What is the instantaneous velocity of an object with position at time t equal to # f(t)= (1/(t-4),sqrt(5t-3)) # at # t=2 #?

Question

What is the instantaneous velocity of an object with position at time t equal to #              f(t)=   (1/(t-4),sqrt(5t-3))  # at #  t=2 #?

Charles Anctil · Accepted Answer

#=(-1/4,5/(2sqrt7))#

Since velocity is defined as the rate of change in position, we may find the velocity by differentiating the position function with respect to time as follows :

#v(t)=dx/dt=f'(t)=(-1/(t-4)^2 , 5/(2sqrt(5t-3)))#

#therefore v(2)=f'(2)=(-1/(2-4)^2,5/(2sqrt(5*2-3)))#

#=(-1/4,5/(2sqrt7))#.

Assuming this is a standard notation 2-dimensional problem, it means that we may then express the instantaneous velocity in terms of standard basis unit vectors and SI units as #v(2)=-1/4 hati+5/(2sqrt7) hatj# #m//s#.

William Abdalla · Answer

undefined To find the instantaneous velocity of an object at $ t = 2 $ given the position function $ f(t) = \left(\frac{1}{t - 4}, \sqrt{5t - 3}\right) $, we need to find the derivative of the position function with respect to time $ t $ and then evaluate it at $ t = 2 $.

The position function $ f(t) $ can be expressed as $ f(t) = \left(\frac{1}{t - 4}, \sqrt{5t - 3}\right) $.

The derivative of $ f(t) $ with respect to $ t $ is $ f'(t) = \left(\frac{d}{dt}\left(\frac{1}{t - 4}\right), \frac{d}{dt}\left(\sqrt{5t - 3}\right)\right) $.

Now, let's find the derivatives of each component function:

1. For the first component, $ \frac{d}{dt}\left(\frac{1}{t - 4}\right) $:
   $$ \frac{d}{dt}\left(\frac{1}{t - 4}\right) = -\frac{1}{{(t - 4)}^2} $$

2. For the second component, $ \frac{d}{dt}\left(\sqrt{5t - 3}\right) $:
   $$ \frac{d}{dt}\left(\sqrt{5t - 3}\right) = \frac{1}{2\sqrt{5t - 3}} \cdot \frac{d}{dt}(5t - 3) = \frac{5}{2\sqrt{5t - 3}} $$

So, the derivative of $ f(t) $ with respect to $ t $ is $ f'(t) = \left(-\frac{1}{{(t - 4)}^2}, \frac{5}{2\sqrt{5t - 3}}\right) $.

Now, we evaluate $ f'(t) $ at $ t = 2 $:
$$ f'(2) = \left(-\frac{1}{{(2 - 4)}^2}, \frac{5}{2\sqrt{5(2) - 3}}\right) = \left(-\frac{1}{(-2)^2}, \frac{5}{2\sqrt{10 - 3}}\right) = \left(-\frac{1}{4}, \frac{5}{2\sqrt{7}}\right) $$

So, the instantaneous velocity of the object at $ t = 2 $ is $ \left(-\frac{1}{4}, \frac{5}{2\sqrt{7}}\right) $.