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put this solution on YOUR website!Sol:
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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.
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