At time t t\,t seconds a particle P P\,P has acceleration a=2ti−6j\mathbf{a}=2t\mathbf{i}-6\mathbf{j}a=2ti−6j m s−2^{-2}−2.
Initially its velocity is (3i+4j)(3\mathbf{i}+4\mathbf{j})(3i+4j) m s−1^{-1}−1 and its position vector is (i−2j)(\mathbf{i}-2\mathbf{j})(i−2j) m.
Find the velocity of P P\,P in terms of ttt.
Find the position vector of P P\,P when t=2t=2t=2.
163 exam-style questions on WJEC A Level Maths 4.8 Kinematics (A-level only), covering 4.8.1 Kinematics (A-level only), 4.8.2 Kinematics (A-level only), and 4.8.3 Kinematics (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.