A particle has an initial velocity of (i−5j) ms−1(\mathbf{i} - 5\mathbf{j}) \text{ ms}^{-1}(i−5j) ms−1 and is accelerating uniformly in the direction (2i+j)(2\mathbf{i} + \mathbf{j})(2i+j) where i\mathbf{i}i and j\mathbf{j}j are perpendicular unit vectors. Given that the magnitude of the acceleration is 35 ms−23\sqrt{5} \text{ ms}^{-2}35 ms−2,
show that, after t t\,t seconds, the velocity vector of the particle is [(6t+1)i+(3t−5)j] ms−1[(6t + 1)\mathbf{i} + (3t - 5)\mathbf{j}] \text{ ms}^{-1}[(6t+1)i+(3t−5)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.