A particle moves along a straight line with velocity defined by v(t)=t^2-3t-18, where 0≤t≤6 (in meters per second). Find the displacement at time t and the total distance traveled up to t=6.

Question

Suppose that a particle moves along a straight line with velocity defined by

, where  (in meters per second). Find the displacement at time  and the total distance traveled up to .

Answer 1

t^2 -3t -18 = (t-6)(t+3) = v(t)

s(t) = integral of v(t) = t^3/3 - 3t^2/2 - 18t + so

0<t<6

evaluate the integral from 0 to 6

6^3/3 -3(6^2)/2 -18(6) + so - so

= 54 meters

t^2 -3t -18 =0

(t-6)(t+3) = 0

t =-3, 6 are times when the direction changes

neither affect the partical in the problem, as it always moves backwards from t=0 to t=6

it stops moving at t=6

then goes forwards for t>6

it's moving forwards up to time t=-3

then stops at time t=-3

then moves backswards when -3 < t < 6

then moves forward when t>6

total distance traveled = +54 meters backwards

total displacement = negative or -54 meters