JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2020
- Medium
- Numerical
A particle is moving along the \(x-\)axis with its coordinate with the time '\(t\)' given be \(\mathrm{x}(\mathrm{t})=10+8 \mathrm{t}-3 \mathrm{t}^{2} .\) Another particle is moving the \(y-\)axis with its coordinate as a function of time given by \(\mathrm{y}(\mathrm{t})=5-8 \mathrm{t}^{3} .\) At \(\mathrm{t}=1\; \mathrm{s},\) the speed of the second particle as measured in the frame of the first particle is given as \(\sqrt{\mathrm{v}} .\) Then \(\mathrm{v}\) (in \(\mathrm{m} / \mathrm{s})\) is
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Answer & Solution
Correct Answer
(C) \(580\)
Step-by-step Solution
Detailed explanation
\(x=10+8 t-3 t^{2}\)
\(v_{x}=8-6 t\)
\(\left(\mathrm{v}_{\mathrm{x}}\right)_{\mathrm{t}=1}=2 \hat{\mathrm{i}}\)
\(y=5-8 t^{3}\)
\(\mathrm{v}_{\mathrm{y}}=-24 \mathrm{t}^{2}\)
\(\left(v_{y}\right)_{t=1}=-24 \hat{j}\)
\(\mathrm{Now}\)
\(\sqrt{\mathrm{v}}=\sqrt{(24)^{2}+(2)^{2}}=\sqrt{580}\)
\(\therefore \mathrm{v}=580 \mathrm{m}^{2} / \mathrm{s}^{2}\)
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