A particle moves in a straight line with an initial velocity of 5 m s−15 \text{ m s}^{-1}5 m s−1.
The acceleration a m s−2a \text{ m s}^{-2}a m s−2 of the particle at time ttt seconds is given by
a=6kt2−4kt+2 a = 6kt^2 - 4kt + 2 a=6kt2−4kt+2where kkk is a constant.
When t=2t = 2t=2, the velocity of the particle is 13 m s−113 \text{ m s}^{-1}13 m s−1.
Show that k=12k = \frac{1}{2}k=21.