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PHXI04:MOTION IN A PLANE

361809 The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2}$ and $y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s$ is

1

5 m / s^{2}

2

- 4 m / s^{2}

3

- 8 m / s^{2}

4

0

Explanation:

$v_{x} = 5 - 4 t, v_{y} = 10$
$a_{x} = - 4, a_{y} = 0$
$\vec{a} = a_{x} \hat{i} + a_{y} \hat{j}$
$\vec{a} = - 4 \hat{i} m / s^{2}$

NEET - 2017

PHXI04:MOTION IN A PLANE

361810 A particle begins accelerating from rest (at the origin) with a constant acceleration $a = 2 \hat{i} - 4 \hat{j}$ (where $a$ has the units $m / s^{2}$ ). How far is the particle from the origin at time $t = 1 s$ ?

1

1 m

2

2 m

3

5 m

4 None of these

Explanation:

$\Rightarrow \vec{r} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$
$\Rightarrow \vec{r} = 0 + \frac{1}{2} (2 \hat{i} - 4 \hat{j}) (1^{2}) = \hat{i} - 2 \hat{j}$
$\Rightarrow | \vec{r} | = \sqrt{1^{2} + 2^{2}} = \sqrt{5} m$
Displacement from origin is $\sqrt{5} m$ .

PHXI04:MOTION IN A PLANE

361811 A particle’s velocity changes from $(2 \hat{i} + 3 \hat{j}) m / s$ in to $(3 \hat{i} - 2 \hat{j}) m / s$ in 2 $s$ . If its mass is 1 $k g$ , the acceleration $(m / s^{2})$ is

1

- (\hat{i} + 5 \hat{j})

2

(\hat{i} + 5 \hat{j}) / 2

3

Z e r o

4

(\hat{i} - 5 \hat{j}) / 2

Explanation:

$a = \frac{\vec{v_{2}} - \vec{v_{1}}}{t} = \frac{(3 \hat{i} - 2 \hat{j}) - (2 \hat{i} + 3 \hat{j})}{2}$
$= \frac{\hat{i} - 5 \hat{j}}{2}$

PHXI04:MOTION IN A PLANE

361812 A bead is free to slide down on a smooth wire rightly stretched between points $A$ and $B$ on a vertical circle of radius 10 $m$ . Find the time taken by the bead to reach point $B$ , if the bead slides from rest from the highest point $A$ on the circle.
supporting img

1

1 s

2

5 s

3

2 s

4

7 s

Explanation:

$A B = 2 R \cos θ$ $\begin{aligned} A B = \frac{1}{2} g \cos θ t^{2} \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2} \\ 2 \sqrt{\frac{R}{g}} = t \Rightarrow 2 \sqrt{\frac{10}{10}} = t = 2 s \end{aligned}$

PHXI04:MOTION IN A PLANE

361809 The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2}$ and $y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s$ is

1

5 m / s^{2}

2

- 4 m / s^{2}

3

- 8 m / s^{2}

4

0

Explanation:

$v_{x} = 5 - 4 t, v_{y} = 10$
$a_{x} = - 4, a_{y} = 0$
$\vec{a} = a_{x} \hat{i} + a_{y} \hat{j}$
$\vec{a} = - 4 \hat{i} m / s^{2}$

NEET - 2017

PHXI04:MOTION IN A PLANE

361810 A particle begins accelerating from rest (at the origin) with a constant acceleration $a = 2 \hat{i} - 4 \hat{j}$ (where $a$ has the units $m / s^{2}$ ). How far is the particle from the origin at time $t = 1 s$ ?

1

1 m

2

2 m

3

5 m

4 None of these

Explanation:

$\Rightarrow \vec{r} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$
$\Rightarrow \vec{r} = 0 + \frac{1}{2} (2 \hat{i} - 4 \hat{j}) (1^{2}) = \hat{i} - 2 \hat{j}$
$\Rightarrow | \vec{r} | = \sqrt{1^{2} + 2^{2}} = \sqrt{5} m$
Displacement from origin is $\sqrt{5} m$ .

PHXI04:MOTION IN A PLANE

361811 A particle’s velocity changes from $(2 \hat{i} + 3 \hat{j}) m / s$ in to $(3 \hat{i} - 2 \hat{j}) m / s$ in 2 $s$ . If its mass is 1 $k g$ , the acceleration $(m / s^{2})$ is

1

- (\hat{i} + 5 \hat{j})

2

(\hat{i} + 5 \hat{j}) / 2

3

Z e r o

4

(\hat{i} - 5 \hat{j}) / 2

Explanation:

$a = \frac{\vec{v_{2}} - \vec{v_{1}}}{t} = \frac{(3 \hat{i} - 2 \hat{j}) - (2 \hat{i} + 3 \hat{j})}{2}$
$= \frac{\hat{i} - 5 \hat{j}}{2}$

PHXI04:MOTION IN A PLANE

361812 A bead is free to slide down on a smooth wire rightly stretched between points $A$ and $B$ on a vertical circle of radius 10 $m$ . Find the time taken by the bead to reach point $B$ , if the bead slides from rest from the highest point $A$ on the circle.
supporting img

1

1 s

2

5 s

3

2 s

4

7 s

Explanation:

$A B = 2 R \cos θ$ $\begin{aligned} A B = \frac{1}{2} g \cos θ t^{2} \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2} \\ 2 \sqrt{\frac{R}{g}} = t \Rightarrow 2 \sqrt{\frac{10}{10}} = t = 2 s \end{aligned}$

PHXI04:MOTION IN A PLANE

361809 The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2}$ and $y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s$ is

1

5 m / s^{2}

2

- 4 m / s^{2}

3

- 8 m / s^{2}

4

0

Explanation:

$v_{x} = 5 - 4 t, v_{y} = 10$
$a_{x} = - 4, a_{y} = 0$
$\vec{a} = a_{x} \hat{i} + a_{y} \hat{j}$
$\vec{a} = - 4 \hat{i} m / s^{2}$

NEET - 2017

PHXI04:MOTION IN A PLANE

361810 A particle begins accelerating from rest (at the origin) with a constant acceleration $a = 2 \hat{i} - 4 \hat{j}$ (where $a$ has the units $m / s^{2}$ ). How far is the particle from the origin at time $t = 1 s$ ?

1

1 m

2

2 m

3

5 m

4 None of these

Explanation:

$\Rightarrow \vec{r} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$
$\Rightarrow \vec{r} = 0 + \frac{1}{2} (2 \hat{i} - 4 \hat{j}) (1^{2}) = \hat{i} - 2 \hat{j}$
$\Rightarrow | \vec{r} | = \sqrt{1^{2} + 2^{2}} = \sqrt{5} m$
Displacement from origin is $\sqrt{5} m$ .

PHXI04:MOTION IN A PLANE

361811 A particle’s velocity changes from $(2 \hat{i} + 3 \hat{j}) m / s$ in to $(3 \hat{i} - 2 \hat{j}) m / s$ in 2 $s$ . If its mass is 1 $k g$ , the acceleration $(m / s^{2})$ is

1

- (\hat{i} + 5 \hat{j})

2

(\hat{i} + 5 \hat{j}) / 2

3

Z e r o

4

(\hat{i} - 5 \hat{j}) / 2

Explanation:

$a = \frac{\vec{v_{2}} - \vec{v_{1}}}{t} = \frac{(3 \hat{i} - 2 \hat{j}) - (2 \hat{i} + 3 \hat{j})}{2}$
$= \frac{\hat{i} - 5 \hat{j}}{2}$

PHXI04:MOTION IN A PLANE

361812 A bead is free to slide down on a smooth wire rightly stretched between points $A$ and $B$ on a vertical circle of radius 10 $m$ . Find the time taken by the bead to reach point $B$ , if the bead slides from rest from the highest point $A$ on the circle.
supporting img

1

1 s

2

5 s

3

2 s

4

7 s

Explanation:

$A B = 2 R \cos θ$ $\begin{aligned} A B = \frac{1}{2} g \cos θ t^{2} \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2} \\ 2 \sqrt{\frac{R}{g}} = t \Rightarrow 2 \sqrt{\frac{10}{10}} = t = 2 s \end{aligned}$

PHXI04:MOTION IN A PLANE

361809 The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2}$ and $y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s$ is

1

5 m / s^{2}

2

- 4 m / s^{2}

3

- 8 m / s^{2}

4

0

Explanation:

$v_{x} = 5 - 4 t, v_{y} = 10$
$a_{x} = - 4, a_{y} = 0$
$\vec{a} = a_{x} \hat{i} + a_{y} \hat{j}$
$\vec{a} = - 4 \hat{i} m / s^{2}$

NEET - 2017

PHXI04:MOTION IN A PLANE

361810 A particle begins accelerating from rest (at the origin) with a constant acceleration $a = 2 \hat{i} - 4 \hat{j}$ (where $a$ has the units $m / s^{2}$ ). How far is the particle from the origin at time $t = 1 s$ ?

1

1 m

2

2 m

3

5 m

4 None of these

Explanation:

$\Rightarrow \vec{r} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$
$\Rightarrow \vec{r} = 0 + \frac{1}{2} (2 \hat{i} - 4 \hat{j}) (1^{2}) = \hat{i} - 2 \hat{j}$
$\Rightarrow | \vec{r} | = \sqrt{1^{2} + 2^{2}} = \sqrt{5} m$
Displacement from origin is $\sqrt{5} m$ .

PHXI04:MOTION IN A PLANE

361811 A particle’s velocity changes from $(2 \hat{i} + 3 \hat{j}) m / s$ in to $(3 \hat{i} - 2 \hat{j}) m / s$ in 2 $s$ . If its mass is 1 $k g$ , the acceleration $(m / s^{2})$ is

1

- (\hat{i} + 5 \hat{j})

2

(\hat{i} + 5 \hat{j}) / 2

3

Z e r o

4

(\hat{i} - 5 \hat{j}) / 2

Explanation:

$a = \frac{\vec{v_{2}} - \vec{v_{1}}}{t} = \frac{(3 \hat{i} - 2 \hat{j}) - (2 \hat{i} + 3 \hat{j})}{2}$
$= \frac{\hat{i} - 5 \hat{j}}{2}$

PHXI04:MOTION IN A PLANE

361812 A bead is free to slide down on a smooth wire rightly stretched between points $A$ and $B$ on a vertical circle of radius 10 $m$ . Find the time taken by the bead to reach point $B$ , if the bead slides from rest from the highest point $A$ on the circle.
supporting img

1

1 s

2

5 s

3

2 s

4

7 s

Explanation:

$A B = 2 R \cos θ$ $\begin{aligned} A B = \frac{1}{2} g \cos θ t^{2} \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2} \\ 2 \sqrt{\frac{R}{g}} = t \Rightarrow 2 \sqrt{\frac{10}{10}} = t = 2 s \end{aligned}$

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