I was given this solution.But i have two questions.. 1.Do the units not matter while we calculate related rates? If it does, is the answer provided correct? 2.How did he/she end up with the 5mi there? Thanks alot for your time..

  • $\begingroup$ I would always recommend converting all the units to SI units, especially in this case where there are mixed units. $\endgroup$ – Matti P. Feb 13 '18 at 10:03

First, let's convert the units to SI units:

  • 4 mi = 6437 m
  • 3 mi = 4828 m
  • 5 mi = 8047 m
  • 60 km/h = 16.67 m/s
  • 50 km/h = 13.89 m/s

Let's denote the car's distance (towards south) from the intersection by $s(t)$ and the truck's distance (towards east) by $r(t)$. Their distance from each other is $$ u^2(t) = s^2(t) + r^2(t) $$ Differentiating both sides, we get $$ 2u u^{'} = 2s s^{'} + 2 r r^{'} $$ where the prime denotes differentiation. Now, the rate of change of the distance between the two cars can be solved easily: $$ u^{'} = \frac{s s^{'} + r r^{'}}{u} $$ Now we just have to plug in the values, remembering that the car is heading north, so that $s^{'}<0$: $$ u^{'} = \frac{\left(6437~\text{m}\right)\left(-16.67~\frac{\text{m}}{\text{s}}\right) + \left( 4828~\text{}\right) \left( 13.89~\frac{\text{m}}{\text{s}}\right)}{8047~\text{m}} \approx -5~\frac{\text{m}}{\text{s}} $$

  • $\begingroup$ how did you get the 5miles? $\endgroup$ – Aby Feb 13 '18 at 15:36
  • $\begingroup$ Pythagoras: $\sqrt{3^2 + 4^2} = 5$. $\endgroup$ – Matti P. Feb 14 '18 at 6:36

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