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Motion in a Lane MCQs for NEET — Physics Questions with Answers

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A particle travels with speed 50 m/s from the point (3, 7) in a direction $7 \hat{i} - 24 \hat{j}$ . Find its position vector after 3 seconds.

$151 \hat{i} + 45 \hat{j}$

$45 \hat{i} - 137 \hat{j}$

$151 \hat{i} + 45 \hat{j}$

$4.5 \hat{i} - 151 \hat{j}$

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If $\vec{A} = 2 \hat{i} + \hat{j} & \vec{B} = \hat{i} - \hat{j}$ . Find component of $\vec{A}$ along $\vec{B}$ & perpendicular to $\vec{B}$ .

$\frac{\hat{i} - \hat{j}}{2}, \frac{3}{2} (\hat{i} + \hat{j})$

$\frac{\hat{i} - \hat{j}}{2}, - \frac{2}{3} (\hat{i} + \hat{j})$

$\frac{\hat{i} - \hat{j}}{2}, - \frac{3}{2} (\hat{i} - \hat{j})$

$\frac{\hat{i} - \hat{j}}{2}, \frac{2}{3} (\hat{i} - \hat{j})$

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Given: $\vec{a} + \vec{b} + \vec{c} = 0$ . Out of the three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ two are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. The angles between the vectors are:

90 $°$ , 135 $°$ , 135 $°$

30 $°$ , 60 $°$ , 90 $°$

45 $°$ , 45 $°$ , 90 $°$

45 $°$ , 60 $°$ , 90 $°$

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Vector $\vec{A}$ is of length 2 cm and is 60 $°$ above the x-axis in the first quadrant. Vector $\vec{B}$ is of length 2 cm and 60 $°$ below the x-axis in the fourth quadrant. The sum $\vec{A}$ + $\vec{B}$ is a vector of magnitude -

2 along + y-axis

2 along + x-axis

1 along - x-axis

2 along - x-axis

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Explanation

Vector A has a magnitude of 2 cm at an angle of 60Â° above the x-axis. Vector B has the same magnitude of 2 cm but at 60Â° below the x-axis. Their sum A + B will be a vector along the positive x-axis with a magnitude of 2âˆš3 â‰ˆ 3.46 cm.

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A particle moves in the XY plane and at time t is at the point whose coordinates are $(t^{2}, t^{3} - 2 t)$ . Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

1/3 sec

2/3 sec

3/2 sec

never

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Explanation

(B)

$\vec{r} = t^{2} \hat{i} + (t^{3} - 2 t) \hat{j} \vec{v} = \frac{d \vec{r}}{dt} = 2 t \hat{i} + (3 t^{2} - 2) \hat{j} \vec{a} = 2 \hat{i} + (6 t) \hat{j} When \vec{v} ⊥ \vec{a}, \vec{v} . \vec{a} = 0 [2 t \hat{i} + (3 t^{2} - 2) \hat{j}] . [2 \hat{i} + 6 t \hat{j}] = 0 4 t + 6 t (3 t^{2} - 2) = 0 2 t [2 + 3 (3 t^{2} - 2)] = 0 t = 0 or 2 + 9 t^{2} - 6 = 0 t^{2} = \frac{4}{9} or t = \frac{2}{3} \sec$

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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 sec. The equation of the path is

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

$x = y + 2$

$x = y^{2} + 2$

$x = y^{2} + y + 2$

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Explanation

$v_{x} = 8 t - 2 v_{y} = 2 \frac{dx}{dt} = 8 t - 2 \frac{dy}{dt} = 2 \int dx = \int (8 t - 2) dt \int dy = \int 2 dt x = 4 t^{2} - 2 t + c y = 2 t + c_{1} At t = 2 s, x = 14 At t = 2 s, y = 4 14 = 4 {(2)}^{2} - 2 (2) + c 4 = 2 (2) + c_{1} c = 2 c_{1} = 0 x = 4 t^{2} - 2 t + 2 y = 2 t Putting t = \frac{y}{2} in x x = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2 x = y^{2} - y + 2$

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Two forces ${\vec{F}}_{1} = 2 \hat{i} + 2 \hat{j} N$ and ${\vec{F}}_{2} = 3 \hat{j} + 4 \hat{k} N$ are acting on a particle.

The resultant force acting on particle is:

$2 \hat{i} + 5 \hat{j} + 4 \hat{k}$

$2 \hat{i} - 5 \hat{j} - 4 \hat{k}$

$\hat{i} - 3 \hat{j} - 2 \hat{k}$

$\hat{i} - \hat{j} - \hat{k}$

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Explanation

Resultant force = $\vec{F_{1}} + \vec{F_{2}} = 2 \hat{i} + 5 \hat{j} + 4 \hat{k}$

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Six particles situated at the corners of a regular hexagon of side a move at constant speed v. Each particle maintains a direction towards the particle at the next. The time which the particle will take to meet each other is

$\frac{2 a}{v} \sec$

$\frac{a}{v} \sec$

$\frac{2 a}{3 v} \sec$

$\frac{3 a}{v} \sec$

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The position of a particle moving in the xy-plane at any time t is given by $x = (3 t^{2} - 6 t)$ metres, $y = (t^{2} - 2 t)$ metres. Select the correct statement about the moving particle from the following

The acceleration of the particle is zero at t = 0 second

The velocity of the particle is zero at t = 0 second

The velocity of the particle is zero at t = 1 second

The velocity and acceleration of the particle are never zero

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Explanation

$v_{x} = \frac{d x}{d t} = \frac{d}{d t} (3 t^{2} - 6 t) = 6 t - 6$ . At $t = 1, v_{x} = 0$

$v_{y} = \frac{d y}{d t} = \frac{d}{d t} (t^{2} - 2 t) = 2 t - 2$ . At $t = 1, v_{y} = 0$

Hence $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 0$

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A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the x-axis and $8 m / s^{2}$ along the y-axis. Its distance from the origin after 4 seconds will be

56 m

64 m

80 m

128 m

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Explanation

$S_{x} = u_{x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow S_{x} = \frac{1}{2} \times 6 \times 16 = 48 m$

$S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow S_{y} = \frac{1}{2} \times 8 \times 16 = 64 m$

$S = \sqrt{S_{x}^{2} + S_{y}^{2}} = 80 m$

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