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Question Number 14246 by tawa tawa last updated on 30/May/17

Commented by ajfour last updated on 30/May/17

Commented by ajfour last updated on 30/May/17

$$a=\frac{v-u}{\mathrm{\Delta }t}=\frac{25-30}{3}=-\frac{5}{3}m/s^2$$$$lettotaltimetostopbeT,then$$$$from\mathit{v}=\mathit{u}+\mathit{a}\mathit{t}$$$$0=30-\frac{5}{3}T\Rightarrow T=18s$$$$andfrom:\mathit{s}=\mathit{u}\mathit{t}+\frac{1}{2}{\mathit{a}\mathit{t}}^2$$$$distance=30\times 18-\left(\frac{1}{2}\right)\left(\frac{5}{3}\right){\left(18\right)}^2$$$$=540-270$$$$=270m.$$

Commented by tawa tawa last updated on 30/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by ajfour last updated on 30/May/17

Commented by ajfour last updated on 30/May/17

$${\left(2A\right)}^2+A^2=64$$$$5A^2=64$$$$A=\frac{8}{\sqrt{5}}.$$

Commented by tawa tawa last updated on 30/May/17

$$\mathrm{Thanks}\mathrm{for}\mathrm{everytime}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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