Question Number 13478 by Tinkutara last updated on 20/May/17
$$\mathrm{The}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is}\:{u} \\ $$$$\left(\mathrm{at}\:{t}\:=\:\mathrm{0}\right)\:\mathrm{and}\:\mathrm{acceleration}\:{f}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${f}\:=\:{at}\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time}\:\mathrm{and}\:'{a}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relations}\:\mathrm{is} \\ $$$$\mathrm{valid}? \\ $$$$\left(\mathrm{1}\right)\:{v}\:=\:{u}\:+\:{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{v}\:=\:{u}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:{v}\:=\:{u}\:+\:{at} \\ $$$$\left(\mathrm{4}\right)\:{v}\:=\:{u} \\ $$
Answered by ajfour last updated on 20/May/17
$$\frac{{dv}}{{dt}}={f}={at} \\ $$$$\Rightarrow\:\int_{{u}} ^{\:\:{v}} {dv}\:=\int_{\mathrm{0}} ^{\:\:{t}} {atdt} \\ $$$${v}−{u}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \: \\ $$$${v}={u}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:. \\ $$$$\left(\mathrm{2}\right)\:{is}\:{valid}\:. \\ $$
Commented by ajfour last updated on 20/May/17
$${is}\:{it}\:{fine}\:{now}? \\ $$
Commented by Tinkutara last updated on 20/May/17
$$\mathrm{OK}.\:\mathrm{Thanks}! \\ $$