = 1200$
$72 t_{2}^{2} = 1200$
$t_{2} = \sqrt{\frac{100}{6}} = 4.08 \sec$
$t_{1} = 8 t_{2}$
$= 8 (4.08)$
$= 32.65$
Total time to complete the journey
$t_{1} + t_{2} = 32.65 + 4.08$
$= 36.73 \sec$

Assam CEE-2014

Motion in One Dimensions

141376 An object moving with a speed of $6.25 m / s$ , is decelerated at a rate given by $dv / dt = - 2.5 \sqrt{v}$ where $v$ is the instantaneous speed(d) The time taken by the object to come to rest, would be

1

2 s

2

4 s

3

8 s

4

1 s

Explanation:

A Given,
$t = 0, v = 6.25 m / s$
$t = t, v = 0$
$\frac{dv}{dt} = - 2.5 \sqrt{v}$
$\frac{dv}{\sqrt{v}} = - 2.5 dt \Rightarrow \int_{6.25}^{0} (V)^{- 1 / 2} = - 2.5 \int_{0}^{t} dt$
$| 2 \sqrt{v} |_{625}^{0} = - 2.5 | t |_{0}^{t}$
$2 \sqrt{6.25} = 2.5 t \Rightarrow t = \frac{2 \times 2.5}{2.5} = 2 \sec$

AIEEE 2011

Motion in One Dimensions

141377 The equation of motion of a body is $\frac{d v (t)}{d t} = 9 - 3 v (t)$ , where $v (t)$ , where $v (t)$ is the speed (in $m / s$ ) at time $t$ (in second). If the body was at rest at $t = 0$ , then which of the following is correct?

1 The terminal speed is

3 m / s

2 Initial acceleration is

9 m / s^{2}

3

v (t) = 3 (1 - e^{- 3 t})

4 All of these are correct

Explanation:

D Given,
$\frac{d v (t)}{d t} = 9 - 3 v (t)$
$d v = [9 - 3 v (t)] d t$
$\frac{d v}{9 - 3 v (t)} = d t$
Integrating both side of above equation, we get
$\int_{0}^{v} \frac{dv}{9 - 3 v (t)} = \int_{0}^{t} dt$
${[\frac{\ln [9 - 3 v]}{- 3}]}_{0}^{v} = t$
$\ln (9 - 3 v) - \ln 9 = - 3 t$
$\ln [\frac{9 - 3 v}{9}] = - 3 t$
$\ln (1 - \frac{v}{3}) = - 3 t$
$1 - \frac{v}{3} = e^{- 3 t}$
$\frac{v}{3} = 1 - e^{- 3 t}$
$v = 3 (1 - e^{- 3 t})$
(ii) The terminal speed I the constant speed when acceleration is zero.
Thus, $0 = 9 - 3 v$
$v = 3 m / s$
(iii) at $t = 0, v = 0$
$∴$ initial acceleration $a = 9 - 0 = 9 m / s^{2}$

Assam CEE-2016

Motion in One Dimensions

141378 A particle moves in the $x$ -y plane with velocity $v_{x} = 8 t - 2$ and $v_{y} = 2$ . If it passes through the point $x = 14$ and $y = 4$ at $t = 2 s$ , then the equation of the path will be

1

x = y^{2} - y + 2

2

x = 2 y^{2} + 2 y - 3

3

x = 3 y^{2} + 5

4 None of the above

Explanation:

D Given,
$v_{x} = 8 t - 2$
$\frac{dx (t)}{dt} = 8 t - 2$
$x (t) = \frac{8 t^{2}}{2} - 2 t + x (0)$
$x (t) = 4 t^{2} - 2 t + x (0)$
Condition, $x = 14$ at $t = 2$
$x (2) = 4 \times (2)^{2} - 2 \times 2 + x (0)$
$14 = 16 - 4 + x (0)$
$\Rightarrow x (0) = 2$
$∴$ equation (i) becomes, we get
$x (t) = 4 t^{2} - 2 t + 2$
Now,
$v_{y} = 2$
$\frac{d y (t)}{d t} = 2$
$y (t) = 2 t + y (0)$
Condition, $y = 4$ at $t = 2 s$ ,
$y (4) = 2 \times 2 + y (0)$
$4 = 4 + y (0)$
$\Rightarrow y (0) = 0$
$∴$ Equation (iii) becomes, wet get,
$y (t) = 2 t$
$t = y / 2$
$∴$ equation (ii) becomes
$x (y / 2) = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2$
$x (t) = y^{2} - y + 2$

Assam CEE-2014

Motion in One Dimensions

141379 Engine of a vehicle can give it an acceleration of $1 m / s^{2}$ , while the brake of the vehicle can retard it at $8 m / s^{2}$ . Then the minimum time in which the vehicle can complete a journey of $600 m$ will be.

1

80 s

2

60 s

3

40 s

4 None of the above

Explanation:

C Area of the curve gives total distance
original image
$\frac{1}{2} \times v_{max} \times t_{1} + \frac{1}{2} \times v_{max} \times t_{2} = 600$
$v_{max} = {at}_{1}$
$v_{max} = 1 \times t_{1} = t_{1}$
$\frac{1}{2} \times t_{1} \times t_{1} + \frac{1}{2} t_{1} t_{2} = 600$
$\frac{1}{2} \times t_{1}^{2} + \frac{1}{2} t_{1} t_{2} = 600$
$v = v_{max} - β t_{2}$
$0 = v_{max} - 8 t_{2}$
$v_{max} = 8 t_{2}$
$t_{1} = 8 t_{2}$
From equation (i) and (ii)
$\frac{1}{2} {(8 t_{2})}^{2} + \frac{1}{2} \times (8 t_{2}) (t_{2}) = 600$
$64 t_{2}^{2} + 8 t_{2}^{2} = 1200$
$72 t_{2}^{2} = 1200$
$t_{2} = \sqrt{\frac{100}{6}} = 4.08 \sec$
$t_{1} = 8 t_{2}$
$= 8 (4.08)$
$= 32.65$
Total time to complete the journey
$t_{1} + t_{2} = 32.65 + 4.08$
$= 36.73 \sec$

Assam CEE-2014

Motion in One Dimensions

141376 An object moving with a speed of $6.25 m / s$ , is decelerated at a rate given by $dv / dt = - 2.5 \sqrt{v}$ where $v$ is the instantaneous speed(d) The time taken by the object to come to rest, would be

1

2 s

2

4 s

3

8 s

4

1 s

Explanation:

A Given,
$t = 0, v = 6.25 m / s$
$t = t, v = 0$
$\frac{dv}{dt} = - 2.5 \sqrt{v}$
$\frac{dv}{\sqrt{v}} = - 2.5 dt \Rightarrow \int_{6.25}^{0} (V)^{- 1 / 2} = - 2.5 \int_{0}^{t} dt$
$| 2 \sqrt{v} |_{625}^{0} = - 2.5 | t |_{0}^{t}$
$2 \sqrt{6.25} = 2.5 t \Rightarrow t = \frac{2 \times 2.5}{2.5} = 2 \sec$

AIEEE 2011

Motion in One Dimensions

141377 The equation of motion of a body is $\frac{d v (t)}{d t} = 9 - 3 v (t)$ , where $v (t)$ , where $v (t)$ is the speed (in $m / s$ ) at time $t$ (in second). If the body was at rest at $t = 0$ , then which of the following is correct?

1 The terminal speed is

3 m / s

2 Initial acceleration is

9 m / s^{2}

3

v (t) = 3 (1 - e^{- 3 t})

4 All of these are correct

Explanation:

D Given,
$\frac{d v (t)}{d t} = 9 - 3 v (t)$
$d v = [9 - 3 v (t)] d t$
$\frac{d v}{9 - 3 v (t)} = d t$
Integrating both side of above equation, we get
$\int_{0}^{v} \frac{dv}{9 - 3 v (t)} = \int_{0}^{t} dt$
${[\frac{\ln [9 - 3 v]}{- 3}]}_{0}^{v} = t$
$\ln (9 - 3 v) - \ln 9 = - 3 t$
$\ln [\frac{9 - 3 v}{9}] = - 3 t$
$\ln (1 - \frac{v}{3}) = - 3 t$
$1 - \frac{v}{3} = e^{- 3 t}$
$\frac{v}{3} = 1 - e^{- 3 t}$
$v = 3 (1 - e^{- 3 t})$
(ii) The terminal speed I the constant speed when acceleration is zero.
Thus, $0 = 9 - 3 v$
$v = 3 m / s$
(iii) at $t = 0, v = 0$
$∴$ initial acceleration $a = 9 - 0 = 9 m / s^{2}$

Assam CEE-2016

Motion in One Dimensions

141378 A particle moves in the $x$ -y plane with velocity $v_{x} = 8 t - 2$ and $v_{y} = 2$ . If it passes through the point $x = 14$ and $y = 4$ at $t = 2 s$ , then the equation of the path will be

1

x = y^{2} - y + 2

2

x = 2 y^{2} + 2 y - 3

3

x = 3 y^{2} + 5

4 None of the above

Explanation:

D Given,
$v_{x} = 8 t - 2$
$\frac{dx (t)}{dt} = 8 t - 2$
$x (t) = \frac{8 t^{2}}{2} - 2 t + x (0)$
$x (t) = 4 t^{2} - 2 t + x (0)$
Condition, $x = 14$ at $t = 2$
$x (2) = 4 \times (2)^{2} - 2 \times 2 + x (0)$
$14 = 16 - 4 + x (0)$
$\Rightarrow x (0) = 2$
$∴$ equation (i) becomes, we get
$x (t) = 4 t^{2} - 2 t + 2$
Now,
$v_{y} = 2$
$\frac{d y (t)}{d t} = 2$
$y (t) = 2 t + y (0)$
Condition, $y = 4$ at $t = 2 s$ ,
$y (4) = 2 \times 2 + y (0)$
$4 = 4 + y (0)$
$\Rightarrow y (0) = 0$
$∴$ Equation (iii) becomes, wet get,
$y (t) = 2 t$
$t = y / 2$
$∴$ equation (ii) becomes
$x (y / 2) = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2$
$x (t) = y^{2} - y + 2$

Assam CEE-2014

Motion in One Dimensions

141379 Engine of a vehicle can give it an acceleration of $1 m / s^{2}$ , while the brake of the vehicle can retard it at $8 m / s^{2}$ . Then the minimum time in which the vehicle can complete a journey of $600 m$ will be.

1

80 s

2

60 s

3

40 s

4 None of the above

Explanation:

C Area of the curve gives total distance
original image
$\frac{1}{2} \times v_{max} \times t_{1} + \frac{1}{2} \times v_{max} \times t_{2} = 600$
$v_{max} = {at}_{1}$
$v_{max} = 1 \times t_{1} = t_{1}$
$\frac{1}{2} \times t_{1} \times t_{1} + \frac{1}{2} t_{1} t_{2} = 600$
$\frac{1}{2} \times t_{1}^{2} + \frac{1}{2} t_{1} t_{2} = 600$
$v = v_{max} - β t_{2}$
$0 = v_{max} - 8 t_{2}$
$v_{max} = 8 t_{2}$
$t_{1} = 8 t_{2}$
From equation (i) and (ii)
$\frac{1}{2} {(8 t_{2})}^{2} + \frac{1}{2} \times (8 t_{2}) (t_{2}) = 600$
$64 t_{2}^{2} + 8 t_{2}^{2} = 1200$
$72 t_{2}^{2} = 1200$
$t_{2} = \sqrt{\frac{100}{6}} = 4.08 \sec$
$t_{1} = 8 t_{2}$
$= 8 (4.08)$
$= 32.65$
Total time to complete the journey
$t_{1} + t_{2} = 32.65 + 4.08$
$= 36.73 \sec$

Assam CEE-2014