Question Number 170411 by Mastermind last updated on 23/May/22
$${The}\:{position}\:{of}\:{a}\:{body}\:{at}\:{time}\:{t}\:\left({in}\right. \\ $$$$\left.{seconds}\right)\:{is}\:{given}\:{by}\:{s}={t}^{\mathrm{3}} −\mathrm{4},\:{what}\:{is} \\ $$$${the}\:{velocity}\:{and}\:{acceleration}\:{of}\:{the} \\ $$$${object}\:{at}\:{time}\:{t}=\mathrm{5}\:? \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by alephzero last updated on 23/May/22
$$\mathrm{velocity}\:=\:\frac{{ds}}{{dt}} \\ $$$$\mathrm{acceleration}\:=\:\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} } \\ $$$$\frac{{ds}}{{dt}}\:=\:\frac{{d}\left({t}^{\mathrm{3}} −\mathrm{4}\right)}{{dt}}\:=\:\frac{{d}\left({t}^{\mathrm{3}} \right)}{{dt}}−\frac{{d}\left(\mathrm{4}\right)}{{dt}}\:= \\ $$$$=\:\mathrm{3}{t}^{\mathrm{2}} −\mathrm{0}\:=\:\mathrm{3}{t}^{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} }\:=\:\frac{{d}\left(\frac{{ds}}{{dt}}\right)}{{dt}}\:=\:\frac{{d}\left(\mathrm{3}{t}^{\mathrm{2}} \right)}{{dt}}\:=\:\mathrm{3}\frac{{d}\left({t}^{\mathrm{2}} \right)}{{dt}}\:=\:\mathrm{3}×\mathrm{2}{t} \\ $$$$=\:\mathrm{6}{t} \\ $$$$\mathrm{velocity}\:\mathrm{at}\:\mathrm{time}\:\mathrm{5}\:=\:\frac{{ds}}{{dt}}\mid_{\mathrm{5}} \:=\:\mathrm{3}\left(\mathrm{5}\right)^{\mathrm{2}} \:= \\ $$$$=\:\mathrm{3}×\mathrm{25}\:=\:\mathrm{75} \\ $$$$\mathrm{acceleration}\:\mathrm{at}\:\mathrm{time}\:\mathrm{5}\:=\:\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} }\mid_{\mathrm{5}} \:=\:\mathrm{6}×\mathrm{5} \\ $$$$=\:\mathrm{30} \\ $$
Commented by Mastermind last updated on 24/May/22
$${Thanks} \\ $$