Question

An object's two dimensional velocity is given by `v(t) = ( 3t^2 - 5t , t )`. What is the object's rate and direction of acceleration at `t=2`?

Answer 1

`veca(t)=(7hati+hatj)m/s^2`

Explanation:

I'm guessing they're using normal convention replaced by brackets.

So, in vectorial form, `vec{v(t)}=(3t^2-5t)hati+thatj`

Acceleration is the time derivative of velocity, so you go that.
`veca=d/dt(vecv(t))=d/dt(3t^2-5t)hati+d/dt(t)\hatj`

I'm sure you know differentiation, hence we'll get acceleration
`veca(t)=(6t-5)hati+1hatj`

We need to find acceleration at time `t=2`, substitute,
`veca(t)_{t=2}=(6*2-5)hati+hatj`

In the end, you'll get why the answer is as given in the box.