A particle starts from origin at t = 0 with a velocity 5i m/s and moves in x - y plane under action of a force which produces a constant acceleration of (3i + 2j) m/s². If the x -coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is √(α) m/s. The value of α is _________. Numerical answer type. Work it out, then open the answer.

Answer: 673. x = 5t + 32t² = 84 ⇒ 3t² + 10t - 168 = 0 ⇒ t = 6 s vx = 5 + 3(6) = 23 m/s, vy = 2(6) = 12 m/s v² = 529 + 144 = 673 So α = 673.

A particle starts from origin at t = 0 with a velocity 5i m/s and… | JEE Main 2024 (27 Jan, Shift 1)

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Motion in a Plane (2-D Kinematics)

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

Numerical answer type. Work it out, then open the answer.

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Answer: 673

$x = 5t + \frac32t^2 = 84 \Rightarrow 3t^2 + 10t - 168 = 0 \Rightarrow t = 6$ s
$v_x = 5 + 3(6) = 23$ m/s, $v_y = 2(6) = 12$ m/s
$v^2 = 529 + 144 = 673$
So $\alpha = 673$ .

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