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Tutoring and Teaching on Trisphee
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Coda
Developer
#
353
So the part that's confusing you -- and a small mistake Serra made -- is that the formula is specifically x = v
mean
t. 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:
v
mean
= (v
0
+ v
f
)/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 = v
mean
t
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/sec
2
EDIT:
If you want to use v
f
2
= v
i
2
+ 2a(x
f
-x
i
) you can, and it gives you the same result:
0 = (88 ft/sec)
2
+ 2a(146 ft)
-7744 ft
2
/sec
2
= 2a(146 ft)
-53.0 ft/sec
2
= 2a
-26.5 ft/sec
2
= a
EDIT 2:
Also your initial answer of 6.50x10
4
mi/h
2
was in fact correct:
(65000 mi/h
2
) / (3600 sec/h) / (3600 sec/h) * (5280 ft/mi) = 26.48 ft/sec
2
Dimensional analysis is a wonderful thing. :P
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Posted 08-15-2017, 10:02 AM