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hboardz133 hboardz133
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13 years ago
Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and undergo identical decelerations. How does the distance required for car A to stop compare with that for car B?
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#1
13 years ago
The stopping distance of car A would be four time of Car B.

Ex.
Use Vf^2=vi^2 +2ad

lets say the deceleration is -2m/s
car A has speed of 6

0=(6)^2+2(-2)(d)
d=9

Car B has speed of 3
0=(3)^2+2(-2)(d)
d=2.25

9/2.25 =4
wrote...
#2
13 years ago
If Car A is driving twice as fast, meaning it's velocity is 2x(Car B), then I would conclude that it would take twice as long to decelerate to a stop (velocity)

so I would say 1/2 times Car A's stopping distance.
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