The speed of the vehicle before the brake were applied was about 83.7 feet/s
<h3>Further explanation</h3>
Acceleration is rate of change of velocity.
![\large {\boxed {v^2 = u^2 + 2ad}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7Bv%5E2%20%3D%20u%5E2%20%2B%202ad%7D)
![\large {\boxed {a = \frac{v - u}{t} } }](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7Ba%20%3D%20%5Cfrac%7Bv%20-%20u%7D%7Bt%7D%20%7D%20%7D)
![\large {\boxed {d = \frac{v + u}{2}~t } }](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7Bd%20%3D%20%5Cfrac%7Bv%20%2B%20u%7D%7B2%7D~t%20%7D%20%7D)
<em>a = acceleration ( m/s² )</em>
<em>v = final velocity ( m/s )</em>
<em>u = initial velocity ( m/s )</em>
<em>t = time taken ( s )</em>
<em>d = distance ( m )</em>
Let us now tackle the problem!
<em>This problem is about </em><em>accelerated motion.</em>
<u>Given:</u>
final speed of the vehicle = v = 0 feet/s
length of skid marks = d = 175 feet
maximum deceleration = a = 20 feet/s² <em>( assumption )</em>
<u>Unknown:</u>
initial speed = u = ?
<u>Solution:</u>
![v^2 = u^2 - 2ad](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20-%202ad)
![0^2 = u^2 - 2(20)(175)](https://tex.z-dn.net/?f=0%5E2%20%3D%20u%5E2%20-%202%2820%29%28175%29)
![u^2 = 7000](https://tex.z-dn.net/?f=u%5E2%20%3D%207000)
![u = \sqrt{7000}](https://tex.z-dn.net/?f=u%20%3D%20%5Csqrt%7B7000%7D)
![u = 10\sqrt{70} \texttt{ feet/s}](https://tex.z-dn.net/?f=u%20%3D%2010%5Csqrt%7B70%7D%20%5Ctexttt%7B%20feet%2Fs%7D)
<em>(rounded to the nearest tenth)</em>
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Conclusion:</h3>
The speed of the vehicle before the brake were applied was about 83.7 feet/s if we assume that maximum deceleration of the vehicle is about 20 feet/s²
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle