Safe distance

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 20 ft/s².

Determine the minimum distance d between the cars so as to avoid a collision. Express your answer to three significant figures and include the appropriate units.

This problem is somewhat complicated... (at least to me it took me a while to get it...)

The first thing that you need to remember is that if you have a graph of velocity over time, the area under the curve would be the position, because the integral of velocity with respect to time is the position. And that is what we are given. A description of the cars velocities.

How long does it take both cars to come to a stop?

Car A:

V = V₀ + a*t

60 = 0 - 20 * t

t = 3 seconds

Car B:

V = V₀ + a*t

60 = 0 - 12 * t

t = 5 seconds

Now here is the part that took me a while to understand: the cars cannot be at the same place at the same time within this time frame, so their velocities cannot meet. Why can't their velocities meet? Because the area under the velocity is the position. If the velocities meet they are at the same spot and an accident occurs. But we need to find where they meet to increase the distance to a minimum that the cars won't collide...

So, when do the speeds meet?

V_A = V_B

60 - 12*t = 60 - 20 (t - 0.75) Notice that we substract the .75 seconds it takes the driver to react which move all the velocity graph to the right

Solve for t to get t = 1.875 seconds

At what speed do they meet?

60 - 12*t when t = 1.875 is 37.5 ft/s

Graphically,

Now, the area under the velocity curve up until 37.5 ft/s is the position of the car. I calculated the area looking at the graph and geometry but you can also calculate the value of the functions and do an integral, which is right but harder. With geometry,

Car A:

The area for car A is,

Area of the rectangle from 0 to .75 + area of the big triangle from 0.75 to 3.75 - area of the small triangle from 1.875 to 3.75,

60 * 0.75 + ((3.75-0.75)*60/2) - ((3.75-1.875)*37.5/2)= 99.8438 ft

Don't think that the units should be ft squared because you are calculating area. The area under the velocity over time is position which is ft.

Car B:

The are for car B is,

Area of the big triangle from 0 to 5 - area of the small triangle from 1.875 to 5,

60*5/2 - (5-1.875)*37.5/2 = 91.4063 ft

All you need to do to calculate the minimum distance is calculate the difference between these two,

99.8438 - 91.4063 = 8.44 ft