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Physics 6 Online
OpenStudy (pallidine):

Please help Physics!!! Determine the stopping distances for a car with an initial speed of 95 km/h and human reaction time of 0.80 s Part 1: a=−3.8m/s^2 Part 2: a=−8.5m/s2

OpenStudy (raffle_snaffle):

I need more information.

OpenStudy (festinger):

This question has 2 parts: the first is the distance the car travels due to human reaction, and the distance that car covers during braking (deceleration). The first part is fairly straight forward. It is simply \(0.80s\times 26.39m/s=21.1m\) (I leave to you verify the conversion from \(95km/h\) to \(m/s\)). For the next part, use the kinematics equation:\[v_{f}^{2}=v_{i}^{2}+2ad\] Since \(v_{f}=0\) (complete stop) and \(a=-3.8ms^{-2}\) (deceleration), we have: \[d=\frac{26.39^{2}}{2\times3.8}=91.64m\] So, the total stopping distance is: \[21.1+91.6=112.7m\] I will leave it to you to figure out how to calculate the case when \(a=-8.5m/s\).

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