A woman driving at a speed of 23 m/s sees a deer on the road ahead and applies the brakes when she is 210 m from the deer. If the deer does not move and the car stops right before it hits the deer, what is the acceleration provided by the car's brakes?

To find the acceleration provided by the car's brakes, we need to use the equation of motion that relates distance, initial velocity, final velocity, and acceleration. The equation is:

\(v_f^2 = v_i^2 + 2a d\)

Where:
\(v_f\) is the final velocity (which is 0 m/s in this case, since the car stops)
\(v_i\) is the initial velocity of the car (23 m/s)
\(a\) is the acceleration
\(d\) is the distance traveled (210 m)

We can rearrange the equation to solve for \(a\):

\(a = \frac{{v_f^2 - v_i^2}}{{2d}}\)

Since \(v_f\) is 0 m/s, the equation simplifies to:

\(a = \frac{{- v_i^2}}{{2d}}\)

Substituting the given values into the equation:

\(a = \frac{{- (23\, \text{m/s})^2}}{{2(210\, \text{m})}}\)

Now, we can calculate:

\(a = \frac{{- 529\, \text{m}^2/\text{s}^2}}{{420\, \text{m}}}\)

Finally, we can simplify:

\(a = -1.2595\, \text{m/s}^2\)

Therefore, the acceleration provided by the car's brakes is approximately -1.2595 m/s². The negative sign indicates that the car is decelerating.

average speed (Va) is __ (23 + 0) / 2

time (t) is distance/speed __ 210 / Va

acceleration is speed change/time
__ (0 - 23) / t

23

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