At time t t\,t seconds a particle P P\,P has acceleration a=[(3t2−4)i+(2t2+4t)j]\mathbf{a} = \left[(3t^2 - 4)\mathbf{i} + (2t^2 + 4t)\mathbf{j}\right]a=[(3t2−4)i+(2t2+4t)j] m s−2^{-2}−2.
When t=3t = 3t=3 the velocity of P P\,P is (10i+20j)(10\mathbf{i} + 20\mathbf{j})(10i+20j) m s−1^{-1}−1.
Find an expression for the velocity of P P\,P in terms of ttt.
Initially P P\,P is at the point with position vector (2i−5j)(2\mathbf{i} - 5\mathbf{j})(2i−5j) m. Find an expression for the position vector of P P\,P in terms of ttt.
265 exam-style questions on AQA A Level Maths 3.2 Q: Kinematics, covering 3.2.1 Language of kinematics, 3.2.2 Graphs in kinematics, 3.2.3 Constant acceleration formulae, 3.2.4 Calculus in kinematics (A-level only), and 3.2.5 Motion under gravity and projectiles (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.