A runner jogs a along a straight road (in the `+x` direction) for `30 min`, travelling a distance of `6 km`. She then turns around and walks back towards her starting point for `20 min`, travelling `2 km` during this time.
State true/ false:
a. The final displacement of the entire trip is `0.16 km min^(-1)` .
b. Her average speed for the entire is `0.16 km min^(-1)`.
c. The average velocity for the entire trip is `0.4 km min^(-1)`.
d. The runner's average velocity while jogging is `0.4 km min^(-1)`.
e. Her average velocity while walking is `0.1 km min^(-1)`.

Answer

True ;
True ; False,
False;
True

Explanation

We first sketch `S4.37` a to depict the motion and indicate our chice of locating the origin of the coordinate system where the runner starts. We also show significant positions of the runner during the motion.
,
`{:(Position" "Time),(x_(0)=0" "t_(0)=0), ("Jogging"" "x_(1) =6 km" "t_(1) =30 min),("Walking"" "x_(2)=4 km" "t_(2) =50 min):}`
a. True,`Delta x=(x_(2)-x_(0))=3 km-0) =4 km`
b. True , average =`(Total distance)/(Total time) =(8km)/(50 mi n)`
c. False, the average velocity for the entire trip is
` v_(av) =(Delta x)/(Delta t) =((x_(2)-x_(0)))/((t_(2)-t_(0)))=(4-0)/(50-0) =0.080 km min^(-2)` ltbrtgt d. False, while jogging
`v_(av)=(Deltax)/(Deltat) =((x_(1)-x_(0)))/((t_(1)-t_(0))) =(6-0)/(30-0) =0.2000 km min^(-1)`
e. True, while walking
`v_(av)=(Delta x)/(Delta t) =((x_(2)-x_(1)))/((t_(2)-t_(1))) =((4 km -6 km))/((50 min-30 mi n))`
`=(-2 km)/(20 mi n) =-0.100 km min^(-1) (walki ng)`
The negative value indicates a velocity in the -`x` direction. A word of caution: Please note that the runner`s average velocity for the entire trip is not simply the average of her jogging and walking velocities, because the times she spends in the these two different motions are not equal.