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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




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.
Aparticlemovesalongastraightlinesuchthatitsvelocity,vms1,isgivenbyv=t34t2+3t,wheretistime,inseconds,afterpassingthroughfixedpointO.Findthetotaldistance,inm,travelledbytheparticleuntiltheparticlereturnedtothefixedpointOforthesecondtime.
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(t)=0tvdt=t444t33+3t22s(t)=t444t33+3t22smaxiswhenv=0,t34t2+3t=0(t24t+3)t=0(t1)(t3)t=0t=0t=1t=3att=1:smax=1444×133+3×122=512totaldistancetravelled:s=2×smax=56m
Commented by ZiYangLee last updated on 17/Sep/20
oops sorry the answer is 5.335m instead
oopssorrytheansweris5.335minstead
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
itsuponwhatismeantwithreturntopointOforthesecondtime.whatimeantisthattheobjectleavespointOandcomesback.butifitmeansthattheobjectleavespointOandcomesbackandleavesagainandcomesbackagain,thentheresultis:att=3:smin=3444×333+3×322=94totaldistancemoved:2×512+2×94=163=5.333m
Commented by mr W last updated on 17/Sep/20

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