# When Maggie applies the brakes of her car, the car slows uniformly from 15.6 m/s to 0 m/s in 2.30 s. How far ahead of a stop sign must she apply her brakes in order to stop at the sign?

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To answer this question one must be familiar with the equations of one dimensional motion for an object undergoing uniform acceleration.

Acceleration is defined as the rate of change in an object's velocity. That is, acceleration occurs when an object's velocity changes and the amount of acceleration is proportional to the change in velocity and inversely proportional to the time it takes to cause the change. In one dimensional motion, the acceleration will result in either an increase in speed or a decrease in speed (commonly called "deceleration").

The equation for acceleration in one-D motion is

a = (Vf - Vi)/T where Vf is the final speed, Vi is the initial speed, and T is the time it takes to cause the change.

From this we can calculate the acceleration of the car required to cause it to stop:

a =(0 m/s - 15.6m/s)/2.30s = -6.7826 m/s^2

With the acceleration it is now possible to calculate the distance it will take the car to stop. There is more than one equation that will work:

D = ViT + (1/2)aT^2

or

Vf^2 = Vi^2 + 2aD

The first equation requires less algebra.

D = (0m/s)X2.30s + (1/2)(-6.7826m/s^2)(2.30s)^2

D = -17.94 m

At this point one should round the answer to the proper number of significant digits. Rounding should be done as the last step before stating the final answer. From the given information, we see that each measured value had only 3 significant digits. Therefore the answer needs to be rounded to three digits.

D = -17.9 m or one could say the driver must apply her brakes 17.9 meters before the stop sign in order to get the car to stop in time.

If one chooses to use the second equation to determine the distance it is first necessary to do algebra an isolate the variable D.

Vf^2 = Vi^2 + 2aD recalling that Vf = 0m/s

0 m/s =Vi^2^ +2aD moving Vi^2 to the other side of the equality

2aD = -Vi^2 dividing both sides by 2a

D = -Vi^2/2a

D = -(15.6m/s)^2/(2X-6.7826)

D = 17.94 m = 17.9 m ahead of the stop sign (as before).

A third way can be used to calculate the distance the car needs to stop.

For one dimensional motion with a constant acceleration it is also true that

(Vi + Vf)/2 = D/T

Doing one step of algebra produces

D = T(Vi + Vf)/2

D = 2.30s(15.6m/s + 0m/s)/2

D = 17.94 m which again rounds to 17.9 m as the previous answers produced.