Discussion
A particle moves along a semi circular path of radius \(r\) in time \(t\) with constant speed, then its
Answer: E
Its average acceleration is \(\frac{2\pi r}{t^{2}}\).
The speed is \(v = \frac{\pi r}{t}\) (half the circumference in time t). At the end of the semicircle the velocity has reversed, so the change in velocity has magnitude 2v.
\[ \begin{aligned} a_{avg} &= \frac{|\Delta \vec{v}|}{t} = \frac{2v}{t} \\ &= \frac{2\pi r}{t^{2}} \end{aligned} \]
The others are wrong: the average speed is \(\frac{\pi r}{t}\), not \(\frac{\pi r}{2t}\); the displacement is 2r (the diameter), so the average velocity is \(\frac{2r}{t}\); and the distance is \(\pi r\), not 2r.
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