COMEDK · Physics · 4. Motion In Two Dimensions
A particle is moving eastward with a velocity \(5 \mathrm{~ms}^{-1}\). In \(10 \mathrm{~s}\) the velocity changes to \(5 \mathrm{~ms}^{-1}\) northwards. The average acceleration in this time is
- A \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\), towards North-West
- B \(\frac{1}{2} \mathrm{~ms}^{-2}\), towards North-West
- C \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\), towards North-East
- D \(\frac{1}{2} \mathrm{~ms}^{-2}\), towards North-East
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\), towards North-West
Step-by-step Solution
Detailed explanation
Given, \(u=5 \mathrm{~ms}^{-1}\), Eastwards \(v=5 \mathrm{~ms}^{-1}\), Northwards and \(\Delta t=10 \mathrm{~s}\) Change in velocity \(|\Delta v|=\sqrt{u^{2}+v^{2}}\) \[ =\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~ms}^{-1} \] Direction of \(\Delta v\),…
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