A particle has an initial velocity of (i−3j) ms−1(\mathbf{i} - 3\mathbf{j}) \text{ ms}^{-1}(i−3j) 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 5 ms−2\sqrt{5} \text{ ms}^{-2}5 ms−2. It has been shown that, after t t\,t seconds, the velocity vector of the particle is [(2t+1)i+(t−3)j] ms−1[(2t + 1)\mathbf{i} + (t - 3)\mathbf{j}] \text{ ms}^{-1}[(2t+1)i+(t−3)j] ms−1,
Using the velocity vector given above, or otherwise, find the value of t t\,t for which the speed of the particle is at its minimum.
363 exam-style questions on Edexcel A Level Maths Further Kinematics, covering 8.1 Vectors in Kinematics, 8.2 Vector Methods with Projectiles, 8.3 Variable Acceleration in One Dimension, 8.4 Differentiating Vectors, and 8.5 Integrating Vectors. Each one has a worked solution and a mark scheme showing where the marks go.