Asked by revan
The velocity function is v(t)=t2−3t+2 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [−3,6].
displacement = ??
distance traveled = ??
displacement = ??
distance traveled = ??
Answers
Answered by
oobleck
v(t) = t^2 - 3t + 2
displacement is
s = ∫[-3,6] (t^2 - 3t + 2) dt = 117/2
distance is
∫[-3,6] |t^2 - 3t + 2| dt = 353/6
displacement is
s = ∫[-3,6] (t^2 - 3t + 2) dt = 117/2
distance is
∫[-3,6] |t^2 - 3t + 2| dt = 353/6