Question 330860: 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. So, the question arises, how would you calculate the position of each vehicle after t secs? For instance, t = 5 or t = 6. Note: remember to convert from miles per hour to miles per second.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 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. So, the question arises, how would you calculate the position of each vehicle after t secs? For instance, t = 5 or t = 6.
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Nothing is specified otherwise, so assume the acceleration is constant (which is not the case in real life).
60 mph = 88 ft/sec
1/4 mile = 1320 feet
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A goes from 0 to 88 ft/sec in 5 secs --> 17.6 ft/sec/sec
B goes from 0 to 88 ft/sec in 4 secs --> 22 ft/sec/sec
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The position (distance from the starting point) = (1/2)at^2
For A:
At t=5, s = (1/2)*17.6*25 = 220 feet
At t=6, s = (1/2)*17.6*36 = 316.8 feet
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For B:
s(5) = 11*25 = 275 feet
s(6) = 11*36 = 396 feet
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Neither of these cars should bother going to a drag strip.
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