Answer:
Draw a sloped line segment on a v vs t graph with y-intercept of 6. At the middle of the line is 21. Symmetrically then, the value of y at the end of the line segment is 36. Thus the answer.
Here is the problem:
A bicyclist starts a timed race at 6.0 mi/h. In order to win, he must average 21 mi/h. Assuming constant acceleration from the start, how fast must he be traveling at the end of the race?
The answer is 36 mi/h, not 48 mi/h as I answered. Is there a graphical method to this one or am I missing a direct formula method?
Please help and thanks,
B
Well the easiest way would be to Vaverage=(Vstart+Vend)/2 ..
Which is logical to you I hope.
Then you could simply
Vaverage=(Vstart+Vend)/2 *2
2Vave=Vs+Ve
Ve=2Vave-Vs
Ve=42-6=36
And yes, graphically, the average should be achieved exactly in the middle of the racing time.