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Question Number 16621 by Tinkutara last updated on 24/Jun/17

$$\mathrm{A}\mathrm{body}\mathrm{moves}\mathrm{in}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{with}\mathrm{a}$$$$\mathrm{velocity}\mathrm{whose}\mathrm{square}\mathrm{decreases}$$$$\mathrm{linearly}\mathrm{with}\mathrm{displacement}\mathrm{between}$$$$\mathrm{two}\mathrm{points}A\mathrm{and}B\mathrm{as}\mathrm{shown}.$$$$\mathrm{Determine}\mathrm{the}\mathrm{acceleration}\mathrm{of}\mathrm{the}$$$$\mathrm{particle}.[\mathbf{Answer}:\frac{8}{3}{\mathrm{ms}}^{-2}]$$

Commented by Tinkutara last updated on 24/Jun/17

Answered by ajfour last updated on 24/Jun/17

$$△\left({\mathrm{v}}^2\right)\propto \mathrm{s}\Rightarrow \mathrm{constant}\mathrm{acceleration}$$$$\mathrm{a}=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_{\mathrm{i}}^2}{2\mathrm{s}}=\frac{900-2500}{2(400-100)}=-\frac{8}{3}\mathrm{m}/{\mathrm{s}}^2.$$

Commented by Tinkutara last updated on 24/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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