# Physics motion question!?

## Two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at $20 m {s}^{-} 1$ and the second at $28 m {s}^{-} 1$. If the second car is 6km behind the first car, how much time will it need to catch up with the first car?

May 20, 2018

$t = 12.5 \text{minutes}$

#### Explanation:

Speeds are relative. The rotation of the Earth's surface at the equator goes at about 460 m/s to accomplish its full revolution every 24 hours. Check out this article from Scientific American. If you were in the lead car, the speed, relative to you, of the 2nd car would be $28 m {s}^{-} 1 - 20 m {s}^{-} 1 = 8 m {s}^{-} 1$. That is the speed of that car's approach.

The formula of speed, or velocity if we know the direction, is $v = \frac{d}{t}$

(We tend not to use s for speed, perhaps because s is already used for seconds, and some times distance.)

Solving that formula for time

$t = \frac{d}{v}$

We can use that formula to solve our problem, but first, there is a problem with units coming up. The distance is in km and our speed is in ms^-1. So it will be best to convert the 6 km to meters.

$6 \cancel{k m} \cdot \frac{1000 m}{1 \cancel{k m}} = 6000 m$

Now we can find the time for the 2nd car to catch you.

$t = \frac{6000 \cancel{m}}{8 \frac{\cancel{m}}{s}} = 750 s$

The question does not say what units it wants the answer in, but just for fun ...

$750 \cancel{s} \cdot \frac{1 \min}{60 \cancel{s}} = 12.5 \min$

I hope this helps,
Steve