Question

A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 21 t^2 + 60 t where s is measured in feet and t in seconds.

a) Find the velocity (in ft/sec) of the particle at time t=0:

The particle stops moving (i.e. is in a rest) twice, once when t=A and again when t=B where A < B.
b) A is
c) B is

d)What is the position of the particle at time 14?

e)Finally, what is the TOTAL distance the particle travels between time 0 and time 14?

Answer 1

s(t)= 2t^3 - 21 t^2 + 60 t

s'(t) = 6t²-42t+60 = 6(t-5)(t-2)

a) at time t = 0

s'(t)= v(t) = 6t²-42t+60 = 60 m/s

b) v(t) = when t = 2 or 5

so A = 2

c) B = 5

d) S(14) = 2(14)³ - 21(14)²+60(14) = 2212 m

e) as distance and displacement are not equal it is bit trickier to find the distance

D = ∫|v(t)|dt

so from above we can see it is possitive before 2 and negative in [2, 5] and positive when greater than 5

so

D = (S(t))_0² - (S(t))_2⁵ + (S(t))_5¹⁴

D = S(2)-S(0) - S(5)+S(2)+S(14)-S(5)

D = S(14)+2S(2)-S(0)-2S(5)

as we know S(14) = 2212

S(2) = 52

S(0) = 0

S(5) = 25

substituting we get

D = 2266m