A Guide to Solving Problems, Example 2

Now we'll change the last problem just a bit and see how that affects the solution.

You drive 4.00 mi at 30.0 mi/h.  Then you turn around and drive 3.00 mi at 50.0 mi/h.  What is your average velocity for the trip?

In the following, we'll use red font to draw your attention to the changes in the solution from the last problem.

Given:

Dx1 = 4.00 mi 
Dx2 = -3.00 mi  (This value is negative, because xf - xi = 1.00 mi - 4.00 mi = -3.00 mi.)
v1 = 30.0 mi/h
v2 = -50.0 mi/h  (This value is negative, because Dx2 is negative.)

Ignore the amount of time it takes the car to turn around and speed up from 30.0 to 50.0 mi/h.

Goal:  What is the average velocity, vav, for the trip?

Note the change in the drawing.

For the first half of the trip, the elapsed time is

,

and for the second half, the elapsed time is

.

The total elapsed time is

,

and the average velocity for the entire trip is

.

Up to this point, the formulas have been identical to the last problem. We deal with the differences in numbers and signs in the final substitution step. Note that in the numerator, the total displacement is +1.00 mi. That is, the car ends up +1.00 mi from the start. In the denominator, both terms are positive, because the negative displacement for the second part of the trip cancels the negative velocity. That makes sense, because the time interval must be positive.

Units:  mi/h = mi/(mi/(mi/hr))

Sign:  The final result for average velocity is positive.  That makes sense because positive was defined to the right and the car finishes to the right of Start.

Sensibility:  The answer is less than either 30 or 50 mi/h.  That makes sense, because the car turns around and ends up close to its starting point.