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Motion in a Lane NEET Questions — Free MCQ Practice

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If $\vec{A} = \vec{B} + \vec{C}$ and magnitudes of A, B and C are 5, 4 and 3 units respectively, the angle between A and C is -

A.

$\sin^{- 1} (3 / 4)$

B.

$\cos^{- 1} (4 / 5)$

C.

$\cos^{- 1} (3 / 5)$

D.

$π / 2$

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What is the resultant of three coplanar forces: $300 N$ at $0 °$ , $400 N$ at $30 °$ and $400 N$ at $150 °$ ?

A.

$500 N$

B.

$700 N$

C.

$1100 N$

D.

$300 N$

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Two forces, $F_{1}$ and $F_{2}$ are acting on a body. One force is doubled of the other force and the resultant is equal to the greater force. Then the angle between the two forces is -

A.

$\cos^{- 1} (1 / 2)$

B.

$\cos^{- 1} (- 1 / 2)$

C.

$\cos^{- 1} (- 1 / 4)$

D.

$\cos^{- 1} (1 / 4)$

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Explanation

(C)

$F_{2} = 2 F_{1} and R = F_{2} R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cosθ} Squaring both sides {(F_{2})}^{2} = F_{1}^{2} + {(2 F_{1})}^{2} + 2 F_{1} (2 F_{1}) cosθ {(2 F_{1})}^{2} = 5 F_{1}^{2} + 4 F_{1}^{2} cosθ Solving, - F_{1}^{2} = 4 F_{1}^{2} cosθ cosθ = \frac{- 1}{4} θ = \cos^{- 1} (\frac{- 1}{4})$

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What displacement must be added to the displacement $25 \hat{i} - 6 \hat{j} m$ to give a displacement of $7.0 m$ pointing in the x-direction?

A.

$18 \hat{i} - 6 \hat{j}$

B.

$32 \hat{i} - 13 \hat{j}$

C.

$- 18 \hat{i} + 6 \hat{j}$

D.

$- 25 \hat{i} + 13 \hat{j}$

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Explanation

$L e t t h e r e q u i r e d v e c t o r b e x \hat{i} + y \hat{j} S o, 25 \hat{i} + - 6 \hat{j} + (x \hat{i} + y \hat{j}) = 7 \hat{i} x \hat{i} + y \hat{j} = - 18 \hat{i} + 6 \hat{j} C o m p a r i n g, x = - 18, y = 6$

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The value of unit vector in the direction of vector $A = 5 \hat{i} - 12 \hat{j}$ , is -

A.

$\hat{i}$

B.

$\hat{j}$

C.

$(\hat{i} + \hat{j}) / 13$

D.

$(5 \hat{i} - 12 \hat{j}) / 13$

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Explanation

$T h e u n i t v e c t o r = \frac{\vec{A}}{| \vec{A} |} = \frac{5 \hat{i} - 12 \hat{j}}{\sqrt{{(5)}^{2} + {(12)}^{2}}} = \frac{1}{13} (5 \hat{i} - 12 \hat{j})$

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A man rows a boat with a speed of $18 km / hr$ in northwest direction. The shoreline makes an angle of $15 °$ south of west. Obtain the component of the velocity of the boat along the shoreline.

A.

$9 km / hr$

B.

$18 \frac{\sqrt{3}}{2} km / hr$

C.

$18 \cos 15 ° km / hr$

D.

$18 \cos 75 ° km / hr$

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A bird moves from point (1, -2) to (4, 2). If the speed of the bird is 10 m/sec, then the velocity vector of the bird is:

A.

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

B.

$5 (4 \hat{i} + 2 \hat{j})$

C.

$0.6 \hat{i} + 0.8 \hat{j}$

D.

$6 \hat{i} + 8 \hat{j}$

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Explanation

$I n i t i a l p o s i t i o n (x_{1}, y_{1}) = (1, - 2) F i n a l p o s i t i o n (x_{2}, y_{2}) = (4, 2) D i s p l a c e m e n t = F i n a l p o s i t i o n - I n i t i a l p o s i t i o n = (x_{2} - x_{1}, y_{2} - y_{1}) = (3, 4) = 3 \hat{i} + 4 \hat{j} D i r e c t i o n o f v e l o c i t y i s s a m e a s d i p l a c e m e n t S o u n i t v e c t o r a l o n g d i s p l a c e m e n t = \frac{V e c t o r}{M a g n i t u d e} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^{2} + 4^{2}}} = \frac{3 \hat{i} + 4 \hat{j}}{5} V e l o c i t y = M a g n i t u d e o f v e l o c i t y \times U n i t V e c t o r a l o n g v e l o c i t y = 10 \times (\frac{3 \hat{i} + 4 \hat{j}}{5}) = 6 \hat{i} + 8 \hat{j}$

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Angle that the vector $A = 2 \hat{i} + 3 \hat{j}$ makes with y-axis is -

A.

$\tan^{- 1} 3 / 2$

B.

$\tan^{- 1} 2 / 3$

C.

$\sin^{- 1} 2 / 3$

D.

$\cos^{- 1} 3 / 2$

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Explanation

The angle that a vector makes with the y-axis is given by tan^-1(y/x). Here, the vector A = 2i + 3j has x = 2 and y = 3. Therefore, the angle it makes with the y-axis is tan^-1(3/2).

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A body is moving according to the equation $x = at + {bt}^{2} - {ct}^{3}$ where x = displacement and a, b and c are constants. The acceleration of the body is

A.

$a + 2 bt$

B.

$2 b + 6 ct$

C.

$2 b - 6 ct$

D.

$3 b - 6 {ct}^{2}$

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Explanation

Diffrentiating x wrt t,

$v = \frac{d x}{d t} = a + 2 b t - 3 c t^{2}$

$a = 2 b - 6 c t$

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A particle moves along x-axis as $x = 4 (t - 2) + a {(t - 2)}^{2}$

Which of the following is true?

A.

The initial velocity of the particle is 4.

B.

The acceleration of the particle is 2a.

C.

The particle is at the origin at t=0.

D.

None of these

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Explanation

$x = 4 (t - 2) + a {(t - 2)}^{2} At t = 0, x = - 8 + 4 a v = \frac{dx}{dt} = 4 + 2 a (t - 2) At t = 0, v_{in} = 4 - 4 a a = \frac{dv}{dt} = 2 a$

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

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