Question

A particle is moving in the xy plane so that its velocity vector at time t is v(t) = and the particle's position vector at time t=0 is <1,0>. What is the position vector of the particle when t=3?

Answer 1

the position vector is\[r(t) = r(0) + \int\limits_{0}^{t} v(t) dt\]\[r(3) = <1,0> + \int\limits_{0}^{3} <t^2, \sin \left( \pi t \right)> dt\]

Answer 2

but what would the final answer be

Answer 3

evaluate the integral and calculate it by yourself

Answer 4

i got 2t, 2cos (pi t) after integrating

Answer 5

\[r(3) = <1,0> + \left(<\frac{1}{3}t^3,-\frac{1}{\pi }\cos \left( \pi t \right)>\right)\Biggr|_0^3\]I have integrate it for you, just plug the limits of integration and calculate