JEE Mains · Physics · STD 11 - 3.2 motion in plane
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
- A \(441\)
- B \(600\)
- C \(580\)
- D \(484\)
Answer & Solution
Correct Answer
(C) \(580\)
Step-by-step Solution
Detailed explanation
\(x=10+8 t-3 t^{2}\) \(v_{x}=8-6 t\) \(\left(\mathrm{v}_{\mathrm{x}}\right)_{\mathrm{t}=1}=2 \hat{\mathrm{i}}\) \(y=5-8 t^{3}\) \(\mathrm{v}_{\mathrm{y}}=-24 \mathrm{t}^{2}\) \(\left(v_{y}\right)_{t=1}=-24 \hat{j}\) \(\mathrm{Now}\)…
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