Question

Given the position vector of the particle
r(t)=(t+1)i+(t^2−1)j+2t k, find the particle's velocity and acceleration vectors at t=1

Answer 1

With position vector

$\vec r(t)=(t+1)\,\vec\imath+(t^2-1)\,\vec\jmath+2t\,vec k$

the particle then has velocity

$\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\vec\imath+2t\,\vec\jmath+2\,vec k$

and acceleration

$\vec a(t)=\dfrac{\mathrm d\vec v(t)}{\mathrm dt}=\dfrac{\mathrm d^2\vec r(t)}{\mathrm dt^2}=2\,\vec\jmath$

Then $t=1$, then particle's velocity and acceleration are, respectively,

$\vec v=\vec\imath+2\vec\jmath+2\,\vec k$

and

$\vec a=2\,\vec\jmath$