What is the difference between the average speed and average velocity?

Question

A car travels 95km to the north at 70.0 km/h, then turns around and travels 21.9 km at 80.0 km/h. What is the difference between the average speed and the average velocity on this trip?

Answer 1

This is a D-I-R-T (Distance-Is-Rate-times-Time) problem (D=RT;  so, R = D/T, and T = D/R).

 

"average speed and the average velocity on this trip"  are the rate:

     average aped is scalar; average velocity is a vector (includes direction)

 

"on this trip"  means total trip distance and total trip time.

 

Average speed =  (total linear distance) / (total time)

                      = (distance north + distance south) / ( (time north) + (time south) )

                   =  (95 km + 21.9 km) / (  ( (95 km) / (70 km/hr) )   +   ( (21.9 km) / (80 km/hr) )  )

                   = (116.9 km) / (1.63 hr)

                   = 71.7 km/hr

 

Average velocity =  (final difference in position) / (total time)

                 =  (distance north - distance south) / (total time)   [note: consider north positive]

                 =  (95 km - 21.9 km) / (  ( (95 km) / (70 km/hr) )   +   ( (21.9 km) / (80 km/hr) )  )

                 =  44.85 km north     (+44.85 km)

        Note that speed and velocity should not be compared when there is a change of direction involved.

Answer 2

Speed is just magnitude, while velocity includes direction.

 

The average speed in this question is determined as follows:  

recall distance = rate (time)

 

Traveling north 95km / 70 km/h = 1.357 hours

Traveling south 21.9 km / 80 km/h = 0.274 hours

 

So the car traveled

(95 + 21.9) km in (1.357+0.274) hours  north

116.9 km / 1.631 hr = 71.67 km/h south

 

Now with velocity we have to consider direction

The distance becomes

(95-21.9) since the car turns around and travels south shortening the distance from the starting point

The time remains the same

73.1 km / 1.631 hr = 44.82 km/h