A small object starts with an initial velocity u=(−8i+2j) ms−1\mathbf{u} = (-8\mathbf{i} + 2\mathbf{j}) \text{ ms}^{-1}u=(−8i+2j) ms−1. It accelerates at a constant rate in the direction (i+j)(\mathbf{i} + \mathbf{j})(i+j). Given that the magnitude of the acceleration is 52 ms−25\sqrt{2} \text{ ms}^{-2}52 ms−2,
Show that the velocity vector v\mathbf{v}v after t t\,t seconds is given by v=[(5t−8)i+(5t+2)j] ms−1\mathbf{v} = [(5t - 8)\mathbf{i} + (5t + 2)\mathbf{j}] \text{ ms}^{-1}v=[(5t−8)i+(5t+2)j] ms−1.
Using your answer to part (a), or otherwise, find the value of t t\,t for which the speed of the particle is at its minimum.
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.