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.
Show that the velocity of P P\,P is v=(t3−4t−5)i+(23t3+2t2−16)j\mathbf{v}=\left(t^3-4t-5\right)\mathbf{i}+\left(\dfrac{2}{3}t^3+2t^2-16\right)\mathbf{j}v=(t3−4t−5)i+(32t3+2t2−16)j m s−1^{-1}−1.
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.