Given the particle's acceleration is
[tex]\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath[/tex]
with initial velocity
[tex]\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath[/tex]
and starting at the origin, so that
[tex]\vec r(0) = \vec 0[/tex]
you can compute the velocity and position functions by applying the fundamental theorem of calculus:
[tex]\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du[/tex]
[tex]\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du[/tex]
We have
• velocity at time t :
[tex]\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}[/tex]
• position at time t :
[tex]\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}[/tex]
