Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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) ]

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

Commented by Tinkutara last updated on 24/Jun/17

Answered by ajfour last updated on 24/Jun/17

△(v^2 )∝ s  ⇒  constant acceleration  a= ((v_f ^2 −v_i ^2 )/(2s)) = ((900−2500)/(2(400−100))) =−(8/3)m/s^2  .

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

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com