Question

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

(a)the vector position and velocity at any time t

Answer 1

Given the particle's acceleration is

`\vec a(t) = \left(3(\rm m)/(\mathrm s^2)\right)\vec\jmath`

with initial velocity

`\vec v(0) = \left(5(\rm m)/(\rm s)\right)\,\vec\imath`

and starting at the origin, so that

`\vec r(0) = \vec 0`

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

`\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du`

`\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du`

We have

• velocity at time t :

`\vec v(t) = \left(5(\rm m)/(\rm s)\right)\,\vec\imath + \displaystyle \int_0^t \left(3(\rm m)/(\mathrm s^2)\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5(\rm m)/(\rm s)\right)\,\vec\imath + \left(3(\rm m)/(\mathrm s^2)\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5(\rm m)/(\rm s)\right)\,\vec\imath + \left(3(\rm m)/(\mathrm s^2)\right)t\,\vec\jmath}`

• position at time t :

`\vec r(t) = \displaystyle \int_0^t \left(\left(5(\rm m)/(\rm s)\right)\,\vec\imath + \left(3(\rm m)/(\mathrm s^2)\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5(\rm m)/(\rm s)\right)t\,\vec\imath + \frac12 \left(3(\rm m)/(\mathrm s^2)\right)t^2\,\vec\jmath}`