Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 ft>s when the driver of B suddenly applies the brakes, causing his car to decelerate at 12 ft>s2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 15 ft>s2. Determine the minimum distance d be tween the cars so as to avoid a collision.

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lcmendozaf lcmendozaf · Answer 1 · 2019-09-13T01:11:57+02:00

Answer:

The minimum distance has to be 15ft

Explanation:

Since car A is behind, I am fixing the origin in that point.

Now, let's calculate the position, for both cars at the end of everything, measured from the origin.

I am using this formula for [tex]t_{Max}=\frac{V_{f}-V_{o}}{a} =\frac{-V_{o}}{a}[/tex]

For car A:

[tex]t_{aMax}=4s[/tex]

[tex]d_{A}=d_{0.75} + V*t_{aMax}-\frac{a*t_{aMax}^{2}}{2}[/tex]

And we also know that [tex]d_{0.75}=V*(0.75s)=45ft[/tex], So:

[tex]d_{A}=165ft[/tex]

For car B:

[tex]t_{bMax}=5s[/tex]

[tex]d_{B}=d + V*t_{bMax}-\frac{a*t_{bMax}^{2}}{2}[/tex]

Replacing the values we get:

[tex]d_{B}=d+150ft[/tex]

To avoid a collision, [tex]d_{A}≤d_{B}[/tex], so:

165 ≤ d + 150 If we solve for d: d ≥ 15ft