Stanton D. answered • 09/07/21
Tutor to Pique Your Sciences Interest
Hello again, Faith L.,
Normally I wouldn't answer two questions from one student on the same topic area, requiring the same skills because that could mean that the student was trying to "game the system" and just get answers rather than learn the skills required to solve ALL such problems, which requires doing some work on your own!
But, this is your lucky day. What DON'T you know in this problem? You don't know: any times, or any speeds!
So there must be another way of attacking it. There is. You can use the equation v(f) ^ = v(i) ^2 + 2 ad to your benefit. Since the maximum speed at turnaround is some value, which may be regarded as identical for the two segments of the trip, the respective distances (d) for the segments must be in inverse ratio to the accelerations (a) for those respective segments. (Don't worry about "ac"celeration vs "de"celeration terminology, just treat both segments AS IF accelerations to that magic spot took place, from each end of the route). So since 5*a1 = a2 , think of the distance value as portioned out into sixths, with a1 phase getting 5 of those distance portions, and a2 phase getting 1 of them, if you see what I mean?
OK, then, phase 1 takes (5/6)*1.41 km, and phase 2 takes (1/6)*1.41 km. you can do the calculations (and round as appropriate!).
Frankly, this is a "predatory" taxi driver: over 1.41 km he/she should be able to accelerate quickly, run at a constant speed for most of the course, and decelerate quickly. He/she is "running the meter" timewise (meter charges run with both time and distance, usually. Unless you are taking Uber or Lyft, etc.; then they are specified and the driver can't pull such stuff).
-- Cheers, -- Mr. d.
