The velocity function (in meters per second) is given for a particle moving along a line.

Question

v(t) = t² − 2t − 15,    1 ≤ t ≤ 6

(a) Find the displacement.

(b) Find the distance traveled by the particle during the given time interval.

Answer 1

t^2 -2t -15 = (t-5)(t+3)

t=5 is a turning point where the particle changes direction

s(t)= integral of v(t) = t^3/3 -t^2 -15t

its displacement is s(6)-s(1) = 72-36-90 - (1/3-1-15) = - 54 - (-16 1/3) = -37 2/3

take the absolute value = 37 2/3 meters

total distance traveled is more, as it counts back & forth motion

it's going backward from t=1 to t=5

then forward from t=5 to t=6

s(5) = 125/3 -25- 75 = - 175/3

total movement = 2(s(1)-s(5)) + (s(6)-s(5)) = 2(126/3) + (13/3) = 265/3 = 88 1/3 meters