At time t t\,t seconds, where t≥0t \geq 0t≥0, a particle P P\,P moves so that its acceleration a ms−2\mathbf{a} \text{ ms}^{-2}a ms−2 is given by a=(1−2t)i+(5−t2)j\mathbf{a} = (1 - 2t)\mathbf{i} + (5 - t^2)\mathbf{j}a=(1−2t)i+(5−t2)j. At the instant when t=0t = 0t=0, the velocity of P P\,P is 20i ms−120\mathbf{i} \text{ ms}^{-1}20i ms−1
Show that the velocity of P P\,P at time t t\,t is (t−t2+20)i+(5t−13t3)j\displaystyle \left(t-t^2+20\right)\mathbf{i}+\left(5t-\frac{1}{3}t^3\right)\mathbf{j}(t−t2+20)i+(5t−31t3)j
Hence find the velocity of P P\,P when t=4t = 4t=4
Find the value of t t\,t at the instant when P P\,P is moving in a direction perpendicular to i\mathbf{i}i
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.