Airplane Braking Distance Question

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Braking distance increases with the square of the speed when acceleration is constant. When comparing braking from 56 mph to 28 mph, the higher speed results in a stopping distance that is four times greater. This is derived from the relationship between velocity and time during deceleration. The calculations confirm that the correct answer is d, 4.0 times farther. Understanding these principles is crucial for solving similar physics problems.
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Homework Statement


Assuming equal rates of acceleration in both cases how much further would you travel if braking from 56mi/h to rest than from 28mi/h ?
a, 5.2 times farther
b, 4.8 times farther
c, 3.2 times farther
d, 4.0 times farther

Homework Equations



not really sure what i should use.
v=v0+at
x= x0+2a(x-x0)

The Attempt at a Solution


0-56/0-28=56/28=2
 
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V1 (56 m/s) is twice as great as V2 (28 m/s) so the time needed to stop is going to be doubled, due to a the equal acceleration. So you take the difference in the velocity values (twice as great), and in the time values (twice as great), and you come up with the four times farther by multiplying the aforementioned values.
 
ohh. thank you!
 
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