Question

A particle moves in the xy-plane so that its velocity vector at time t is \(v(t)=\langle t^2, \sin(\pi t)\rangle\) and the particle's position vector at time \(t = 0\) is \(\langle 1,0 \rangle\). What is the position vector of the particle when t = 3?

Answer 1

Integrate \(v(t)\) to get our position vector \(s(t)\).$$s(t)=\int(t^2,\sin\pi t)\,dt=\left(\frac13t^3+C_x,-\frac1\pi\cos\pi t+C_y\right)$$Now we merely verify solve for \(C_x,C_y\) so that our initial condition holds:$$s(0)=\left(0+C_x,-\frac1\pi+C_y\right)=\left(0,1\right)$$so \(C_x=0,C_y=1+\frac1{\pi}\pi\)

The question is then \(s(3)=\left(\frac133^3,-\frac1\pi\cos3\pi+1+\frac1\pi\right)=\left(9,\frac2\pi+1\right)\)