What is the instantaneous velocity of an object with position at time t equal to # f(t)= ((t-2)^3,sqrt(5t-3)) # at # t=2 #?

Question

What is the instantaneous velocity of an object with position at time t equal to #              f(t)=   ((t-2)^3,sqrt(5t-3))  # at #  t=2 #?

Emily Ammerman · Accepted Answer

Instantaneous velocity, at t = 2 is #5/14sqrt7=0.9449#, nearly, in (distance)/(time) units.

Velocity #= f'(t)# #=sqrt((x')^2+(y')^2)# #=sqrt(9(t-2)^4+25/4(1/(5t-3))# #=sqrt(0+25/4(1/7))#, at t = 2 #=5/14sqrt7#

Charles Anctil · Answer

undefined To find the instantaneous velocity of an object at $ t = 2 $, we need to take the derivative of the position function $ f(t) $ with respect to time $ t $, and then evaluate it at $ t = 2 $.

The position function $ f(t) $ is given by $ f(t) = ((t-2)^3, \sqrt{5t-3}) $.

Taking the derivative with respect to time $ t $, we get:

$ f'(t) = (\frac{d}{dt}(t-2)^3, \frac{d}{dt}\sqrt{5t-3}) $

$ f'(t) = (3(t-2)^2, \frac{1}{2\sqrt{5t-3}} \cdot 5) $

Evaluating $ f'(t) $ at $ t = 2 $, we have:

$ f'(2) = (3(2-2)^2, \frac{1}{2\sqrt{5(2)-3}} \cdot 5) $

$ f'(2) = (0, \frac{1}{2\sqrt{10-3}} \cdot 5) $

$ f'(2) = (0, \frac{1}{2\sqrt{7}} \cdot 5) $

$ f'(2) = (0, \frac{5}{2\sqrt{7}}) $

So, the instantaneous velocity of the object at $ t = 2 $ is $ (0, \frac{5}{2\sqrt{7}}) $.