QBManager

Motion in One Dimensions

141384 The acceleration of a particle is increasing linearly with time $t$ as bt. The particle starts from the origin with an initial velocity $v_{0}$. The distance travelled by the particle in time $t$ will be

1

v_{0} t + \frac{1}{3} {bt}^{2}

2

v_{0} t + \frac{1}{3} {bt}^{3}

3

v_{0} t + \frac{1}{6} {bt}^{3}

4

v_{0} t + \frac{1}{2} {bt}^{2}

Explanation:

C Given,
Acceleration $a = bt$
time $= t$
Acceleration $= \frac{change in velocity}{change in time}$
$a = \frac{d v}{d t}$
$\frac{dv}{dt} = a = bt$
$\Rightarrow \frac{dv}{dt} = bt$
$dv = bt dt$
Now integrating this we will get
$\int dv = \int btdt$
$v = \frac{{bt}^{2}}{2} + C$
At, $t = 0, v = v_{0}$ , thus $v_{0} = C$
$∴ v = \frac{{bt}^{2}}{2} + v_{0}$
Now, $v = \frac{change is position}{change in time} = \frac{ds}{dt}$
$\Rightarrow \frac{ds}{dt} = \frac{{bt}^{2}}{2} + v_{0}$
$\Rightarrow ds = (\frac{{bt}^{2}}{2} + v_{0}) dt$
Integrating above we get,
$s = \frac{{bt}^{3}}{6} + v_{0} t + C$
At $t = 0, s = 0, C = 0$
i.e. $s = \frac{b t^{3}}{6} + v_{0} t$
or $s = v_{0} t + \frac{1}{6} {bt}^{3}$

UP CPMT-2011

Motion in One Dimensions

141385 A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position. If the car moves to a point where angle is $\frac{π}{3}$ , what is the distance moved by it ?

1

(\frac{100}{\sqrt{3}}) m

2

(200 \sqrt{3}) m

3

(\frac{200}{\sqrt{3}}) m

4

(\frac{150}{\sqrt{3}}) m

Explanation:

C A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position.
If the car moves to a point where angle is $\frac{π}{3}$ Let,
original image
In $△ BCD$
$\tan \frac{π}{3} = \frac{DC}{BC}$
$\Rightarrow \sqrt{3} = \frac{100}{y}$
$\Rightarrow y = \frac{100}{\sqrt{3}}$
Now,
$\tan \frac{π}{6} = \frac{DC}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AB + BC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x + y}$
$\Rightarrow x + y = 100 \sqrt{3}$
$In △ ACD$
Put the value of $y$ in above equation
$x + \frac{100}{\sqrt{3}} = 100 \sqrt{3}$
$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}}$
$x = \frac{100 \times 3 - 100}{\sqrt{3}}$
$x = \frac{300 - 100}{\sqrt{3}}$
$x = \frac{200}{\sqrt{3}} m$

UP CPMT-2003

Motion in One Dimensions

141386 The displacement of a particle moving in a straight line is given by the expression $x = {A t}^{3} +$ $B t^{2} + C t + D$ in meter, where $t$ is in seconds and $A, B, C$ and $D$ are constants.
The ratio between the initial acceleration and initial velocity is

1

\frac{2 C}{B}

2

\frac{2 B}{C}

3

2 C

4

\frac{C}{2 B}

Explanation:

B Displacement of a particle moving in a straight line is given by
$x = A t^{3} + B t^{2} + C t + D$
$Velocity v = \frac{d x}{d t}$
$v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C$
Initial velocity is equal to $\frac{d x}{d t}$ at $t = 0$
Initial velocity $v = (\frac{d x}{d t})$ at $t = 0 = C$
Acceleration $a = \frac{d^{2} x}{d t^{2}} = 6 A t + 2 B$
Initial acceleration a at $t = 0$
$a = {(\frac{d^{2} x}{{dt}^{2}})}_{t = 0} = 2 B$
Ratio between the initial acceleration and initial
Velocity is, $\frac{a}{v} = \frac{{(\frac{d^{2} x}{d t^{2}})}_{t = 0}}{{(\frac{d x}{d t})}_{t = 0}} = \frac{2 B}{C}$
$\frac{a}{v} = \frac{2 B}{C}$

TS EAMCET (Engg.)-2015

Motion in One Dimensions

141387 The position - time (x-t) graph for motion of a body is given below :
original image
Which one among the following is depicted by the above graph?

1 Positive acceleration

2 Negative acceleration

3 Zero acceleration

4 None of the above

Explanation:

C Given,
From the given $(x - t)$ graph it is clear that velocity is constant.
Therefore, acceleration is zero

NDA (II) 2011

Motion in One Dimensions

141388 The displacement of a particle at time $t$ is given by
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{\boldsymbol k}$
Where $a, b$ and $c$ are positive constants. Then the particle is

1 accelerated along

\hat{k}

direction

2 decelerated along

\hat{k}

direction

3 decelerated along

\hat{j}

direction

4 accelerated along

\hat{j}

direction

Explanation:

A Given,
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{k}$
$Velocity (v_{x}) = \frac{dx}{dt}$
$v_{x} = 0$
$v_{y} = \frac{d (bt)}{dt}$
$v_{y} = b$
And, $v_{z} = \frac{dz}{dt} = \frac{d}{dt} (\frac{c}{2} t^{2})$
$v_{z} = ct$
Acceleration
$a_{x} = \frac{{dv}_{x}}{dt} = 0$
$a_{y} = \frac{{dv}_{y}}{dt} = 0$
$a_{z} = \frac{{dv}_{z}}{dt} = c$
Thus, the particle is acceleration along $\hat{k}$ direction.

NDA (I) 2013

Motion in One Dimensions

141384 The acceleration of a particle is increasing linearly with time $t$ as bt. The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

1

v_{0} t + \frac{1}{3} {bt}^{2}

2

v_{0} t + \frac{1}{3} {bt}^{3}

3

v_{0} t + \frac{1}{6} {bt}^{3}

4

v_{0} t + \frac{1}{2} {bt}^{2}

Explanation:

C Given,
Acceleration $a = bt$
time $= t$
Acceleration $= \frac{change in velocity}{change in time}$
$a = \frac{d v}{d t}$
$\frac{dv}{dt} = a = bt$
$\Rightarrow \frac{dv}{dt} = bt$
$dv = bt dt$
Now integrating this we will get
$\int dv = \int btdt$
$v = \frac{{bt}^{2}}{2} + C$
At, $t = 0, v = v_{0}$ , thus $v_{0} = C$
$∴ v = \frac{{bt}^{2}}{2} + v_{0}$
Now, $v = \frac{change is position}{change in time} = \frac{ds}{dt}$
$\Rightarrow \frac{ds}{dt} = \frac{{bt}^{2}}{2} + v_{0}$
$\Rightarrow ds = (\frac{{bt}^{2}}{2} + v_{0}) dt$
Integrating above we get,
$s = \frac{{bt}^{3}}{6} + v_{0} t + C$
At $t = 0, s = 0, C = 0$
i.e. $s = \frac{b t^{3}}{6} + v_{0} t$
or $s = v_{0} t + \frac{1}{6} {bt}^{3}$

UP CPMT-2011

Motion in One Dimensions

141385 A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position. If the car moves to a point where angle is $\frac{π}{3}$ , what is the distance moved by it ?

1

(\frac{100}{\sqrt{3}}) m

2

(200 \sqrt{3}) m

3

(\frac{200}{\sqrt{3}}) m

4

(\frac{150}{\sqrt{3}}) m

Explanation:

C A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position.
If the car moves to a point where angle is $\frac{π}{3}$ Let,
original image
In $△ BCD$
$\tan \frac{π}{3} = \frac{DC}{BC}$
$\Rightarrow \sqrt{3} = \frac{100}{y}$
$\Rightarrow y = \frac{100}{\sqrt{3}}$
Now,
$\tan \frac{π}{6} = \frac{DC}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AB + BC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x + y}$
$\Rightarrow x + y = 100 \sqrt{3}$
$In △ ACD$
Put the value of $y$ in above equation
$x + \frac{100}{\sqrt{3}} = 100 \sqrt{3}$
$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}}$
$x = \frac{100 \times 3 - 100}{\sqrt{3}}$
$x = \frac{300 - 100}{\sqrt{3}}$
$x = \frac{200}{\sqrt{3}} m$

UP CPMT-2003

Motion in One Dimensions

141386 The displacement of a particle moving in a straight line is given by the expression $x = {A t}^{3} +$ $B t^{2} + C t + D$ in meter, where $t$ is in seconds and $A, B, C$ and $D$ are constants.
The ratio between the initial acceleration and initial velocity is

1

\frac{2 C}{B}

2

\frac{2 B}{C}

3

2 C

4

\frac{C}{2 B}

Explanation:

B Displacement of a particle moving in a straight line is given by
$x = A t^{3} + B t^{2} + C t + D$
$Velocity v = \frac{d x}{d t}$
$v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C$
Initial velocity is equal to $\frac{d x}{d t}$ at $t = 0$
Initial velocity $v = (\frac{d x}{d t})$ at $t = 0 = C$
Acceleration $a = \frac{d^{2} x}{d t^{2}} = 6 A t + 2 B$
Initial acceleration a at $t = 0$
$a = {(\frac{d^{2} x}{{dt}^{2}})}_{t = 0} = 2 B$
Ratio between the initial acceleration and initial
Velocity is, $\frac{a}{v} = \frac{{(\frac{d^{2} x}{d t^{2}})}_{t = 0}}{{(\frac{d x}{d t})}_{t = 0}} = \frac{2 B}{C}$
$\frac{a}{v} = \frac{2 B}{C}$

TS EAMCET (Engg.)-2015

Motion in One Dimensions

141387 The position - time (x-t) graph for motion of a body is given below :
original image
Which one among the following is depicted by the above graph?

1 Positive acceleration

2 Negative acceleration

3 Zero acceleration

4 None of the above

Explanation:

C Given,
From the given $(x - t)$ graph it is clear that velocity is constant.
Therefore, acceleration is zero

NDA (II) 2011

Motion in One Dimensions

141388 The displacement of a particle at time $t$ is given by
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{\boldsymbol k}$
Where $a, b$ and $c$ are positive constants. Then the particle is

1 accelerated along

\hat{k}

direction

2 decelerated along

\hat{k}

direction

3 decelerated along

\hat{j}

direction

4 accelerated along

\hat{j}

direction

Explanation:

A Given,
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{k}$
$Velocity (v_{x}) = \frac{dx}{dt}$
$v_{x} = 0$
$v_{y} = \frac{d (bt)}{dt}$
$v_{y} = b$
And, $v_{z} = \frac{dz}{dt} = \frac{d}{dt} (\frac{c}{2} t^{2})$
$v_{z} = ct$
Acceleration
$a_{x} = \frac{{dv}_{x}}{dt} = 0$
$a_{y} = \frac{{dv}_{y}}{dt} = 0$
$a_{z} = \frac{{dv}_{z}}{dt} = c$
Thus, the particle is acceleration along $\hat{k}$ direction.

NDA (I) 2013

Motion in One Dimensions

141384 The acceleration of a particle is increasing linearly with time $t$ as bt. The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

1

v_{0} t + \frac{1}{3} {bt}^{2}

2

v_{0} t + \frac{1}{3} {bt}^{3}

3

v_{0} t + \frac{1}{6} {bt}^{3}

4

v_{0} t + \frac{1}{2} {bt}^{2}

Explanation:

C Given,
Acceleration $a = bt$
time $= t$
Acceleration $= \frac{change in velocity}{change in time}$
$a = \frac{d v}{d t}$
$\frac{dv}{dt} = a = bt$
$\Rightarrow \frac{dv}{dt} = bt$
$dv = bt dt$
Now integrating this we will get
$\int dv = \int btdt$
$v = \frac{{bt}^{2}}{2} + C$
At, $t = 0, v = v_{0}$ , thus $v_{0} = C$
$∴ v = \frac{{bt}^{2}}{2} + v_{0}$
Now, $v = \frac{change is position}{change in time} = \frac{ds}{dt}$
$\Rightarrow \frac{ds}{dt} = \frac{{bt}^{2}}{2} + v_{0}$
$\Rightarrow ds = (\frac{{bt}^{2}}{2} + v_{0}) dt$
Integrating above we get,
$s = \frac{{bt}^{3}}{6} + v_{0} t + C$
At $t = 0, s = 0, C = 0$
i.e. $s = \frac{b t^{3}}{6} + v_{0} t$
or $s = v_{0} t + \frac{1}{6} {bt}^{3}$

UP CPMT-2011

Motion in One Dimensions

141385 A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position. If the car moves to a point where angle is $\frac{π}{3}$ , what is the distance moved by it ?

1

(\frac{100}{\sqrt{3}}) m

2

(200 \sqrt{3}) m

3

(\frac{200}{\sqrt{3}}) m

4

(\frac{150}{\sqrt{3}}) m

Explanation:

C A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position.
If the car moves to a point where angle is $\frac{π}{3}$ Let,
original image
In $△ BCD$
$\tan \frac{π}{3} = \frac{DC}{BC}$
$\Rightarrow \sqrt{3} = \frac{100}{y}$
$\Rightarrow y = \frac{100}{\sqrt{3}}$
Now,
$\tan \frac{π}{6} = \frac{DC}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AB + BC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x + y}$
$\Rightarrow x + y = 100 \sqrt{3}$
$In △ ACD$
Put the value of $y$ in above equation
$x + \frac{100}{\sqrt{3}} = 100 \sqrt{3}$
$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}}$
$x = \frac{100 \times 3 - 100}{\sqrt{3}}$
$x = \frac{300 - 100}{\sqrt{3}}$
$x = \frac{200}{\sqrt{3}} m$

UP CPMT-2003

Motion in One Dimensions

141386 The displacement of a particle moving in a straight line is given by the expression $x = {A t}^{3} +$ $B t^{2} + C t + D$ in meter, where $t$ is in seconds and $A, B, C$ and $D$ are constants.
The ratio between the initial acceleration and initial velocity is

1

\frac{2 C}{B}

2

\frac{2 B}{C}

3

2 C

4

\frac{C}{2 B}

Explanation:

B Displacement of a particle moving in a straight line is given by
$x = A t^{3} + B t^{2} + C t + D$
$Velocity v = \frac{d x}{d t}$
$v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C$
Initial velocity is equal to $\frac{d x}{d t}$ at $t = 0$
Initial velocity $v = (\frac{d x}{d t})$ at $t = 0 = C$
Acceleration $a = \frac{d^{2} x}{d t^{2}} = 6 A t + 2 B$
Initial acceleration a at $t = 0$
$a = {(\frac{d^{2} x}{{dt}^{2}})}_{t = 0} = 2 B$
Ratio between the initial acceleration and initial
Velocity is, $\frac{a}{v} = \frac{{(\frac{d^{2} x}{d t^{2}})}_{t = 0}}{{(\frac{d x}{d t})}_{t = 0}} = \frac{2 B}{C}$
$\frac{a}{v} = \frac{2 B}{C}$

TS EAMCET (Engg.)-2015

Motion in One Dimensions

141387 The position - time (x-t) graph for motion of a body is given below :
original image
Which one among the following is depicted by the above graph?

1 Positive acceleration

2 Negative acceleration

3 Zero acceleration

4 None of the above

Explanation:

C Given,
From the given $(x - t)$ graph it is clear that velocity is constant.
Therefore, acceleration is zero

NDA (II) 2011

Motion in One Dimensions

141388 The displacement of a particle at time $t$ is given by
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{\boldsymbol k}$
Where $a, b$ and $c$ are positive constants. Then the particle is

1 accelerated along

\hat{k}

direction

2 decelerated along

\hat{k}

direction

3 decelerated along

\hat{j}

direction

4 accelerated along

\hat{j}

direction

Explanation:

A Given,
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{k}$
$Velocity (v_{x}) = \frac{dx}{dt}$
$v_{x} = 0$
$v_{y} = \frac{d (bt)}{dt}$
$v_{y} = b$
And, $v_{z} = \frac{dz}{dt} = \frac{d}{dt} (\frac{c}{2} t^{2})$
$v_{z} = ct$
Acceleration
$a_{x} = \frac{{dv}_{x}}{dt} = 0$
$a_{y} = \frac{{dv}_{y}}{dt} = 0$
$a_{z} = \frac{{dv}_{z}}{dt} = c$
Thus, the particle is acceleration along $\hat{k}$ direction.

NDA (I) 2013

Motion in One Dimensions

141384 The acceleration of a particle is increasing linearly with time $t$ as bt. The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

1

v_{0} t + \frac{1}{3} {bt}^{2}

2

v_{0} t + \frac{1}{3} {bt}^{3}

3

v_{0} t + \frac{1}{6} {bt}^{3}

4

v_{0} t + \frac{1}{2} {bt}^{2}

Explanation:

C Given,
Acceleration $a = bt$
time $= t$
Acceleration $= \frac{change in velocity}{change in time}$
$a = \frac{d v}{d t}$
$\frac{dv}{dt} = a = bt$
$\Rightarrow \frac{dv}{dt} = bt$
$dv = bt dt$
Now integrating this we will get
$\int dv = \int btdt$
$v = \frac{{bt}^{2}}{2} + C$
At, $t = 0, v = v_{0}$ , thus $v_{0} = C$
$∴ v = \frac{{bt}^{2}}{2} + v_{0}$
Now, $v = \frac{change is position}{change in time} = \frac{ds}{dt}$
$\Rightarrow \frac{ds}{dt} = \frac{{bt}^{2}}{2} + v_{0}$
$\Rightarrow ds = (\frac{{bt}^{2}}{2} + v_{0}) dt$
Integrating above we get,
$s = \frac{{bt}^{3}}{6} + v_{0} t + C$
At $t = 0, s = 0, C = 0$
i.e. $s = \frac{b t^{3}}{6} + v_{0} t$
or $s = v_{0} t + \frac{1}{6} {bt}^{3}$

UP CPMT-2011

Motion in One Dimensions

141385 A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position. If the car moves to a point where angle is $\frac{π}{3}$ , what is the distance moved by it ?

1

(\frac{100}{\sqrt{3}}) m

2

(200 \sqrt{3}) m

3

(\frac{200}{\sqrt{3}}) m

4

(\frac{150}{\sqrt{3}}) m

Explanation:

C A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position.
If the car moves to a point where angle is $\frac{π}{3}$ Let,
original image
In $△ BCD$
$\tan \frac{π}{3} = \frac{DC}{BC}$
$\Rightarrow \sqrt{3} = \frac{100}{y}$
$\Rightarrow y = \frac{100}{\sqrt{3}}$
Now,
$\tan \frac{π}{6} = \frac{DC}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AB + BC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x + y}$
$\Rightarrow x + y = 100 \sqrt{3}$
$In △ ACD$
Put the value of $y$ in above equation
$x + \frac{100}{\sqrt{3}} = 100 \sqrt{3}$
$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}}$
$x = \frac{100 \times 3 - 100}{\sqrt{3}}$
$x = \frac{300 - 100}{\sqrt{3}}$
$x = \frac{200}{\sqrt{3}} m$

UP CPMT-2003

Motion in One Dimensions

141386 The displacement of a particle moving in a straight line is given by the expression $x = {A t}^{3} +$ $B t^{2} + C t + D$ in meter, where $t$ is in seconds and $A, B, C$ and $D$ are constants.
The ratio between the initial acceleration and initial velocity is

1

\frac{2 C}{B}

2

\frac{2 B}{C}

3

2 C

4

\frac{C}{2 B}

Explanation:

B Displacement of a particle moving in a straight line is given by
$x = A t^{3} + B t^{2} + C t + D$
$Velocity v = \frac{d x}{d t}$
$v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C$
Initial velocity is equal to $\frac{d x}{d t}$ at $t = 0$
Initial velocity $v = (\frac{d x}{d t})$ at $t = 0 = C$
Acceleration $a = \frac{d^{2} x}{d t^{2}} = 6 A t + 2 B$
Initial acceleration a at $t = 0$
$a = {(\frac{d^{2} x}{{dt}^{2}})}_{t = 0} = 2 B$
Ratio between the initial acceleration and initial
Velocity is, $\frac{a}{v} = \frac{{(\frac{d^{2} x}{d t^{2}})}_{t = 0}}{{(\frac{d x}{d t})}_{t = 0}} = \frac{2 B}{C}$
$\frac{a}{v} = \frac{2 B}{C}$

TS EAMCET (Engg.)-2015

Motion in One Dimensions

141387 The position - time (x-t) graph for motion of a body is given below :
original image
Which one among the following is depicted by the above graph?

1 Positive acceleration

2 Negative acceleration

3 Zero acceleration

4 None of the above

Explanation:

C Given,
From the given $(x - t)$ graph it is clear that velocity is constant.
Therefore, acceleration is zero

NDA (II) 2011

Motion in One Dimensions

141388 The displacement of a particle at time $t$ is given by
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{\boldsymbol k}$
Where $a, b$ and $c$ are positive constants. Then the particle is

1 accelerated along

\hat{k}

direction

2 decelerated along

\hat{k}

direction

3 decelerated along

\hat{j}

direction

4 accelerated along

\hat{j}

direction

Explanation:

A Given,
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{k}$
$Velocity (v_{x}) = \frac{dx}{dt}$
$v_{x} = 0$
$v_{y} = \frac{d (bt)}{dt}$
$v_{y} = b$
And, $v_{z} = \frac{dz}{dt} = \frac{d}{dt} (\frac{c}{2} t^{2})$
$v_{z} = ct$
Acceleration
$a_{x} = \frac{{dv}_{x}}{dt} = 0$
$a_{y} = \frac{{dv}_{y}}{dt} = 0$
$a_{z} = \frac{{dv}_{z}}{dt} = c$
Thus, the particle is acceleration along $\hat{k}$ direction.

NDA (I) 2013

Motion in One Dimensions

141384 The acceleration of a particle is increasing linearly with time $t$ as bt. The particle starts from the origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time $t$ will be

1

v_{0} t + \frac{1}{3} {bt}^{2}

2

v_{0} t + \frac{1}{3} {bt}^{3}

3

v_{0} t + \frac{1}{6} {bt}^{3}

4

v_{0} t + \frac{1}{2} {bt}^{2}

Explanation:

C Given,
Acceleration $a = bt$
time $= t$
Acceleration $= \frac{change in velocity}{change in time}$
$a = \frac{d v}{d t}$
$\frac{dv}{dt} = a = bt$
$\Rightarrow \frac{dv}{dt} = bt$
$dv = bt dt$
Now integrating this we will get
$\int dv = \int btdt$
$v = \frac{{bt}^{2}}{2} + C$
At, $t = 0, v = v_{0}$ , thus $v_{0} = C$
$∴ v = \frac{{bt}^{2}}{2} + v_{0}$
Now, $v = \frac{change is position}{change in time} = \frac{ds}{dt}$
$\Rightarrow \frac{ds}{dt} = \frac{{bt}^{2}}{2} + v_{0}$
$\Rightarrow ds = (\frac{{bt}^{2}}{2} + v_{0}) dt$
Integrating above we get,
$s = \frac{{bt}^{3}}{6} + v_{0} t + C$
At $t = 0, s = 0, C = 0$
i.e. $s = \frac{b t^{3}}{6} + v_{0} t$
or $s = v_{0} t + \frac{1}{6} {bt}^{3}$

UP CPMT-2011

Motion in One Dimensions

141385 A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position. If the car moves to a point where angle is $\frac{π}{3}$ , what is the distance moved by it ?

1

(\frac{100}{\sqrt{3}}) m

2

(200 \sqrt{3}) m

3

(\frac{200}{\sqrt{3}}) m

4

(\frac{150}{\sqrt{3}}) m

Explanation:

C A man is at a height of $100 m$ . He sees a car which makes an angle of $\frac{π}{6}$ with man's position.
If the car moves to a point where angle is $\frac{π}{3}$ Let,
original image
In $△ BCD$
$\tan \frac{π}{3} = \frac{DC}{BC}$
$\Rightarrow \sqrt{3} = \frac{100}{y}$
$\Rightarrow y = \frac{100}{\sqrt{3}}$
Now,
$\tan \frac{π}{6} = \frac{DC}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AB + BC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{x + y}$
$\Rightarrow x + y = 100 \sqrt{3}$
$In △ ACD$
Put the value of $y$ in above equation
$x + \frac{100}{\sqrt{3}} = 100 \sqrt{3}$
$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}}$
$x = \frac{100 \times 3 - 100}{\sqrt{3}}$
$x = \frac{300 - 100}{\sqrt{3}}$
$x = \frac{200}{\sqrt{3}} m$

UP CPMT-2003

Motion in One Dimensions

141386 The displacement of a particle moving in a straight line is given by the expression $x = {A t}^{3} +$ $B t^{2} + C t + D$ in meter, where $t$ is in seconds and $A, B, C$ and $D$ are constants.
The ratio between the initial acceleration and initial velocity is

1

\frac{2 C}{B}

2

\frac{2 B}{C}

3

2 C

4

\frac{C}{2 B}

Explanation:

B Displacement of a particle moving in a straight line is given by
$x = A t^{3} + B t^{2} + C t + D$
$Velocity v = \frac{d x}{d t}$
$v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C$
Initial velocity is equal to $\frac{d x}{d t}$ at $t = 0$
Initial velocity $v = (\frac{d x}{d t})$ at $t = 0 = C$
Acceleration $a = \frac{d^{2} x}{d t^{2}} = 6 A t + 2 B$
Initial acceleration a at $t = 0$
$a = {(\frac{d^{2} x}{{dt}^{2}})}_{t = 0} = 2 B$
Ratio between the initial acceleration and initial
Velocity is, $\frac{a}{v} = \frac{{(\frac{d^{2} x}{d t^{2}})}_{t = 0}}{{(\frac{d x}{d t})}_{t = 0}} = \frac{2 B}{C}$
$\frac{a}{v} = \frac{2 B}{C}$

TS EAMCET (Engg.)-2015

Motion in One Dimensions

141387 The position - time (x-t) graph for motion of a body is given below :
original image
Which one among the following is depicted by the above graph?

1 Positive acceleration

2 Negative acceleration

3 Zero acceleration

4 None of the above

Explanation:

C Given,
From the given $(x - t)$ graph it is clear that velocity is constant.
Therefore, acceleration is zero

NDA (II) 2011

Motion in One Dimensions

141388 The displacement of a particle at time $t$ is given by
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{\boldsymbol k}$
Where $a, b$ and $c$ are positive constants. Then the particle is

1 accelerated along

\hat{k}

direction

2 decelerated along

\hat{k}

direction

3 decelerated along

\hat{j}

direction

4 accelerated along

\hat{j}

direction

Explanation:

A Given,
$\vec{x} = a \hat{i} + b t \hat{j} + \frac{c}{2} t^{2} \hat{k}$
$Velocity (v_{x}) = \frac{dx}{dt}$
$v_{x} = 0$
$v_{y} = \frac{d (bt)}{dt}$
$v_{y} = b$
And, $v_{z} = \frac{dz}{dt} = \frac{d}{dt} (\frac{c}{2} t^{2})$
$v_{z} = ct$
Acceleration
$a_{x} = \frac{{dv}_{x}}{dt} = 0$
$a_{y} = \frac{{dv}_{y}}{dt} = 0$
$a_{z} = \frac{{dv}_{z}}{dt} = c$
Thus, the particle is acceleration along $\hat{k}$ $\hat{k}$ $\hat{k}$ direction.

NDA (I) 2013