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So the part that's confusing you -- and a small mistake Serra made -- is that the formula is specifically x = vmeant. It's not supposed to be the initial speed for the whole distance, it's the average speed.

For uniform acceleration, you should remember this from last semester:
vmean = (v0 + vf)/2.

So Serra's math is off by a factor of 2, because the average speed during braking is not 88 ft/sec, but 44 ft/sec.

The rest of the process is correct:

x = vmeant
146 ft = (44 ft/sec) * t
t = (146 ft) / (44 ft/sec) ~= 3.32 sec

a = Δv/t
a = (-88 ft/sec) / (3.32 sec) -- this one is still 88 because it's the total change in velocity, not the average
a = -26.5 ft/sec2

EDIT: If you want to use vf2 = vi2 + 2a(xf-xi) you can, and it gives you the same result:

0 = (88 ft/sec)2 + 2a(146 ft)
-7744 ft2/sec2 = 2a(146 ft)
-53.0 ft/sec2 = 2a
-26.5 ft/sec2 = a

EDIT 2: Also your initial answer of 6.50x104 mi/h2 was in fact correct:

(65000 mi/h2) / (3600 sec/h) / (3600 sec/h) * (5280 ft/mi) = 26.48 ft/sec2

Dimensional analysis is a wonderful thing. :P
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Old Posted 08-15-2017, 10:02 AM Reply With Quote