What is the instantaneous velocity of an object moving in accordance to # f(t)= (t/(t-5),3t-2) # at # t=-2 #?

Question

What is the instantaneous velocity of an object moving in accordance to #              f(t)=   (t/(t-5),3t-2)     # at #  t=-2   #?

Charles Arocha · Accepted Answer

About #3.002# at an angle of #-1.536#

Take the derivative of #f(t)#. Don't' forget quotient rule:

#f'(t)=(((1)(t-5)-(t)(1))/(t-5)^2,3)#

Simplify:

#f'(t)=((t-5-t)/(t-5)^2,3)#

#f'(t)=(-5/(t-5)^2,3)#

Now plug in #t=--2#

#f'(-2)=(-5/(-2-5)^2,3)#

#f'(-2)=(-5/49,3)#

Now we have the velocity in the #x# direction and #y# direction. Use Pythagorean theorem.

#(-5/49)^2+(3)^2=c^2#

Solve for #c#:

#c~~3.002#

We can find the angle of this velocity with tangents:

#tan(theta)=3/(-5/49)#

#theta=-1.536#

Samantha Adkison · Answer

undefined To find the instantaneous velocity of the object at $ t = -2 $, you need to find the derivative of the function $ f(t) $ with respect to $ t $, which represents the velocity vector. Then, evaluate this derivative function at $ t = -2 $.

Given $ f(t) = \left(\frac{t}{t - 5}, 3t - 2\right) $, the derivative of $ f(t) $ with respect to $ t $ is:

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

To compute the derivatives:

$ \frac{d}{dt}\left(\frac{t}{t - 5}\right) = \frac{(t - 5) - t(1)}{(t - 5)^2} $

$ \frac{d}{dt}(3t - 2) = 3 $

Now, you have the derivative function:

$ f'(t) = \left(\frac{-5}{(t - 5)^2}, 3\right) $

To find the instantaneous velocity at $ t = -2 $, substitute $ t = -2 $ into $ f'(t) $:

$ f'(-2) = \left(\frac{-5}{(-2 - 5)^2}, 3\right) $

$ f'(-2) = \left(\frac{-5}{(-7)^2}, 3\right) $

$ f'(-2) = \left(\frac{-5}{49}, 3\right) $

So, the instantaneous velocity of the object at $ t = -2 $ is $ \left(\frac{-5}{49}, 3\right) $.