# 22-24 One car from a city starts north, another car from nearby city starts east

**Problem 22**

One city C, is 30 miles north and 35 miles east from another city, D. At noon, a car starts north from C at 40 miles per hour, at 12:10 PM, another car starts east from D at 60 miles per hour. Find when the cars will be nearest together.

**Solution 22**

$x = 35 - 60(t - \frac{10}{60})$

$x = 45 - 60t$

$s = \sqrt{(45 - 60t)^2 + (30 + 40t)^2}$

$\dfrac{ds}{dt} = \dfrac{2(45 - 60t)(-60) + 2(30 + 40t)(40)}{2\sqrt{(45 - 60t)^2 + (30 + 40t)^2}} = 0$

$-60(45 - 60t) + 40(30 + 40t) = 0$

$-3(45 - 60t) + 2(30 + 40t) = 0$

$-135 + 180t + 60 + 80t = 0$

$260t = 75$

$t = 0.2885 \, \text{ hr}$

Time = 12.2885 PM or 12:17:18 PM *answer*

**Problem 23**

For the condition of Problem 22, draw the appropriate figure for times before 12:45 PM and after that time. Show that in terms of time after noon, the formulas for distance between the two cars (one formula associated with each figure) are equivalent.

**Solution 23**

**Before 12:45 PM**

For time before 12:45 PM, see the figure in Solution 22.

$x = 45 - 60t$

$s = \sqrt{(45 - 60t)^2 + (30 + 40t)^2}$

**After 12:45 PM**

$s = \sqrt{x^2 + (30 + 40t)^2}$

$x = 60(t - \frac{10}{60}) - 35 = -(45 - 60t)$

$x^2 = (45 - 60t)^2$

Thus,

$s = \sqrt{(45 - 60t)^2 + (30 + 40t)^2}$ (*okay!*)

**Problem 24**

For Problem 22, compute the time-rate of change of the distance between the cars at (a) 12:15 PM, (b) 12:45 PM.

**Solution 24**

$s = \sqrt{(45 - 60t)^2 + (30 + 40t)^2}$

$\dfrac{ds}{dt} = \dfrac{2(45 - 60t)(-60) + 2(30 + 40t)(40)}{2\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$ *See Solution 22*

$\dfrac{ds}{dt} = \dfrac{-60(45 - 60t) + 40(30 + 40t)}{\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$

$\dfrac{ds}{dt} = \dfrac{-2700 + 3600t + 1200 + 1600t}{\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$

$\dfrac{ds}{dt} = \dfrac{5200t - 1500}{\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$

(a) at 12:15 PM, t = 15/60 = 0.25 hr

$\dfrac{ds}{dt} = \dfrac{5200(0.25) - 1500}{\sqrt{[45 - 60(0.25)]^2 + [30 + 40(0.25)]^2}}$

$\dfrac{ds}{dt} = -4 \, \text{ mi/hr}$ *answer*

(b) at 12:45 PM, t = 45/60 = 0.75 hr

$\dfrac{ds}{dt} = \dfrac{5200(0.75) - 1500}{\sqrt{[45 - 60(0.75)]^2 + [30 + 40(0.75)]^2}}$

$\dfrac{ds}{dt} = 40 \, \text{ mi/hr}$ *answer*

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