Question related to average velocity 
In this problem, P.A. Tipler writes

this is not the average of running and jogging speed because she ran for 10 s but jogged for 30 s.

Why must the time interval for running and jogging speeds be equal for this to be the average of those speds?
 A: It's implicit in the way the author is using the word 'average'.  The author is implicitly defining 'average' to mean 'average over time', rather than 'average over distance'.  By this definition, the time intervals must be the same in order for the numerical average (of 10 m/s and -1.67 m/s) to be equal to average velocity.
In fairness, this is the way most people use the word 'average': if you were in a car going 50 mph for 1 min and then 60 mph for 1 min, most people would say your average speed was 55 mph.  However, given the nature of the problem, the definition of 'average' should probably have been made explicit.
A: You can't average values that have different units/denominators, because those values mean different things. Same concept applies when you say that the average of 10 feet and 20 meters isn't 15. Units of 10s and 20s are no different.
A: i think according to definition of average velocity;
V(avg)=(u+v)/2
u=100/10 =10m/s   and  v=-50/30 =-1.67m/s.
:- V(avg)=(10+-1.67)/2 =4.17m/s.
Because the definition of average velocity is not (average displacement over total time)
