A particle is moving along the x-axis with its coordinate with the… | JEE Main 2020 (8 Jan, Shift 1)

Q8JEE Main 2020 · 8 JAN · SHIFT 1Medium

Motion in a Plane (2-D Kinematics)

A particle is moving along the x-axis with its coordinate with the time 't' given be x(t) = 10 + 8t – 3t2. Another particle is moving the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t3.

At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$ . Then v (in m/s) is ______.

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

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Answer: 580

First particle: $v_1 = \frac{dx}{dt} = 8 - 6t = 2$ m/s along $x$ at $t = 1$ s
Second particle: $v_2 = \frac{dy}{dt} = -24t^2 = -24$ m/s along $y$ at $t = 1$ s
$\vec v_{21} = \vec v_2 - \vec v_1 = -2\hat i - 24\hat j$
$|\vec v_{21}|^2 = 4 + 576 = 580$ , so $v = 580$ m/s (in the sense $\sqrt v$ m/s).

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