A particle moves in the direction AB with velocity v m/s given by v = 4t - 4, where t is the time....?

Question

A particle moves in the direction AB with velocity v m/s given by v = 4t - 4, where t is the time in seconds after passing the point B.
(a) If AB = 9m, find the displacement, sm, from A in terms of t.
(b) Explain why initially the particle is moving in the direction BA.
(c) Find the total distance travelled from t=o to t=3.

Answer 1

(a) Since ds/dt = v, we integrate to find s(t) = 2t^2 - 4t + C. At t = 0, s = 9, so C = 9 and s(t) = 2t^2 - 4t + 9.

(b) Because at t = 0, the velocity is negative.

(c) As noted above, v(0) < 0, so the particle is moving from B to A, and continues until v = 0 (at t = 1). Then v becomes positive and the particle is moving away from A. Therefore, the total distance is [s(1) - s(0)] + [s(9) - s(1)] = 17.