The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t2 − t − 110, 1 ≤ t ≤ 12

Question

The velocity function, in feet per second, is given for a particle moving along a straight line.

v(t) = t² − t − 110, 1 ≤ t ≤ 12

A)Find the displacement.

B)Find the total distance that the particle travels over the given interval.

Answer 1

v=t^2 -t-110 v 1<t<12

s= integral of v = t^3/3 -t^2/2 -110t

s(1) = 1/3 -1/2 -110 = -110 1/6

s(12) = 12^3/3 - 12^2/2 -110(12) = -888

displacement = s(12)-s(1) = -777 5/6 feet

s(11) = -887 1/3

it moved 2/3 forward, from t=11 to t=12

v(11) = 0 when it changed direction

it moved 777 5/6 backwards from t=1 to t=11

add backwards to forwards distances =

total distance traveled = 777 5/6 + 2/3 = 778 1/2 feet

it's moving backwards from time t=1 to t=11

then forwards from t=11 to t=12

Answer 2

A) Displaycement s= ∫₁¹²(t² - t - 110)dt = (t³/3 - t²/2 - 110t)₁¹² = - 816 +110.1667 = - 705.83

B) Total Distance S = ∫₁¹²It² - t - 110Idt = 727.5