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.
163 exam-style questions on WJEC A Level Maths 4.8 Kinematics (A-level only), covering 4.8.1 Kinematics (A-level only), 4.8.2 Kinematics (A-level only), and 4.8.3 Kinematics (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.