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The function s = f(t) gives the position of a moving object 9) A particle moves according to a law of motion s = f(t) fort > 0 where t is measured in seconds and s in feet. f(t) = 13 - 912 + 150 (a) Find the velocity at time t. v(t) = (b) What is the velocity after 3 seconds ? (3) = (c) When is the particle at rest? (d) When is the particle moving in the positive direction? (e) Find the total distance traveled during the first 10 seconds (0) Find the acceleration at timet. (h) When, for Osts 10 when is the particle speeding up or slowing down ? Explain.

Answer 1

alt) = a Cult) = ( 3+ 18+ +15) alt) = Gt-18 ® alt) = 0 - ot=18 ita = 3] tttt at t=1 alt) = 601-18 = - 12 LO at t=u alt) = 6(4)-18= 6 • speeding up slowing down 0 tasy (3,10) toxal (0,3)

Given f(t) = t3 at +15t Veloci VCD - @ velocity v(t) = 2 [+1)] [t-4+ +157 [VIE) = 31² - 18 + +15 at +15t de 1(3) = 3(3)- 1813) +15 ve 3) = -127mis vlt) = 0 (particle is at rest) - 3t² 18t+15=0 2² - 6t +5=0 - 2²-St-t+5=0 t(t-5)-1(t-5)=0 > (t-1)(7-5) = 0 - at t=1, t= 5 particle is at rest. ---- ++++ + ++ 10 at t=0.5 110:51 = 3 (0.5) - 18 (0.5) +15= 6.750 0.

at t=2 v (2) = 3 (2) – 18 (2) +15= -920 at t=5 1667= 3(6) 18 (6) +15= 1570 Therefore, the particle moving in positive direction on (0,1), (5, 10). Total distance travelled + fitldt 5. t-18tt = 'vt at a l Fillat + f Feldt SGR=18+15) ot - J&t=1515)dte + Spez notar)dt the most be entity = 13.9013150) (0529657715 (5) - (033_9003 271500) + (10) 2.(10)715010) – (53290)150 18+ 15 % -18t+15t + - 7- (-25 - 7) + (250- (-2 = 7+32 +250 +25 = 314