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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?

The initial velocity of the particle is 4.

The acceleration of the particle is 2a.

The particle is at the origin at t=0.

None of these

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Explanation

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

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The maximum value of $7 + 6 x - 9 x^{2}$ is

-8

-4

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Explanation

$\frac{dx}{dt} = 6 - 18 x = 0 \Rightarrow x = \frac{1}{3} \frac{d^{2} x}{{dt}^{2}} = - 18 < 0 . So x = \frac{1}{3} is a point of maxima . Maximum value of the function is at x = \frac{1}{3} .$

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If $f (x) = x^{2} - 2 x + 4$ , then f(x) has

a minimum at x=1

a maximum at x=1

no extreme point

no minimum

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Explanation

$f (x) = x^{2} - 2 x + 4 \frac{df}{dx} = 2 x - 2 At x = 1, \frac{df}{dx} = 0 Differentiating again, \frac{d^{2} f}{{dx}^{2}} = 2 At x = 1, \frac{d^{2} f}{{dx}^{2}} > 0 So, x = 1 is a point of minima$

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A particle is moving along x-axis. The velocity v of particle varies with its position x as $v = \frac{1}{x}$ . Find velocity of particle as a function of time t given that at t=0, x=1 .

$v = \sqrt{2 t + 1}$

$v = \frac{1}{\sqrt{2 t + 1}}$

$v = \frac{1}{t}$

None of these

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Explanation

$v = \frac{1}{x} \frac{d x}{d t} = \frac{1}{x}$

$x d x = d t \int_{1}^{x} x d x = \int_{0}^{t} d t \frac{x^{2} - 1}{2} = t x = \sqrt{2 t + 1} v = \frac{d x}{d t} = \frac{1}{\sqrt{2 t + 1}}$

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A vector $\vec{A}$ is directed along $30 °$ west of north direction and another vector $\vec{B}$ along $15 °$ south of east. Their resultant cannot be in ____________ direction.

North

East

North-East

South

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Explanation

The resultant of two vectors A and B will be in a direction between the directions of A and B. Since A is directed 30Â° west of north and B is directed 15Â° south of east, their resultant cannot be in the south direction.

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ABCD is a quadrilateral. Forces $\vec{BA}, \vec{BC}, \vec{CD} & \vec{DA}$ act at a point. Their resultant is

$2 \vec{AB}$

$2 \vec{DA}$

zero vector

$2 \vec{BA}$

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The resultant of the forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ . If $\vec{Q}$ is double then the resultant also doubles in magnitude. Find the angle between $\vec{P}$ and $\vec{Q}$ .

$\cos θ = \frac{Q}{2 P}$

$\cos θ = \frac{- 4 Q}{3 P}$

$\cos θ = \frac{- 2 Q}{3 P}$

$\cos θ = \frac{- 3 P}{4 Q}$

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Explanation

(D)

$Resultant of \vec{P} & \vec{Q} is \vec{R} R^{2} = P^{2} + Q^{2} + 2 PQ cosθ . . . . . (i) If Q is doubled, resultant is 2 R {(2 R)}^{2} = P^{2} + {(2 Q)}^{2} + 2 P (2 Q) cosθ . . . . . . (ii) From (i) & (ii) 4 (P^{2} + Q^{2} + 2 PQ cosθ) = P^{2} + 4 Q^{2} + 4 PQ cosθ 4 P^{2} + 8 PQ cosθ = P^{2} + 4 QP cosθ 4 Q cosθ = - 3 P cosθ = \frac{- 3 P}{4 Q}$

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The maximum and minimum magnitude of the resultant of two vectors are 17 units and 7 units respectively. Then the magnitude of resultant of the vectors when they act perpendicular to each other is:

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Explanation

When two vectors are perpendicular, their resultant magnitude is given by the Pythagorean theorem: âˆš(a^2 + b^2), where a and b are the magnitudes of the vectors. If the maximum and minimum resultants are 17 and 7 units, then âˆš(17^2 + 7^2) = 13 units.

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A vector $\vec{A}$ makes an angle of $20 °$ and $\vec{B}$ makes an angle of $110 °$ with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the magnitude of the resultant.

$1$

$5 \sqrt{2}$

$5$

$7$

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Explanation

Angle between

$\vec{A} a n d \vec{B} = 90 ° R = \sqrt{A^{2} + B^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = 5 m$

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A particle is moving westward with a velocity ${\vec{v}}_{1} = 5 m / s .$ Its velocity changed to ${\vec{v}}_{2} = 5 m / s$ northward. The change in velocity vector $(∆ \vec{V} = {\vec{v}}_{2} - {\vec{v}}_{1})$ is:

$5 \sqrt{2} m / s$ towards north east

$5 m / s$ towards north west

zero

$5 \sqrt{2} m / s$ towards north west

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