Asked by glenn
a driver traveling 42 mph speeds up to 58 mph over distance of 660 feet how long does this increase in speed take what is the driver's acceleration
Answers
Answered by
Damon
Well, now we seem to be back in the dark ages of some king's feet.
v1 = 42*5280/3600 ft/s
v2 = 58*5280/3600 ft/s
assuming constant acceleration we can use average v
va = 50*5280/3600 ft/s
so t = 660/va
a = (v2-v1)/t
v1 = 42*5280/3600 ft/s
v2 = 58*5280/3600 ft/s
assuming constant acceleration we can use average v
va = 50*5280/3600 ft/s
so t = 660/va
a = (v2-v1)/t
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