Expert: Scotto - 4/15/2009

Question
A car is traveling at 50 mph when the breaks are fully applied, producing a constant deceleration of 40 ft/sec^2. What is the distance covered before the car comes to a stop?

Answer
To start with, convert 50 mph to ft/sc.
To do this, take 50 miles/hour and do the following:
(50 mile)/(1 hour)(1 hour/3600 sec)(5280 feet/1 mile).
Hours cancel and miles cances, leaving feet/sec.
In units, we have 50*5280/3600.

For the 5280/3600, the following can be done:
5280/3600 = 528/360 = 132/90 =66/45 = 22/15.
That's as far as it can be reduced and is our conversion of
miles per hour to feet per second.

50 * 22/15 = 220/3, and that was really (220/3)ft/sec.

So what to say is the initial speed was (220/3)ft/sc.

We know the acceleration A = -40 ft / sc².
Given this, when we're traveling at 220/3 ft/sc,
the time required to stop is 220/(3*40) = 11/6 sc.

Integrating with respect to time on that constant gives velocity
V = -40t ft/sc + C.

Now C is the original speed (220/3 ft/sc) so we have
V = (220/3 - 40t) ft/sc.

Integrating with respect to time again gives the distance
D = (220t/3 - 40t²/2) ft = (220t/3 - 20t²) ft.

Now farther back, we said that t was 11/6,
so put this into the equation for D and get
D = [2420/18 - 20(11/6)²]ft = [1210/9 - 20*(121/36)]ft.
Now we know that 20*121/36 is 605/9, so this is
D = [(1210 - 605)/9]ft = [605/9]ft, which is a little over 66 ft.