QBManager

Motion in One Dimensions

141696 A particle moves along a straight line $O X$ . At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

1

24 m

2

40 m

3

56 m

4

16 m

Explanation:

C Given, $x = 40 + 12 t - t^{3}$
Velocity $(v) = \frac{dx}{dt} = 0 + 12 - 3 t^{2}$
At coming to rest $(v = 0)$ ,
$0 = 12 - 3 t^{2}$
$t^{2} = 4 \Rightarrow t = 2 \sec$
The distance travelled by particle before coming to rest
$x = 40 + 12 (2) - (2)^{3}$
$x = 56 m$

AIPMT 2006

Motion in One Dimensions

141697 The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} m / s^{2}$ , in the third second is.

1

10 / 3 m

2

19 / 3 m

3

6 m

4

4 m

Explanation:

A Given,
Acceleration $(a) = \frac{4}{3} m / s^{2}$
Distance covered in $n^{th}$ second
$s_{n} = u + \frac{1}{2} a (2 n - 1)$
Distance covered by particle in third second is-
$s_{3} = 0 + \frac{1}{2} \times \frac{4}{3} (2 \times 3 - 1) (initial velocity u = 0)$
$s_{3} = \frac{10}{3} m$

CG PET-22.05.2022

Motion in One Dimensions

141698 A ball under uniform acceleration travels $6 m$ in first $2 s$ and $16 m$ in the next $2 s$ . Its initial velocity is

1

\frac{1}{2} m / s

2

1 m / s

3

\frac{8}{3} m / s

4

\frac{1}{4} m / s

Explanation:

A Given,
Distance travelled $(s_{1}) = 6 m$
Distance travelled $(s_{2}) = 16 m$
Time taken by $s_{1} (t_{1}) = 2 s$
Time taken by $s_{2} (t_{2}) = 4 s$
Let the initial velocity be $u$
We know,
$s = ut + \frac{1}{2} {at}^{2}$
When $s_{1} = 6 m, t_{1} = 2 s$
$6 = 2 u + 2 a$
and when $s = s_{1} + s_{2} = 22 m$
$t = t_{1} + t_{2} = 4 s$
$22 = 4 u + 8 a$
On solving equation (1) and equation (2), we get
$u = \frac{1}{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141699 The position of a particle is $r = x i + y j$ where $x$ and $y$ are function of time $t$ and given as $x = (12 + 5 t - t^{2}) m$ and $y = (18 + 5 t - t^{2}) m$ . At $t =$ $1 s$ , the magnitude of the velocity vector of the particle is

1

2 \sqrt{3} m / s

2

3 \sqrt{2} m / s

3

4 \sqrt{2} m / s

4

3 \sqrt{3} m / s

Explanation:

B Given,
Position $(\vec{r}) = xi + yj$
Where, $x = 12 + 5 t - t^{2}$
$y = 18 + 5 t - t^{2}$
We know,
$\vec{v} = \frac{\vec{dr}}{dt}$
Differentiating the $\vec{r}$ ,
$\frac{dr}{dt} = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = 3 \hat{i} + 3 \hat{j}$
$| v | = \sqrt{3^{2} + 3^{2}}$
$| v | = 3 \sqrt{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141700 The velocity-time(v-t) relation of a particle moving in a plane is $v = 3 t^{2} m / s$ . At $t = 0$ ; displacement $x = 8 m$ . The velocity of the particle at $x = 16 m$ is

1

12 m / s

2

14 m / s

3

18 m / s

4

10 m / s

Explanation:

A Give that, $v = 3 t^{2}$
Differentiation of displacement gives velocity i.e.
$\frac{dx}{dt} = v$
$vdt = dx$
Integrating both side, we get
$\int 3 t^{2} d t = \int d x$
$\frac{3 t^{3}}{3} + c = x$
$t^{3} + c = x$
At $t = 0, x = 8$
$(0)^{3} + c = 8 \Rightarrow c = 8$
Putting value of $c$ in ${eq}^{n} (i)$ , we get
$t^{3} + 8 = x$
At $x = 16 m, t =$ ?
$t^{3} + 8 = 16 \Rightarrow t = 2 \sec$
Then, $v = 3 t^{2}$
$= 3 \times (2)^{2}$
$= 12 m / s$

TS EAMCET 31.07.2022

Motion in One Dimensions

141696 A particle moves along a straight line $O X$ . At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

1

24 m

2

40 m

3

56 m

4

16 m

Explanation:

C Given, $x = 40 + 12 t - t^{3}$
Velocity $(v) = \frac{dx}{dt} = 0 + 12 - 3 t^{2}$
At coming to rest $(v = 0)$ ,
$0 = 12 - 3 t^{2}$
$t^{2} = 4 \Rightarrow t = 2 \sec$
The distance travelled by particle before coming to rest
$x = 40 + 12 (2) - (2)^{3}$
$x = 56 m$

AIPMT 2006

Motion in One Dimensions

141697 The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} m / s^{2}$ , in the third second is.

1

10 / 3 m

2

19 / 3 m

3

6 m

4

4 m

Explanation:

A Given,
Acceleration $(a) = \frac{4}{3} m / s^{2}$
Distance covered in $n^{th}$ second
$s_{n} = u + \frac{1}{2} a (2 n - 1)$
Distance covered by particle in third second is-
$s_{3} = 0 + \frac{1}{2} \times \frac{4}{3} (2 \times 3 - 1) (initial velocity u = 0)$
$s_{3} = \frac{10}{3} m$

CG PET-22.05.2022

Motion in One Dimensions

141698 A ball under uniform acceleration travels $6 m$ in first $2 s$ and $16 m$ in the next $2 s$ . Its initial velocity is

1

\frac{1}{2} m / s

2

1 m / s

3

\frac{8}{3} m / s

4

\frac{1}{4} m / s

Explanation:

A Given,
Distance travelled $(s_{1}) = 6 m$
Distance travelled $(s_{2}) = 16 m$
Time taken by $s_{1} (t_{1}) = 2 s$
Time taken by $s_{2} (t_{2}) = 4 s$
Let the initial velocity be $u$
We know,
$s = ut + \frac{1}{2} {at}^{2}$
When $s_{1} = 6 m, t_{1} = 2 s$
$6 = 2 u + 2 a$
and when $s = s_{1} + s_{2} = 22 m$
$t = t_{1} + t_{2} = 4 s$
$22 = 4 u + 8 a$
On solving equation (1) and equation (2), we get
$u = \frac{1}{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141699 The position of a particle is $r = x i + y j$ where $x$ and $y$ are function of time $t$ and given as $x = (12 + 5 t - t^{2}) m$ and $y = (18 + 5 t - t^{2}) m$ . At $t =$ $1 s$ , the magnitude of the velocity vector of the particle is

1

2 \sqrt{3} m / s

2

3 \sqrt{2} m / s

3

4 \sqrt{2} m / s

4

3 \sqrt{3} m / s

Explanation:

B Given,
Position $(\vec{r}) = xi + yj$
Where, $x = 12 + 5 t - t^{2}$
$y = 18 + 5 t - t^{2}$
We know,
$\vec{v} = \frac{\vec{dr}}{dt}$
Differentiating the $\vec{r}$ ,
$\frac{dr}{dt} = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = 3 \hat{i} + 3 \hat{j}$
$| v | = \sqrt{3^{2} + 3^{2}}$
$| v | = 3 \sqrt{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141700 The velocity-time(v-t) relation of a particle moving in a plane is $v = 3 t^{2} m / s$ . At $t = 0$ ; displacement $x = 8 m$ . The velocity of the particle at $x = 16 m$ is

1

12 m / s

2

14 m / s

3

18 m / s

4

10 m / s

Explanation:

A Give that, $v = 3 t^{2}$
Differentiation of displacement gives velocity i.e.
$\frac{dx}{dt} = v$
$vdt = dx$
Integrating both side, we get
$\int 3 t^{2} d t = \int d x$
$\frac{3 t^{3}}{3} + c = x$
$t^{3} + c = x$
At $t = 0, x = 8$
$(0)^{3} + c = 8 \Rightarrow c = 8$
Putting value of $c$ in ${eq}^{n} (i)$ , we get
$t^{3} + 8 = x$
At $x = 16 m, t =$ ?
$t^{3} + 8 = 16 \Rightarrow t = 2 \sec$
Then, $v = 3 t^{2}$
$= 3 \times (2)^{2}$
$= 12 m / s$

TS EAMCET 31.07.2022

Motion in One Dimensions

141696 A particle moves along a straight line $O X$ . At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

1

24 m

2

40 m

3

56 m

4

16 m

Explanation:

C Given, $x = 40 + 12 t - t^{3}$
Velocity $(v) = \frac{dx}{dt} = 0 + 12 - 3 t^{2}$
At coming to rest $(v = 0)$ ,
$0 = 12 - 3 t^{2}$
$t^{2} = 4 \Rightarrow t = 2 \sec$
The distance travelled by particle before coming to rest
$x = 40 + 12 (2) - (2)^{3}$
$x = 56 m$

AIPMT 2006

Motion in One Dimensions

141697 The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} m / s^{2}$ , in the third second is.

1

10 / 3 m

2

19 / 3 m

3

6 m

4

4 m

Explanation:

A Given,
Acceleration $(a) = \frac{4}{3} m / s^{2}$
Distance covered in $n^{th}$ second
$s_{n} = u + \frac{1}{2} a (2 n - 1)$
Distance covered by particle in third second is-
$s_{3} = 0 + \frac{1}{2} \times \frac{4}{3} (2 \times 3 - 1) (initial velocity u = 0)$
$s_{3} = \frac{10}{3} m$

CG PET-22.05.2022

Motion in One Dimensions

141698 A ball under uniform acceleration travels $6 m$ in first $2 s$ and $16 m$ in the next $2 s$ . Its initial velocity is

1

\frac{1}{2} m / s

2

1 m / s

3

\frac{8}{3} m / s

4

\frac{1}{4} m / s

Explanation:

A Given,
Distance travelled $(s_{1}) = 6 m$
Distance travelled $(s_{2}) = 16 m$
Time taken by $s_{1} (t_{1}) = 2 s$
Time taken by $s_{2} (t_{2}) = 4 s$
Let the initial velocity be $u$
We know,
$s = ut + \frac{1}{2} {at}^{2}$
When $s_{1} = 6 m, t_{1} = 2 s$
$6 = 2 u + 2 a$
and when $s = s_{1} + s_{2} = 22 m$
$t = t_{1} + t_{2} = 4 s$
$22 = 4 u + 8 a$
On solving equation (1) and equation (2), we get
$u = \frac{1}{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141699 The position of a particle is $r = x i + y j$ where $x$ and $y$ are function of time $t$ and given as $x = (12 + 5 t - t^{2}) m$ and $y = (18 + 5 t - t^{2}) m$ . At $t =$ $1 s$ , the magnitude of the velocity vector of the particle is

1

2 \sqrt{3} m / s

2

3 \sqrt{2} m / s

3

4 \sqrt{2} m / s

4

3 \sqrt{3} m / s

Explanation:

B Given,
Position $(\vec{r}) = xi + yj$
Where, $x = 12 + 5 t - t^{2}$
$y = 18 + 5 t - t^{2}$
We know,
$\vec{v} = \frac{\vec{dr}}{dt}$
Differentiating the $\vec{r}$ ,
$\frac{dr}{dt} = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = 3 \hat{i} + 3 \hat{j}$
$| v | = \sqrt{3^{2} + 3^{2}}$
$| v | = 3 \sqrt{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141700 The velocity-time(v-t) relation of a particle moving in a plane is $v = 3 t^{2} m / s$ . At $t = 0$ ; displacement $x = 8 m$ . The velocity of the particle at $x = 16 m$ is

1

12 m / s

2

14 m / s

3

18 m / s

4

10 m / s

Explanation:

A Give that, $v = 3 t^{2}$
Differentiation of displacement gives velocity i.e.
$\frac{dx}{dt} = v$
$vdt = dx$
Integrating both side, we get
$\int 3 t^{2} d t = \int d x$
$\frac{3 t^{3}}{3} + c = x$
$t^{3} + c = x$
At $t = 0, x = 8$
$(0)^{3} + c = 8 \Rightarrow c = 8$
Putting value of $c$ in ${eq}^{n} (i)$ , we get
$t^{3} + 8 = x$
At $x = 16 m, t =$ ?
$t^{3} + 8 = 16 \Rightarrow t = 2 \sec$
Then, $v = 3 t^{2}$
$= 3 \times (2)^{2}$
$= 12 m / s$

TS EAMCET 31.07.2022

Motion in One Dimensions

141696 A particle moves along a straight line $O X$ . At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

1

24 m

2

40 m

3

56 m

4

16 m

Explanation:

C Given, $x = 40 + 12 t - t^{3}$
Velocity $(v) = \frac{dx}{dt} = 0 + 12 - 3 t^{2}$
At coming to rest $(v = 0)$ ,
$0 = 12 - 3 t^{2}$
$t^{2} = 4 \Rightarrow t = 2 \sec$
The distance travelled by particle before coming to rest
$x = 40 + 12 (2) - (2)^{3}$
$x = 56 m$

AIPMT 2006

Motion in One Dimensions

141697 The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} m / s^{2}$ , in the third second is.

1

10 / 3 m

2

19 / 3 m

3

6 m

4

4 m

Explanation:

A Given,
Acceleration $(a) = \frac{4}{3} m / s^{2}$
Distance covered in $n^{th}$ second
$s_{n} = u + \frac{1}{2} a (2 n - 1)$
Distance covered by particle in third second is-
$s_{3} = 0 + \frac{1}{2} \times \frac{4}{3} (2 \times 3 - 1) (initial velocity u = 0)$
$s_{3} = \frac{10}{3} m$

CG PET-22.05.2022

Motion in One Dimensions

141698 A ball under uniform acceleration travels $6 m$ in first $2 s$ and $16 m$ in the next $2 s$ . Its initial velocity is

1

\frac{1}{2} m / s

2

1 m / s

3

\frac{8}{3} m / s

4

\frac{1}{4} m / s

Explanation:

A Given,
Distance travelled $(s_{1}) = 6 m$
Distance travelled $(s_{2}) = 16 m$
Time taken by $s_{1} (t_{1}) = 2 s$
Time taken by $s_{2} (t_{2}) = 4 s$
Let the initial velocity be $u$
We know,
$s = ut + \frac{1}{2} {at}^{2}$
When $s_{1} = 6 m, t_{1} = 2 s$
$6 = 2 u + 2 a$
and when $s = s_{1} + s_{2} = 22 m$
$t = t_{1} + t_{2} = 4 s$
$22 = 4 u + 8 a$
On solving equation (1) and equation (2), we get
$u = \frac{1}{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141699 The position of a particle is $r = x i + y j$ where $x$ and $y$ are function of time $t$ and given as $x = (12 + 5 t - t^{2}) m$ and $y = (18 + 5 t - t^{2}) m$ . At $t =$ $1 s$ , the magnitude of the velocity vector of the particle is

1

2 \sqrt{3} m / s

2

3 \sqrt{2} m / s

3

4 \sqrt{2} m / s

4

3 \sqrt{3} m / s

Explanation:

B Given,
Position $(\vec{r}) = xi + yj$
Where, $x = 12 + 5 t - t^{2}$
$y = 18 + 5 t - t^{2}$
We know,
$\vec{v} = \frac{\vec{dr}}{dt}$
Differentiating the $\vec{r}$ ,
$\frac{dr}{dt} = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = 3 \hat{i} + 3 \hat{j}$
$| v | = \sqrt{3^{2} + 3^{2}}$
$| v | = 3 \sqrt{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141700 The velocity-time(v-t) relation of a particle moving in a plane is $v = 3 t^{2} m / s$ . At $t = 0$ ; displacement $x = 8 m$ . The velocity of the particle at $x = 16 m$ is

1

12 m / s

2

14 m / s

3

18 m / s

4

10 m / s

Explanation:

A Give that, $v = 3 t^{2}$
Differentiation of displacement gives velocity i.e.
$\frac{dx}{dt} = v$
$vdt = dx$
Integrating both side, we get
$\int 3 t^{2} d t = \int d x$
$\frac{3 t^{3}}{3} + c = x$
$t^{3} + c = x$
At $t = 0, x = 8$
$(0)^{3} + c = 8 \Rightarrow c = 8$
Putting value of $c$ in ${eq}^{n} (i)$ , we get
$t^{3} + 8 = x$
At $x = 16 m, t =$ ?
$t^{3} + 8 = 16 \Rightarrow t = 2 \sec$
Then, $v = 3 t^{2}$
$= 3 \times (2)^{2}$
$= 12 m / s$

TS EAMCET 31.07.2022

Motion in One Dimensions

141696 A particle moves along a straight line $O X$ . At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x = 40 + 12 t - t^{3}$ . How long would the particle travel before coming to rest?

1

24 m

2

40 m

3

56 m

4

16 m

Explanation:

C Given, $x = 40 + 12 t - t^{3}$
Velocity $(v) = \frac{dx}{dt} = 0 + 12 - 3 t^{2}$
At coming to rest $(v = 0)$ ,
$0 = 12 - 3 t^{2}$
$t^{2} = 4 \Rightarrow t = 2 \sec$
The distance travelled by particle before coming to rest
$x = 40 + 12 (2) - (2)^{3}$
$x = 56 m$

AIPMT 2006

Motion in One Dimensions

141697 The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} m / s^{2}$ , in the third second is.

1

10 / 3 m

2

19 / 3 m

3

6 m

4

4 m

Explanation:

A Given,
Acceleration $(a) = \frac{4}{3} m / s^{2}$
Distance covered in $n^{th}$ second
$s_{n} = u + \frac{1}{2} a (2 n - 1)$
Distance covered by particle in third second is-
$s_{3} = 0 + \frac{1}{2} \times \frac{4}{3} (2 \times 3 - 1) (initial velocity u = 0)$
$s_{3} = \frac{10}{3} m$

CG PET-22.05.2022

Motion in One Dimensions

141698 A ball under uniform acceleration travels $6 m$ in first $2 s$ and $16 m$ in the next $2 s$ . Its initial velocity is

1

\frac{1}{2} m / s

2

1 m / s

3

\frac{8}{3} m / s

4

\frac{1}{4} m / s

Explanation:

A Given,
Distance travelled $(s_{1}) = 6 m$
Distance travelled $(s_{2}) = 16 m$
Time taken by $s_{1} (t_{1}) = 2 s$
Time taken by $s_{2} (t_{2}) = 4 s$
Let the initial velocity be $u$
We know,
$s = ut + \frac{1}{2} {at}^{2}$
When $s_{1} = 6 m, t_{1} = 2 s$
$6 = 2 u + 2 a$
and when $s = s_{1} + s_{2} = 22 m$
$t = t_{1} + t_{2} = 4 s$
$22 = 4 u + 8 a$
On solving equation (1) and equation (2), we get
$u = \frac{1}{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141699 The position of a particle is $r = x i + y j$ where $x$ and $y$ are function of time $t$ and given as $x = (12 + 5 t - t^{2}) m$ and $y = (18 + 5 t - t^{2}) m$ . At $t =$ $1 s$ , the magnitude of the velocity vector of the particle is

1

2 \sqrt{3} m / s

2

3 \sqrt{2} m / s

3

4 \sqrt{2} m / s

4

3 \sqrt{3} m / s

Explanation:

B Given,
Position $(\vec{r}) = xi + yj$
Where, $x = 12 + 5 t - t^{2}$
$y = 18 + 5 t - t^{2}$
We know,
$\vec{v} = \frac{\vec{dr}}{dt}$
Differentiating the $\vec{r}$ ,
$\frac{dr}{dt} = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = (5 - 2 t) \hat{i} + (5 - 2 t) \hat{j}$
$v_{t} = 1 s = 3 \hat{i} + 3 \hat{j}$
$| v | = \sqrt{3^{2} + 3^{2}}$
$| v | = 3 \sqrt{2} m / s$

TS EAMCET 30.07.2022

Motion in One Dimensions

141700 The velocity-time(v-t) relation of a particle moving in a plane is $v = 3 t^{2} m / s$ . At $t = 0$ ; displacement $x = 8 m$ . The velocity of the particle at $x = 16 m$ is

1

12 m / s

2

14 m / s

3

18 m / s

4

10 m / s

Explanation:

A Give that, $v = 3 t^{2}$
Differentiation of displacement gives velocity i.e.
$\frac{dx}{dt} = v$
$vdt = dx$
Integrating both side, we get
$\int 3 t^{2} d t = \int d x$
$\frac{3 t^{3}}{3} + c = x$
$t^{3} + c = x$
At $t = 0, x = 8$
$(0)^{3} + c = 8 \Rightarrow c = 8$
Putting value of $c$ in ${eq}^{n} (i)$ , we get
$t^{3} + 8 = x$
At $x = 16 m, t =$ ?
$t^{3} + 8 = 16 \Rightarrow t = 2 \sec$
Then, $v = 3 t^{2}$
$= 3 \times (2)^{2}$
$= 12 m / s$

TS EAMCET 31.07.2022