Question Number 13475 by Tinkutara last updated on 20/May/17
$$\mathrm{An}\:\mathrm{object}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{acceleration}\:\mathrm{4}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} ,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{travelled}\:\mathrm{by}\:\mathrm{object}\:\mathrm{in}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{half}\:\mathrm{second}. \\ $$
Answered by ajfour last updated on 20/May/17
$${s}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:\:{with}\:\boldsymbol{{u}}=\mathrm{0} \\ $$$$\mathrm{5}^{\boldsymbol{{th}}} \:{half}\:{second}\:{is}\:{from} \\ $$$${t}=\mathrm{2}{s}\:\rightarrow\:{t}=\frac{\mathrm{5}}{\mathrm{2}}{s} \\ $$$${s}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)\left(\mathrm{4}\right)=\mathrm{8}{m} \\ $$$${s}_{\mathrm{5}/\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)\left(\frac{\mathrm{25}}{\mathrm{4}}\right)=\mathrm{12}.\mathrm{5}{m} \\ $$$${s}_{\mathrm{5}/\mathrm{2}} −{s}_{\mathrm{2}} =\left(\mathrm{12}.\mathrm{5}−\mathrm{8}\right){m}\:=\:\mathrm{4}.\mathrm{5}{m}\:\:. \\ $$
Commented by Tinkutara last updated on 20/May/17
$$\mathrm{Thanks}! \\ $$