A particle has an initial velocity of (2i−3j) ms−1(2\mathbf{i} - 3\mathbf{j}) \text{ ms}^{-1}(2i−3j) ms−1 and is accelerating uniformly in the direction (i+2j)(\mathbf{i} + 2\mathbf{j})(i+2j) where i\mathbf{i}i and j\mathbf{j}j are perpendicular unit vectors. Given that the magnitude of the acceleration is 25 ms−22\sqrt{5} \text{ ms}^{-2}25 ms−2,
show that the acceleration vector of the particle is (2i+4j) ms−2(2\mathbf{i} + 4\mathbf{j}) \text{ ms}^{-2}(2i+4j) ms−2.
show that, after t t\,t seconds, the velocity vector of the particle is [(2t+2)i+(4t−3)j] ms−1[(2t + 2)\mathbf{i} + (4t - 3)\mathbf{j}] \text{ ms}^{-1}[(2t+2)i+(4t−3)j] ms−1.
Using your answer to part (b), 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.