A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by \(a = (3t – 2) ms^{-2}\). Find the other time when the velocity would be zero.
- \(\frac{1}{3} seconds\)
- \(\frac{3}{4} seconds\)
-
\(\frac{4}{3} seconds\)
- \(2 seconds\)
The correct answer is: C
Explanation
Given \(a(t) = 3t - 2\), \(v(t) = \int (a(t)) \mathrm {d} t\)
= \(\int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)
\(\frac{3}{2}t^{2} - 2t = 0 \implies t(\frac{3}{2}t - 2) = 0\)
\(t = \text{0 or t} = \frac{3}{2}t - 2 = 0 \implies t = \frac{4}{3}\)
The time t = 0 was the starting point, The next time v = 0 m/s is at \(t = \frac{4}{3} seconds\).