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 #?

1 Answer
Jul 3, 2016

Answer:

#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.