Question 11:
easy
A particle moves along a straight line with velocity given by \( v = (6 – 3t) \) where \( v \) is in \( \text{m/s} \) and \( t \) in seconds. Determine when the particle returns to its starting point.
Displacement is \( S = \int v \, dt = \int_0^t (6 - 3t) \, dt = 6t - 1.5t^2 \). Returning to the starting point means \( S = 0 ⇒ 6t - 1.5t^2 = 0 ⇒ t = 4\text{ s} \).