_______________ A particle moves according to a law of motion s = f ( t ), t les
ID: 2867154 • Letter: #
Question
_______________
A particle moves according to a law of motion s = f(t), t less than or equal to 0, where t is measured in seconds and s in feet.
f(t) = f(t) = 0.01t4 - 0.06t3
(a) Find the velocity at time t (in ft/s).
v(t) = __________
(b) What is the velocity after 4 s?
v(4) = __________ ft/s
(c) When is the particle at rest?
(d) When is the particle moving in the positive direction? (Enter your answer using interval notation.)
_____________
When, for
0 less than or equal to t < infinite,
is the particle speeding up? (Enter your answer using interval notation.)
_______________
When, for
0 less than or equal to t < infinite,
is it slowing down? (Enter your answer using interval notation.)
_________________
(e) Find the total distance traveled during the first 11 s. (Round your answer to two decimal places.)
______________ ft
(f) Find the acceleration at time t (in ft/s2).
a(t) = _____________
Find the acceleration after 4 s.
a(4) = _____________ ft/s2
t = _________ s (smaller value) t = _________ s (larger value)Explanation / Answer
f(t) = f(t) = 0.01t4 - 0.06t3
v(t) = df/dt = 0.04t3 - 0.18t2
a). v(t) = df/dt = 0.04t3 - 0.18t2 Ans.
b). v(4) = -0.32 ft/s Ans.
c). v(t) = 0 = 0.04t3 - 0.18t2
hence 0.04t = 0.18
t = 4.5 sec.Ans. is same for both
d). 0.04t3 - 0.18t2 must be greater than 0
i.e. after time 4.5 sec velocity will be positive.
e). total distance traveled during the first 11 s.
f(t) = 0.01t4 - 0.06t3 = 0.01(11)4 - 0.06(11)3 = 66.55 ft ans.
f). Acceleration at time t (in ft/s2).
v(t) = 0.04t3 - 0.18t2
a(t) = dv/dt = 0.12t2 - 0.36t Ans.
a(4) = 0.12(4)2 - 0.36*4 = 0.48 ft/s2 Ans.
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