Question Number 14291 by Tinkutara last updated on 30/May/17
$$\mathrm{Is}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{correct}\:\mathrm{for}\:\mathrm{uniformly} \\ $$$$\mathrm{accelerated}\:\mathrm{motion}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{line}? \\ $$$${s}\:=\:{vt}\:−\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{No}....\:\mathrm{that}\:\mathrm{is}\:\mathrm{for}\:\mathrm{motion}\:\mathrm{under}\:\mathrm{gravity} \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\:+ \\ $$
Commented by Tinkutara last updated on 30/May/17
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{says}\:\mathrm{it}\:\mathrm{is}\:\mathrm{correct}.\:\mathrm{I}\:\mathrm{am} \\ $$$$\mathrm{posting}\:\mathrm{full}\:\mathrm{question}. \\ $$
Answered by prakash jain last updated on 30/May/17
$${v}={u}+{at} \\ $$$${vt}−\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} ={ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 30/May/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$