An object's two dimensional velocity is given by #v(t) = ( 1/t, t^2)#. What is the object's rate and direction of acceleration at #t=2 #?

1 Answer

Answer:

The acceleration at #t=2# is #a(t) = (-1/4, 4)#, whose absolute value is #sqrt257/4# and the direction is #93.576^@# w.r.t. #x#-axis.

Explanation:

Acceleration is the first derivative of velocity, so if we differentiate the velocity function, we get a function for the acceleration:

#(dv)/(dt) = (-1/t^2, 2t)#

Inserting #t=2# gives us the acceleration at #t= 2s#, namely

#a(t) = (-1/4, 4)#, whose absolute value is #sqrt((-1/4)^2+4^2)=sqrt(257/16)=sqrt257/4#

The direction is given by #alpha=tan^(-1)(4/(-1/4))=tan^(-1)(-16)=93.576^@# w.r.t. #x#-axis.