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Question Number 171332 by Tawa11 last updated on 12/Jun/22

A bicycle has a constant velocity of 10 m/s. A person starts from rest and runs to catch up to the bicycle in 30 s. What is the acceleration of the person?\n

Commented byTawa11 last updated on 12/Jun/22

$$\mathrm{I}\mathrm{work}\mathrm{towards}\mathrm{answer}\mathrm{to}\mathrm{get}$$ $$300=450\mathrm{a},\mathrm{a}=\frac{2}{3}\mathrm{m}/{\mathrm{s}}^2.$$ $$\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{know}\mathrm{why}\mathrm{the}\mathrm{bicycle}\mathrm{time}\mathrm{too}\mathrm{is}30\sec .$$ $$\mathrm{Since}\mathrm{the}\mathrm{question}\mathrm{did}\mathrm{not}\mathrm{indicate}\mathrm{equal}\mathrm{time}.$$ $$$$ $$\mathrm{For}\mathrm{bicycle}:\mathrm{s}=\mathrm{vt}=10\left(30\right)=300\mathrm{m}$$ $$\mathrm{For}\mathrm{person}:\mathrm{s}=\frac{1}{2}{\mathrm{at}}^2=\frac{1}{2}\times \mathrm{a}\times {30}^2=\frac{1}{2}\times \mathrm{a}\times 900=450\mathrm{a}$$ $$\mathrm{Equal}\mathrm{distance}$$ $$300=450\mathrm{a}$$

Commented byTawa11 last updated on 12/Jun/22

$$\mathrm{Or}\mathrm{the}\mathrm{question}\mathrm{means}\mathrm{that}\mathrm{they}\mathrm{both}\mathrm{start}\mathrm{from}\mathrm{the}\mathrm{beginning}.$$ $$\mathrm{And}\mathrm{the}\mathrm{person}\mathrm{still}\mathrm{be}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{position}\mathrm{as}\mathrm{the}\mathrm{bicycle}\mathrm{after}30\mathrm{seconds}?$$

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