Question

ellie runs twice as fast as matt. if ellie gives matt a head start of 50m, how far should ellie run to catch up with matthew? solve algebraically.

Answer 1

rate * time = distance
Let ellie be ellie's rate or speed, length/time. If ellie is twice as fast as matt, then, matt is 1/2 as fast as ellie.\[\text{ellie } t = \frac{\text{ellie}}{2} t + 50m \]\[\frac{\text{ellie}}{2} t = 50m \]\[t=\frac{100m}{\text{ellie}} \]rate * time = distance\[\text{ellie}*t=\frac{100m}{\text{ellie}}*\text{ellie} \]\[\text{ellie}=100m \]This is an unusual problem in that there are no "dimensional units." Matt's head start is "50m" but m is not clearly defined in the problem. The problem author was probably thinking meters.

In the LHS of the second equation above, the symbol ellie, a rate at this point, length divided by unit time, is multiplied by time. In dimensional analysis, the result is length because the time dimensions cancel, divide out. The point is that the dimensions associated with a symbol can change throughout the solution to a problem.

An introduction to Dimensional Analysis:

http://en.wikipedia.org/wiki/Dimensional_analysis