Kinematics

JEE Main level, numerical answer, medium

147 of 308

Question 147 of 308

A particle is moving along the xx -axis with its coordinate with time tt given by x(t)=10+8t−3t2.x\left(t\right)=10+8t-3{t}^{2}. Another particle is moving along the yy -axis with its coordinate as a function of time given by y(t)=5−8t3.y\left(t\right)=5-8{t}^{3}. At t=1s,t=1s, the speed of the second particle as measured in the frame of the first particle is given as v.\sqrt{v}. Then v(inms−1)v\left(inm{s}^{-1}\right) is ___________.

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Answer: 580

v1⃗=(8−6t)i^\vec{{v}_{1}}=\left(8-6t\right)\hat{i}
v2⃗=(−24t2)j^\vec{{v}_{2}}=\left(-24{t}^{2}\right)\hat{j}
∣v21⃗∣=580\left|\vec{{v}_{21}}\right|=\sqrt{580}

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