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Question Number 114145 by ZiYangLee last updated on 17/Sep/20
A particle moves along a straight line such that its velocity, v m s^(−1) , is given by v=t^3 −4t^2 +3t, where t is time, in seconds, after passing through fixed point O. Find the total distance, in m, travelled by the particle until the particle returned to the fixed point O for the second time.
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{its}\:\mathrm{velocity},\:{v}\:\mathrm{m}\:\mathrm{s}^{−\mathrm{1}} ,\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by}\:{v}={t}^{\mathrm{3}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}{t},\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time},\:\mathrm{in}\: \\ $$$$\mathrm{seconds},\:\mathrm{after}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{fixed} \\ $$$$\mathrm{point}\:\mathrm{O}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{total}\:\mathrm{distance},\:\mathrm{in}\:\mathrm{m},\:\mathrm{travelled} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{until}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{returned} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{fixed}\:\mathrm{point}\:\mathrm{O}\:\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{time}. \\ $$
Answered by mr W last updated on 17/Sep/20
s(t)=∫_0 ^t vdt=(t^4 /4)−((4t^3 )/3)+((3t^2 )/2) s(t)=(t^4 /4)−((4t^3 )/3)+((3t^2 )/2) s_(max) is when v=0, t^3 −4t^2 +3t=0 (t^2 −4t+3)t=0 (t−1)(t−3)t=0 t=0 t=1 t=3 at t=1: s_(max) =(1^4 /4)−((4×1^3 )/3)+((3×1^2 )/2)=(5/(12)) total distance travelled: s=2×s_(max) =(5/6) m
$${s}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} {vdt}=\frac{{t}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${s}\left({t}\right)=\frac{{t}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${s}_{{max}} \:{is}\:{when}\:{v}=\mathrm{0}, \\ $$$${t}^{\mathrm{3}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}{t}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{3}\right){t}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}−\mathrm{3}\right){t}=\mathrm{0} \\ $$$${t}=\mathrm{0} \\ $$$${t}=\mathrm{1} \\ $$$${t}=\mathrm{3} \\ $$$${at}\:{t}=\mathrm{1}:\:{s}_{{max}} =\frac{\mathrm{1}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{4}×\mathrm{1}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{3}×\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${total}\:{distance}\:{travelled}: \\ $$$${s}=\mathrm{2}×{s}_{{max}} =\frac{\mathrm{5}}{\mathrm{6}}\:{m} \\ $$
Commented by ZiYangLee last updated on 17/Sep/20
oops sorry the answer is 5.335m instead
$$\mathrm{oops}\:\mathrm{sorry}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{5}.\mathrm{335m}\:\mathrm{instead} \\ $$
Commented by mr W last updated on 17/Sep/20
it′s upon what is meant with “return to point O for the second time”. what i meant is that the object leaves point O and comes back. but if it means that the object leaves point O and comes back and leaves again and comes back again, then the result is: at t=3: s_(min) =(3^4 /4)−((4×3^3 )/3)+((3×3^2 )/2)=−(9/4) ⇒total distance moved: 2×(5/(12))+2×(9/4)=((16)/3)=5.333 m
$${it}'{s}\:{upon}\:{what}\:{is}\:{meant}\:{with}\:“{return} \\ $$$${to}\:{point}\:{O}\:{for}\:{the}\:{second}\:{time}''. \\ $$$${what}\:{i}\:{meant}\:{is}\:{that}\:{the}\:{object}\:{leaves} \\ $$$${point}\:{O}\:{and}\:{comes}\:{back}.\:{but}\:{if}\:{it} \\ $$$${means}\:{that}\:{the}\:{object}\:{leaves}\:{point}\:{O} \\ $$$${and}\:{comes}\:{back}\:{and}\:{leaves}\:{again}\:{and} \\ $$$${comes}\:{back}\:{again},\:{then}\:{the}\:{result}\:{is}: \\ $$$${at}\:{t}=\mathrm{3}:\:{s}_{{min}} =\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{4}×\mathrm{3}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{3}×\mathrm{3}^{\mathrm{2}} }{\mathrm{2}}=−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$ \\ $$$$\Rightarrow{total}\:{distance}\:{moved}: \\ $$$$\mathrm{2}×\frac{\mathrm{5}}{\mathrm{12}}+\mathrm{2}×\frac{\mathrm{9}}{\mathrm{4}}=\frac{\mathrm{16}}{\mathrm{3}}=\mathrm{5}.\mathrm{333}\:{m} \\ $$
Commented by mr W last updated on 17/Sep/20

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