AP EAMCET · PHYSICS · Motion In One Dimension
The speed of a particle changes from \(\sqrt{5} \mathrm{~ms}^{-1}\) to \(2 \sqrt{5} \mathrm{~ms}^{-1}\) in a time \(t\). If the magnitude of change in its velocity is \(5 \mathrm{~ms}^{-1}\), the angle between the initial and final velocities of the particle is
- A \(30^{\circ}\)
- B \(45^{\circ}\)
- C \(60^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(D) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
Given, \(v_i=\sqrt{5} \mathrm{~ms}^{-1}, v_f=2 \sqrt{5} \mathrm{~ms}^{-1}\) and \[ \Delta v=5 \mathrm{~ms}^{-1} \] Since, both \(v_i\) and \(v_f\) are extreme speeds, i.e. at \(t=0\) and \(t=t\). So, they can be considered as magnitude of the velocities at time, \(t=0\) and…
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