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Question

A particle starts from origin at $t = 0$ with a velocity $5\hat{i}$ m/s and moves in x-y plane under action of a force which produces a constant acceleration of $(3\hat{i} + 2\hat{j})\ 	ext{m/s}^2$. If the x-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is $\sqrt{\alpha}$ m/s. The value of $\alpha$ is ______. A). 205 B). 450 C). 673 D). 340

Edvaya · Accepted Answer

The correct answer is 673. Position equations:
$x = 5t + \dfrac{1}{2}(3)t^2 = 5t + 1.5t^2$
$y = \dfrac{1}{2}(2)t^2 = t^2$
When $x = 84$ m:
$1.5t^2 + 5t - 84 = 0 \Rightarrow 3t^2 + 10t - 168 = 0$
Using the quadratic formula:
$t = \dfrac{-10 + \sqrt{100 + 2016}}{6} = \dfrac{-10 + \sqrt{2116}}{6} = \dfrac{-10 + 46}{6} = 6\ 	ext{s}$
Velocity components at $t = 6$ s:
$v_x = 5 + 3(6) = 23\ 	ext{m/s}$
$v_y = 0 + 2(6) = 12\ 	ext{m/s}$
Speed:
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{529 + 144} = \sqrt{673}\ 	ext{m/s}$
Therefore $\alpha = 673$.

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Physics > Mechanics > Motion In A Straight Line And Plane

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