Solution:
Displacement is \(s = \int v \, dt = 6t - 1.5t^2\). To return to the starting point, \(s = 0 \implies 6t - 1.5t^2 = 0 \implies t = 4\text{ s}\).
A particle moves along a straight line with velocity given by v = (6 – 3t) where v is in m/s and t in seconds. Determine when the particle returns to its starting point.
Leave a Reply