Question

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.

Answer 1

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 $t_{Max}=\frac{V_{f}-V_{o}}{a} =\frac{-V_{o}}{a}$

For car A:

$t_{aMax}=4s$

$d_{A}=d_{0.75} + V*t_{aMax}-\frac{a*t_{aMax}^{2}}{2}$

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

$d_{A}=165ft$

For car B:

$t_{bMax}=5s$

$d_{B}=d + V*t_{bMax}-\frac{a*t_{bMax}^{2}}{2}$

Replacing the values we get:

$d_{B}=d+150ft$

To avoid a collision, $d_{A}≤d_{B}$, so:

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