Question Number 16621 by Tinkutara last updated on 24/Jun/17
![A body moves in a straight line with a velocity whose square decreases linearly with displacement between two points A and B as shown. Determine the acceleration of the particle. [Answer: (8/3) ms^(−2) ]](https://www.tinkutara.com/question/Q16621.png)
$$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:\mathrm{whose}\:\mathrm{square}\:\mathrm{decreases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{displacement}\:\mathrm{between} \\ $$$$\mathrm{two}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}.\:\left[\boldsymbol{\mathrm{Answer}}:\:\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$
Commented by Tinkutara last updated on 24/Jun/17
Answered by ajfour last updated on 24/Jun/17
$$\bigtriangleup\left(\mathrm{v}^{\mathrm{2}} \right)\propto\:\mathrm{s}\:\:\Rightarrow\:\:\mathrm{constant}\:\mathrm{acceleration} \\ $$$$\mathrm{a}=\:\frac{\mathrm{v}_{\mathrm{f}} ^{\mathrm{2}} −\mathrm{v}_{\mathrm{i}} ^{\mathrm{2}} }{\mathrm{2s}}\:=\:\frac{\mathrm{900}−\mathrm{2500}}{\mathrm{2}\left(\mathrm{400}−\mathrm{100}\right)}\:=−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 24/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$