QBManager

Motion in One Dimensions

141215 A person sitting in the ground floor of a building notices through the window of height $1.5 m$ , a ball dropped from the roof of the building crosses the window in $0.1 s$ . What is the velocity of the ball when it is at the topmost point of the window? $(g = 10 m / s^{2})$

1

15.5 m / s

2

14.5 m / s

3

4.5 m / s

4

20 m / s

Explanation:

B Given, $g = 10 m / s^{2}$
$s = 1.5 m$
$t = 0.1$ Second
From equation of motion,
$s = ut + \frac{1}{2} {gt}^{2}$
$1.5 = u (0.1) + \frac{1}{2} \times 10 \times (0.1)^{2}$
$1.5 = (0.1) u + 5 \times 0.01$
$1.5 = (0.1) u + 0.05$
$u = \frac{1.45}{0.1} = 14.5 m / s$

NEET (Oct.) 2020

Motion in One Dimensions

141216 A particle starts from the origin with an initial velocity $\vec{u} = (3.0 \hat{j}) m / s$ and a constant acceleration $\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$ . The velocity of the particle when it reaches its maximum y coordinate is:

1

2 \hat{i}

2

- 1.5 \hat{i} + 1.5 \hat{j}

3

- 1.5 \hat{i}

4

1.5 \hat{i} + 1.0 \hat{j}

Explanation:

C $\vec{u} = (3.0 \hat{j}) m / s$
$\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$
This can be treated as 2 motions
One is $x$ directions and other is $y$ direction $y$ direction-
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
Let $s_{y}$ be the displacement in $y$ -direction,
$u_{y} = 3 m / \sec, a_{y} = - 1 m / \sec^{2}$
$∴ s_{y} = 3 t - \frac{t^{2}}{2}$
$v_{y} = u_{y} + a_{y} t$
$= 3 + (- 1) t$
$= (3 - t) m / s$
y component will be maximum, when $\frac{{ds}_{y}}{dt}$ or $v_{y}$ is zero.
$v_{y} = 0 \Rightarrow 3 - t = 0$
Similarly is $x$ -direction,
$u_{x} = 0, a_{x} = - 0.5 m / \sec^{2}$
$v_{x} = u_{x} + a_{y} t$
$v_{x} = 0 + (- 0.5) t$
$v_{x} = - 0.5 t$
Now at $t = 3 \sec$
$v_{x} = - 0.5 \times 3 = - 1.5 m / \sec$
Velocity vector $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = - 1.5 \hat{i} + 0 \hat{j}$
$\vec{v} = - 1.5 \hat{i} m / \sec$

TS EAMCET 29.09.2020

Motion in One Dimensions

141217 An object is moving in a plane and its trajectory $\overset{―}{r} (t)$ (in meter) is given by, $\overset{―}{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 t \hat{j}$ , where $t$ is time (in second). The angle that the velocity vector makes with the $X$ -axis at $t = 0$ .

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

C Trajectory of an object-
$\vec{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 \hat{t} j$
velocity, $\vec{v} = \frac{d \vec{r}}{dt} = (6 t + 2) \hat{i} + 2 \hat{j} m / s$
$\vec{v} = \underset{v_{x}}{\underset{⏟}{(6 t + 2)}} \hat{i} + \underset{v_{y}}{\underset{⏟}{2 \hat{j}}}$
Let $θ$ is that angle which makes with $X$ -axis,
at $t = 0$
$∵ \tan θ = \frac{v_{y}}{v_{x}} = \frac{2}{6 t + 2}$
At $t = 0$ ,
$\tan θ = \frac{2}{0 + 2} = 1 = \tan 45^{\circ}$
$θ = 45^{\circ}$

TS EAMCET 29.09.2020

Motion in One Dimensions

141219 Starting from a point on circumference of a circle of radius ' $r$ ', a person moves to a point which is diametrically opposite its initial position. The distance travelled and displacement are respectively

1

2 π r, 0

2

π r, 2 r

3

2 r, 2 π r

4

2 r, π r

Explanation:

B original image
According to question let a person stand initially at point $A$ and he travels from $A$ to $B$ on circle which is diametrically opposite its initial position and has radius $r$ .
So,
Distance $(D)$ travelled by person $=$ Distance along circumference from A to B.
$D = \frac{2 π r}{2} = π r$
and Displacement $=$ Distance along diameter $(AB)$ Displacement $= AB = 2 r$

AP EAMCET-25.09.2020

Motion in One Dimensions

141220 The displacement is given by $x = 2 t^{2} + t + 5$ , the acceleration at $t = 2 s$ is

1

4 m \cdot s^{- 2}

2

8 m \cdot s^{- 2}

3

10 m \cdot s^{- 2}

4

15 m \cdot s^{- 2}

Explanation:

A Given,
Displacement $(x) = 2 t^{2} + t + 5$
Differentiating the above equation,
$v = \frac{dx}{dt} = 4 t + 1 + 0$
$\frac{dx}{dt} = 4 t + 1$
Again, differentiating the equation (i)
$a = \frac{d^{2} x}{{dt}^{2}} = 4 \times 1$
$\frac{d^{2} x}{{dt}^{2}} = a = 4 m / s^{2} at t = 2 s$
$a = 4 m / s^{2}$

AP EAMCET-07.10.2020

Motion in One Dimensions

141215 A person sitting in the ground floor of a building notices through the window of height $1.5 m$ , a ball dropped from the roof of the building crosses the window in $0.1 s$ . What is the velocity of the ball when it is at the topmost point of the window? $(g = 10 m / s^{2})$

1

15.5 m / s

2

14.5 m / s

3

4.5 m / s

4

20 m / s

Explanation:

B Given, $g = 10 m / s^{2}$
$s = 1.5 m$
$t = 0.1$ Second
From equation of motion,
$s = ut + \frac{1}{2} {gt}^{2}$
$1.5 = u (0.1) + \frac{1}{2} \times 10 \times (0.1)^{2}$
$1.5 = (0.1) u + 5 \times 0.01$
$1.5 = (0.1) u + 0.05$
$u = \frac{1.45}{0.1} = 14.5 m / s$

NEET (Oct.) 2020

Motion in One Dimensions

141216 A particle starts from the origin with an initial velocity $\vec{u} = (3.0 \hat{j}) m / s$ and a constant acceleration $\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$ . The velocity of the particle when it reaches its maximum y coordinate is:

1

2 \hat{i}

2

- 1.5 \hat{i} + 1.5 \hat{j}

3

- 1.5 \hat{i}

4

1.5 \hat{i} + 1.0 \hat{j}

Explanation:

C $\vec{u} = (3.0 \hat{j}) m / s$
$\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$
This can be treated as 2 motions
One is $x$ directions and other is $y$ direction $y$ direction-
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
Let $s_{y}$ be the displacement in $y$ -direction,
$u_{y} = 3 m / \sec, a_{y} = - 1 m / \sec^{2}$
$∴ s_{y} = 3 t - \frac{t^{2}}{2}$
$v_{y} = u_{y} + a_{y} t$
$= 3 + (- 1) t$
$= (3 - t) m / s$
y component will be maximum, when $\frac{{ds}_{y}}{dt}$ or $v_{y}$ is zero.
$v_{y} = 0 \Rightarrow 3 - t = 0$
Similarly is $x$ -direction,
$u_{x} = 0, a_{x} = - 0.5 m / \sec^{2}$
$v_{x} = u_{x} + a_{y} t$
$v_{x} = 0 + (- 0.5) t$
$v_{x} = - 0.5 t$
Now at $t = 3 \sec$
$v_{x} = - 0.5 \times 3 = - 1.5 m / \sec$
Velocity vector $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = - 1.5 \hat{i} + 0 \hat{j}$
$\vec{v} = - 1.5 \hat{i} m / \sec$

TS EAMCET 29.09.2020

Motion in One Dimensions

141217 An object is moving in a plane and its trajectory $\overset{―}{r} (t)$ (in meter) is given by, $\overset{―}{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 t \hat{j}$ , where $t$ is time (in second). The angle that the velocity vector makes with the $X$ -axis at $t = 0$ .

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

C Trajectory of an object-
$\vec{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 \hat{t} j$
velocity, $\vec{v} = \frac{d \vec{r}}{dt} = (6 t + 2) \hat{i} + 2 \hat{j} m / s$
$\vec{v} = \underset{v_{x}}{\underset{⏟}{(6 t + 2)}} \hat{i} + \underset{v_{y}}{\underset{⏟}{2 \hat{j}}}$
Let $θ$ is that angle which makes with $X$ -axis,
at $t = 0$
$∵ \tan θ = \frac{v_{y}}{v_{x}} = \frac{2}{6 t + 2}$
At $t = 0$ ,
$\tan θ = \frac{2}{0 + 2} = 1 = \tan 45^{\circ}$
$θ = 45^{\circ}$

TS EAMCET 29.09.2020

Motion in One Dimensions

141219 Starting from a point on circumference of a circle of radius ' $r$ ', a person moves to a point which is diametrically opposite its initial position. The distance travelled and displacement are respectively

1

2 π r, 0

2

π r, 2 r

3

2 r, 2 π r

4

2 r, π r

Explanation:

B original image
According to question let a person stand initially at point $A$ and he travels from $A$ to $B$ on circle which is diametrically opposite its initial position and has radius $r$ .
So,
Distance $(D)$ travelled by person $=$ Distance along circumference from A to B.
$D = \frac{2 π r}{2} = π r$
and Displacement $=$ Distance along diameter $(AB)$ Displacement $= AB = 2 r$

AP EAMCET-25.09.2020

Motion in One Dimensions

141220 The displacement is given by $x = 2 t^{2} + t + 5$ , the acceleration at $t = 2 s$ is

1

4 m \cdot s^{- 2}

2

8 m \cdot s^{- 2}

3

10 m \cdot s^{- 2}

4

15 m \cdot s^{- 2}

Explanation:

A Given,
Displacement $(x) = 2 t^{2} + t + 5$
Differentiating the above equation,
$v = \frac{dx}{dt} = 4 t + 1 + 0$
$\frac{dx}{dt} = 4 t + 1$
Again, differentiating the equation (i)
$a = \frac{d^{2} x}{{dt}^{2}} = 4 \times 1$
$\frac{d^{2} x}{{dt}^{2}} = a = 4 m / s^{2} at t = 2 s$
$a = 4 m / s^{2}$

AP EAMCET-07.10.2020

Motion in One Dimensions

141215 A person sitting in the ground floor of a building notices through the window of height $1.5 m$ , a ball dropped from the roof of the building crosses the window in $0.1 s$ . What is the velocity of the ball when it is at the topmost point of the window? $(g = 10 m / s^{2})$

1

15.5 m / s

2

14.5 m / s

3

4.5 m / s

4

20 m / s

Explanation:

B Given, $g = 10 m / s^{2}$
$s = 1.5 m$
$t = 0.1$ Second
From equation of motion,
$s = ut + \frac{1}{2} {gt}^{2}$
$1.5 = u (0.1) + \frac{1}{2} \times 10 \times (0.1)^{2}$
$1.5 = (0.1) u + 5 \times 0.01$
$1.5 = (0.1) u + 0.05$
$u = \frac{1.45}{0.1} = 14.5 m / s$

NEET (Oct.) 2020

Motion in One Dimensions

141216 A particle starts from the origin with an initial velocity $\vec{u} = (3.0 \hat{j}) m / s$ and a constant acceleration $\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$ . The velocity of the particle when it reaches its maximum y coordinate is:

1

2 \hat{i}

2

- 1.5 \hat{i} + 1.5 \hat{j}

3

- 1.5 \hat{i}

4

1.5 \hat{i} + 1.0 \hat{j}

Explanation:

C $\vec{u} = (3.0 \hat{j}) m / s$
$\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$
This can be treated as 2 motions
One is $x$ directions and other is $y$ direction $y$ direction-
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
Let $s_{y}$ be the displacement in $y$ -direction,
$u_{y} = 3 m / \sec, a_{y} = - 1 m / \sec^{2}$
$∴ s_{y} = 3 t - \frac{t^{2}}{2}$
$v_{y} = u_{y} + a_{y} t$
$= 3 + (- 1) t$
$= (3 - t) m / s$
y component will be maximum, when $\frac{{ds}_{y}}{dt}$ or $v_{y}$ is zero.
$v_{y} = 0 \Rightarrow 3 - t = 0$
Similarly is $x$ -direction,
$u_{x} = 0, a_{x} = - 0.5 m / \sec^{2}$
$v_{x} = u_{x} + a_{y} t$
$v_{x} = 0 + (- 0.5) t$
$v_{x} = - 0.5 t$
Now at $t = 3 \sec$
$v_{x} = - 0.5 \times 3 = - 1.5 m / \sec$
Velocity vector $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = - 1.5 \hat{i} + 0 \hat{j}$
$\vec{v} = - 1.5 \hat{i} m / \sec$

TS EAMCET 29.09.2020

Motion in One Dimensions

141217 An object is moving in a plane and its trajectory $\overset{―}{r} (t)$ (in meter) is given by, $\overset{―}{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 t \hat{j}$ , where $t$ is time (in second). The angle that the velocity vector makes with the $X$ -axis at $t = 0$ .

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

C Trajectory of an object-
$\vec{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 \hat{t} j$
velocity, $\vec{v} = \frac{d \vec{r}}{dt} = (6 t + 2) \hat{i} + 2 \hat{j} m / s$
$\vec{v} = \underset{v_{x}}{\underset{⏟}{(6 t + 2)}} \hat{i} + \underset{v_{y}}{\underset{⏟}{2 \hat{j}}}$
Let $θ$ is that angle which makes with $X$ -axis,
at $t = 0$
$∵ \tan θ = \frac{v_{y}}{v_{x}} = \frac{2}{6 t + 2}$
At $t = 0$ ,
$\tan θ = \frac{2}{0 + 2} = 1 = \tan 45^{\circ}$
$θ = 45^{\circ}$

TS EAMCET 29.09.2020

Motion in One Dimensions

141219 Starting from a point on circumference of a circle of radius ' $r$ ', a person moves to a point which is diametrically opposite its initial position. The distance travelled and displacement are respectively

1

2 π r, 0

2

π r, 2 r

3

2 r, 2 π r

4

2 r, π r

Explanation:

B original image
According to question let a person stand initially at point $A$ and he travels from $A$ to $B$ on circle which is diametrically opposite its initial position and has radius $r$ .
So,
Distance $(D)$ travelled by person $=$ Distance along circumference from A to B.
$D = \frac{2 π r}{2} = π r$
and Displacement $=$ Distance along diameter $(AB)$ Displacement $= AB = 2 r$

AP EAMCET-25.09.2020

Motion in One Dimensions

141220 The displacement is given by $x = 2 t^{2} + t + 5$ , the acceleration at $t = 2 s$ is

1

4 m \cdot s^{- 2}

2

8 m \cdot s^{- 2}

3

10 m \cdot s^{- 2}

4

15 m \cdot s^{- 2}

Explanation:

A Given,
Displacement $(x) = 2 t^{2} + t + 5$
Differentiating the above equation,
$v = \frac{dx}{dt} = 4 t + 1 + 0$
$\frac{dx}{dt} = 4 t + 1$
Again, differentiating the equation (i)
$a = \frac{d^{2} x}{{dt}^{2}} = 4 \times 1$
$\frac{d^{2} x}{{dt}^{2}} = a = 4 m / s^{2} at t = 2 s$
$a = 4 m / s^{2}$

AP EAMCET-07.10.2020

Motion in One Dimensions

141215 A person sitting in the ground floor of a building notices through the window of height $1.5 m$ , a ball dropped from the roof of the building crosses the window in $0.1 s$ . What is the velocity of the ball when it is at the topmost point of the window? $(g = 10 m / s^{2})$

1

15.5 m / s

2

14.5 m / s

3

4.5 m / s

4

20 m / s

Explanation:

B Given, $g = 10 m / s^{2}$
$s = 1.5 m$
$t = 0.1$ Second
From equation of motion,
$s = ut + \frac{1}{2} {gt}^{2}$
$1.5 = u (0.1) + \frac{1}{2} \times 10 \times (0.1)^{2}$
$1.5 = (0.1) u + 5 \times 0.01$
$1.5 = (0.1) u + 0.05$
$u = \frac{1.45}{0.1} = 14.5 m / s$

NEET (Oct.) 2020

Motion in One Dimensions

141216 A particle starts from the origin with an initial velocity $\vec{u} = (3.0 \hat{j}) m / s$ and a constant acceleration $\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$ . The velocity of the particle when it reaches its maximum y coordinate is:

1

2 \hat{i}

2

- 1.5 \hat{i} + 1.5 \hat{j}

3

- 1.5 \hat{i}

4

1.5 \hat{i} + 1.0 \hat{j}

Explanation:

C $\vec{u} = (3.0 \hat{j}) m / s$
$\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$
This can be treated as 2 motions
One is $x$ directions and other is $y$ direction $y$ direction-
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
Let $s_{y}$ be the displacement in $y$ -direction,
$u_{y} = 3 m / \sec, a_{y} = - 1 m / \sec^{2}$
$∴ s_{y} = 3 t - \frac{t^{2}}{2}$
$v_{y} = u_{y} + a_{y} t$
$= 3 + (- 1) t$
$= (3 - t) m / s$
y component will be maximum, when $\frac{{ds}_{y}}{dt}$ or $v_{y}$ is zero.
$v_{y} = 0 \Rightarrow 3 - t = 0$
Similarly is $x$ -direction,
$u_{x} = 0, a_{x} = - 0.5 m / \sec^{2}$
$v_{x} = u_{x} + a_{y} t$
$v_{x} = 0 + (- 0.5) t$
$v_{x} = - 0.5 t$
Now at $t = 3 \sec$
$v_{x} = - 0.5 \times 3 = - 1.5 m / \sec$
Velocity vector $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = - 1.5 \hat{i} + 0 \hat{j}$
$\vec{v} = - 1.5 \hat{i} m / \sec$

TS EAMCET 29.09.2020

Motion in One Dimensions

141217 An object is moving in a plane and its trajectory $\overset{―}{r} (t)$ (in meter) is given by, $\overset{―}{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 t \hat{j}$ , where $t$ is time (in second). The angle that the velocity vector makes with the $X$ -axis at $t = 0$ .

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

C Trajectory of an object-
$\vec{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 \hat{t} j$
velocity, $\vec{v} = \frac{d \vec{r}}{dt} = (6 t + 2) \hat{i} + 2 \hat{j} m / s$
$\vec{v} = \underset{v_{x}}{\underset{⏟}{(6 t + 2)}} \hat{i} + \underset{v_{y}}{\underset{⏟}{2 \hat{j}}}$
Let $θ$ is that angle which makes with $X$ -axis,
at $t = 0$
$∵ \tan θ = \frac{v_{y}}{v_{x}} = \frac{2}{6 t + 2}$
At $t = 0$ ,
$\tan θ = \frac{2}{0 + 2} = 1 = \tan 45^{\circ}$
$θ = 45^{\circ}$

TS EAMCET 29.09.2020

Motion in One Dimensions

141219 Starting from a point on circumference of a circle of radius ' $r$ ', a person moves to a point which is diametrically opposite its initial position. The distance travelled and displacement are respectively

1

2 π r, 0

2

π r, 2 r

3

2 r, 2 π r

4

2 r, π r

Explanation:

B original image
According to question let a person stand initially at point $A$ and he travels from $A$ to $B$ on circle which is diametrically opposite its initial position and has radius $r$ .
So,
Distance $(D)$ travelled by person $=$ Distance along circumference from A to B.
$D = \frac{2 π r}{2} = π r$
and Displacement $=$ Distance along diameter $(AB)$ Displacement $= AB = 2 r$

AP EAMCET-25.09.2020

Motion in One Dimensions

141220 The displacement is given by $x = 2 t^{2} + t + 5$ , the acceleration at $t = 2 s$ is

1

4 m \cdot s^{- 2}

2

8 m \cdot s^{- 2}

3

10 m \cdot s^{- 2}

4

15 m \cdot s^{- 2}

Explanation:

A Given,
Displacement $(x) = 2 t^{2} + t + 5$
Differentiating the above equation,
$v = \frac{dx}{dt} = 4 t + 1 + 0$
$\frac{dx}{dt} = 4 t + 1$
Again, differentiating the equation (i)
$a = \frac{d^{2} x}{{dt}^{2}} = 4 \times 1$
$\frac{d^{2} x}{{dt}^{2}} = a = 4 m / s^{2} at t = 2 s$
$a = 4 m / s^{2}$

AP EAMCET-07.10.2020

Motion in One Dimensions

141215 A person sitting in the ground floor of a building notices through the window of height $1.5 m$ , a ball dropped from the roof of the building crosses the window in $0.1 s$ . What is the velocity of the ball when it is at the topmost point of the window? $(g = 10 m / s^{2})$

1

15.5 m / s

2

14.5 m / s

3

4.5 m / s

4

20 m / s

Explanation:

B Given, $g = 10 m / s^{2}$
$s = 1.5 m$
$t = 0.1$ Second
From equation of motion,
$s = ut + \frac{1}{2} {gt}^{2}$
$1.5 = u (0.1) + \frac{1}{2} \times 10 \times (0.1)^{2}$
$1.5 = (0.1) u + 5 \times 0.01$
$1.5 = (0.1) u + 0.05$
$u = \frac{1.45}{0.1} = 14.5 m / s$

NEET (Oct.) 2020

Motion in One Dimensions

141216 A particle starts from the origin with an initial velocity $\vec{u} = (3.0 \hat{j}) m / s$ and a constant acceleration $\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$ . The velocity of the particle when it reaches its maximum y coordinate is:

1

2 \hat{i}

2

- 1.5 \hat{i} + 1.5 \hat{j}

3

- 1.5 \hat{i}

4

1.5 \hat{i} + 1.0 \hat{j}

Explanation:

C $\vec{u} = (3.0 \hat{j}) m / s$
$\vec{a} = (- 0.5 \hat{i} - 1.0 \hat{j}) m / s^{2}$
This can be treated as 2 motions
One is $x$ directions and other is $y$ direction $y$ direction-
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
Let $s_{y}$ be the displacement in $y$ -direction,
$u_{y} = 3 m / \sec, a_{y} = - 1 m / \sec^{2}$
$∴ s_{y} = 3 t - \frac{t^{2}}{2}$
$v_{y} = u_{y} + a_{y} t$
$= 3 + (- 1) t$
$= (3 - t) m / s$
y component will be maximum, when $\frac{{ds}_{y}}{dt}$ or $v_{y}$ is zero.
$v_{y} = 0 \Rightarrow 3 - t = 0$
Similarly is $x$ -direction,
$u_{x} = 0, a_{x} = - 0.5 m / \sec^{2}$
$v_{x} = u_{x} + a_{y} t$
$v_{x} = 0 + (- 0.5) t$
$v_{x} = - 0.5 t$
Now at $t = 3 \sec$
$v_{x} = - 0.5 \times 3 = - 1.5 m / \sec$
Velocity vector $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = - 1.5 \hat{i} + 0 \hat{j}$
$\vec{v} = - 1.5 \hat{i} m / \sec$

TS EAMCET 29.09.2020

Motion in One Dimensions

141217 An object is moving in a plane and its trajectory $\overset{―}{r} (t)$ (in meter) is given by, $\overset{―}{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 t \hat{j}$ , where $t$ is time (in second). The angle that the velocity vector makes with the $X$ -axis at $t = 0$ .

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

C Trajectory of an object-
$\vec{r} (t) = (3 t^{2} + 2 t) \hat{i} + 2 \hat{t} j$
velocity, $\vec{v} = \frac{d \vec{r}}{dt} = (6 t + 2) \hat{i} + 2 \hat{j} m / s$
$\vec{v} = \underset{v_{x}}{\underset{⏟}{(6 t + 2)}} \hat{i} + \underset{v_{y}}{\underset{⏟}{2 \hat{j}}}$
Let $θ$ is that angle which makes with $X$ -axis,
at $t = 0$
$∵ \tan θ = \frac{v_{y}}{v_{x}} = \frac{2}{6 t + 2}$
At $t = 0$ ,
$\tan θ = \frac{2}{0 + 2} = 1 = \tan 45^{\circ}$
$θ = 45^{\circ}$

TS EAMCET 29.09.2020

Motion in One Dimensions

141219 Starting from a point on circumference of a circle of radius ' $r$ ', a person moves to a point which is diametrically opposite its initial position. The distance travelled and displacement are respectively

1

2 π r, 0

2

π r, 2 r

3

2 r, 2 π r

4

2 r, π r

Explanation:

B original image
According to question let a person stand initially at point $A$ and he travels from $A$ to $B$ on circle which is diametrically opposite its initial position and has radius $r$ .
So,
Distance $(D)$ travelled by person $=$ Distance along circumference from A to B.
$D = \frac{2 π r}{2} = π r$
and Displacement $=$ Distance along diameter $(AB)$ Displacement $= AB = 2 r$

AP EAMCET-25.09.2020

Motion in One Dimensions

141220 The displacement is given by $x = 2 t^{2} + t + 5$ , the acceleration at $t = 2 s$ is

1

4 m \cdot s^{- 2}

2

8 m \cdot s^{- 2}

3

10 m \cdot s^{- 2}

4

15 m \cdot s^{- 2}

Explanation:

A Given,
Displacement $(x) = 2 t^{2} + t + 5$
Differentiating the above equation,
$v = \frac{dx}{dt} = 4 t + 1 + 0$
$\frac{dx}{dt} = 4 t + 1$
Again, differentiating the equation (i)
$a = \frac{d^{2} x}{{dt}^{2}} = 4 \times 1$
$\frac{d^{2} x}{{dt}^{2}} = a = 4 m / s^{2} at t = 2 s$
$a = 4 m / s^{2}$

AP EAMCET-07.10.2020