SOLUTION: In drag racing, both vehicles begin at a dead stop, then accelerate through 1/4 mi. Let us say, for the sake of discussion, that our two vehicles are not quite matched in power.

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Question 140771: In drag racing, both vehicles begin at a dead stop, then accelerate through 1/4 mi. Let us say, for the sake of discussion, that our two vehicles are not quite matched in power. That is, vehicle A can accelerate from 0 mph to 60 mph in 5 secs., whereas the vehicle B can accelerate from 0 mph to 60 mph in 4 secs.

DQ 3: How would you determine the time, t, for the fastest vehicle to traverse 1/4 mi.?

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
for constant acceleration __ distance equals one half times the acceleration times the square of the time __ d=.5at^2

2d=at^2 __ 2d/a=t^2 __ sqrt(2d/a)=t

60mph=60mph(5280ft/mi)/(3600sec/hr)=88ft/sec
__ a=60mph/4sec=88ft/sec/4sec=22ft/sec^2

t=sqrt(2*1320ft/22ft/sec^2)=sqrt(120sec^2)=slightly under 11sec