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.
Determine the velocity vector of P P\,P as a function 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.
158 exam-style questions on OCR (MEI) A Level Maths 3.2 Kinematics, covering 3.2.1 Language of kinematics, 3.2.2 Position, displacement, distance and distance travelled, 3.2.3 Velocity, speed and acceleration distinctions, 3.2.4 Draw and interpret kinematics graphs, 3.2.5 Differentiate position and velocity (A-level only), 3.2.6 Integrate acceleration and velocity (A-level only), 3.2.7 When constant acceleration formulae apply, 3.2.8 Solve 1-D kinematics problems, 3.2.9 Language of kinematics in 2 dimensions (A-level only), 3.2.10 Extend 1-D techniques to 2-D using vectors (A-level only), 3.2.11 Cartesian equation of a path (A-level only), and 3.2.12 Vectors to solve kinematics problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.