JEE PYQ: Motion In Two Dimensions Question 12
Question 12 - 2020 (08 Jan Shift 1)
A particle is moving along the $x$-axis with its coordinate with time ‘$t$’ given by $x(t) = 10 + 8t - 3t^2$. Another particle is moving along the $y$-axis with its coordinate as a function of time given by $y(t) = 5 - 8t^3$. At $t = 1$ s, 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 ________.
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Answer: 580
Solution
At $t = 1$: $\vec{v}_A = 2\hat{i}$, $\vec{v}_B = -24\hat{j}$. Relative velocity $= -2\hat{i} - 24\hat{j}$. Speed $= \sqrt{4 + 576} = \sqrt{580}$, so $v = 580$.
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