Asked by Anonymous
A particle moves along the x-axis with velocity v(t) = t^2 − 1, with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = 2 seconds.
Answers
Answered by
Steve
clearly that would be
∫[0,2] t^2-1 dt = 2/3
But that is the algebraic distance (final displacement), since it starts out with negative velocity. To get the actual distance traveled, you need to separate the two parts:
∫[0,1] -(t^2-1) dt + ∫[1,2] t^2-1 dt = 2/3 + 4/3 = 2
∫[0,2] t^2-1 dt = 2/3
But that is the algebraic distance (final displacement), since it starts out with negative velocity. To get the actual distance traveled, you need to separate the two parts:
∫[0,1] -(t^2-1) dt + ∫[1,2] t^2-1 dt = 2/3 + 4/3 = 2
Answered by
Nagaroth
uh my answer is .667
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