Although the formula looks involved, the key here is looking to see where the information goes.
We are given all the pieces but need to convert mph to ft/s to use the formula. Let's do it with 1 mph so that we have a ratio to use. We and solve a unit conversion problem.
[tex]\frac{1mile}{hour} * \frac{hour}{60 minutes} * \frac{minute}{60 seconds} * \frac{5280 feet}{mile} = \frac{5280}{60*60} = \frac{5280}{3600} = 1.46666[/tex]
That ratio tells us that 1 mph is 1.466666 ft/s. Now we solve two proportions.
1 mph / 1.466666 feet per second = 60 mph / x feet per second.
1x = (60)(1.466666)
So x = 88 feet per second.
Next, We repeat for 24 mph.
1 mph / 1.46666 feet per second = 24 mph / x feet per second.
1x = (1.4666666)(24)
x = 35.2 feet per second
Now we have the found appropriate V₁ and V₂. V₁ > V₂, so V₁ is 88 ft/s and V₂ is 35.2 ft/s. The problem tells us θ = 2.3 degrees, K₁ = .4 and K₂ = .06. The rest of the problem is calculator work. Start by substituting our degree measure of 2.3 degrees and the given values in the problem for V₁, V₂, K₁, and K₂
[tex]D = \frac{1.05[(88)^{2}-(35.2)^{2}]}{64.4(.4+.06 + (sin 2.3))}[/tex]
[tex]D = \frac{1.05[(7744-1239.04]}{64.4(.46 + (sin 2.3))}[/tex]
[tex]D = \frac{1.05(6504.96)}{64.4(.46 + 0.04013)}[/tex]
[tex]D = \frac{6830.208}{64.4(0.50013)}[/tex]
D = 6830.208 / 32.208372
D = 212.0631 = 212 (to the nearest foot)
Thus the car needs 212 feet to stop.
