# Velocity of Separation: How fast is the distance between two cars changing?

**Problem**

A Toyota Land Cruiser drives east from point *A* at 30 kph. Another car, Ford Expedition, starting from *B* at the same time, drives S30°W toward *A* at 60 kph. *B* is 30 km from *A*. How fast in kph is the distance between two cars changing after 30 minutes? Hint: Use the Cosine Law.

A. 70 kph | C. 55 kph |

B. 80 kph | D. 60 kph |

**Answer Key**

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**Solution**

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*A*.

$v_\text{separation} = v_{Fx} + v_T$

$v_\text{separation} = 60 \cos 60^\circ + 30$

$v_\text{separation} = 60 ~ \text{kph}$ ← *answer*

**Another Solution (by Calculus)**

$s^2 = (30t)^2 + (30 - 60t)^2 - 2(30t)(30 - 60t)\cos 60^\circ$

$s = \sqrt{900t^2 + (30 - 60t)^2 - 30t(30 - 60t)}$

After 30 minutes, *t* = 0.5 hr

$\dfrac{ds}{dt} = \dfrac{d}{dt}\Big[ \sqrt{900t^2 + (30 - 60t)^2 - 30t(30 - 60t)} \Big]_{\, t = 0.5}$

$\dfrac{ds}{dt} = 60 ~ \text{kph}$ ← *answer*

Below is the details of differentiation if you are using a calculator with no derivative function.

## Differentiation Details

$s = \sqrt{900t^2 + (30 - 60t)^2 - 30t(30 - 60t)}$

$s = \sqrt{900t^2 + (900 - 3600t + 3600t^2) - 900t + 1800t^2}$

$s = \sqrt{6300t^2 - 4500t + 900}$

$\dfrac{ds}{dt} = \dfrac{12,600t - 4500}{2\sqrt{6300t^2 - 4500t + 900}}$

When *t* = 0.5 hr

$\dfrac{ds}{dt} = \dfrac{12,600(0.5) - 4500}{2\sqrt{6300(0.5^2) - 4500(0.5) + 900}}$

$\dfrac{ds}{dt} = 60 ~ \text{kph}$ ← *answer*

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