Tashas school bus usually drops her off in town every day at

Tasha\'s school bus usually drops her off in town every day at 4pm and her mother picks her up at the bus stop. Last Friday was a short day, only Tasha\'s mother forgot so she wasn\'t at the bus stop when Tasha was dropped off an hour early, at 3pm. Since Tasha had forgotten her cell phone at home, she decided to walk home along her mother\'s driving route. Her mother left home at the usual time and picked Tasha up along the way. They arrived home 10 minutes before their usual arrival time. At what time did Tasha\'s mother pick her up? Can you determine how far Tasha walked and/or how far her home is from the bus stop?

Solution

Sol:
<------------*--------------------->
A(Bus Stop) P B(Home)

Assume that her mother left home M minutes before 4 pm and the driving speed of the
mother is V mile/min. Also, as the diagram above, let P be the point of picking up
Tasha last Friday.
Since mother\'s car arrives the bus stop at 4 pm (usually),so the distance(AB) between
the bus stop and home is VM miles.

With constant speed, the time(M min) of driving from B to A is the same as the time
of driving from A to B. Thus,their usual arrival time is M min after 4pm.
But on Friday, the time of their arriving home was M - 10 min after 4pm.
Note, on that day, the total driving time of mother (from B to P and then back to B)
was M + M -10 = 2M -10 (min). And so the driving time from B to P was M - 5(half of
2M -10). Recall that her other left B at M min before 4 pm.
Hence, she picked up Tasha at 4 pm - M + M -5 = 4 pm - 5 min,that is,5 min before 4 pm.

The distance from A to P was the distance that Tasha walked. We see that AP = AB - BP
= VM - V(M-5) = 5 V. And we know that the distance AB between home and the bus stop
is VM.Therefore, to determine the distance AP we have to know the speed V of the are.
To get AB ,we should know both the speed V and the time M.
This problem is not that difficult,but should be considered very
carefully to get to the right track.