Expert: Josh - 1/31/2006

Question
My friend likes to travel everywhere at 60 miles per hour because he knows that he is travelling at a mile a minute.  If he spent 5 minutes travelling at 70mph(overtaking) would he be able to average out his speed by travelling  for the same time at 50 mph?  Common sense tells me he would but I don't know how to demonstrate it mathematically.  Can you help me?

Answer
Hi Brian,

I have been trying to send my reply over the past two days, but was unsuccessful. The system must have been down or something.

Let's look at the variables here.
We have "D(t)", the distance travelled; over a time period of "t" at a speed of "s" miles per hour.

In the first case, we travel at a speed of
s=D/t, where D=60 [miles] over t=1 [hour]

In the second case, the total distance travelled comprises two components. Let's call this D1 and D2, for the period from t=0 to t=5 and t=5 to t=60, respectively.

Remember, distanced travelled is the product of speed and time. So, D1=70[miles/hour]*(5/60)[hour], D1 is approx. 5.833 miles. For the remainder of the hour (the other 55 minutes), he travels at 50 miles per hour. So, we have D2=50[miles/hour]*(55/60)[hour]=45.833 miles.

Thus, the total distance travelled is D=D1+D2, over a period of 1 hour. So, the equivalent speed is (45.833+5.833)/1 [miles/hour], i.e., at about 51.666 miles/hour.

Not sure if I understood your question correctly, if instead, what you meant were travelling at 70 mph for 5 minutes and then, travelling at 50 mph for another 5 minutes, then using the same argument, the total distance travelled is D=D1+D2 over a period of 1/6 hour; where D1=70[miles/hour]*(1/12)[hour], D2=50[miles/hour]*(1/12)[hour].
D=D1+D2=5.833+4.167=10 miles over 10 minutes.

This agrees with our intuition that the average speed is 60 miles/hour, as specified in case one.