When fritz drives to work his trip takes 40 minutes, but when he takes the train it takes 30 minutes. find the distance fritz travels to work if the train travels an average of 15 miles per hour faster than his driving. assume that the train travels the same distance as the car?
convert minutes to hour first because the question talking about 15 mile per hour 40 mins = 40/60 2/3 hrs 30 mins = 30/60 = 1/2hrs
Assume that s be the speed when Fritz driving, so s + 15 will be the speed of the train.
We know the time we know the speed, Next distance that Fritz drive = [tex] \frac{2}{3}s[/tex] distance the train travel = [tex] \frac{1}{2}(s+15)[/tex]
The question: Assume that the train travels the same distance as the car ==> [tex] \frac{2}{3}s = \frac{1}{2}(s+15)[/tex] ==> [tex] \frac{2}{3}s = \frac{1}{2}s + \frac{15}{2}[/tex] ==> [tex] \frac{2}{3}s - \frac{1}{2}s = \frac{15}{2}[/tex] ==> [tex] \frac{1}{6}s = \frac{15}{2}[/tex] ==> [tex] \frac{6}{1} * \frac{1}{6}s = \frac{15}{2} * \frac{6}{1}[/tex] ==> [tex] s = 45 [/tex] Now we know that Fritz drive at 45 mph, distance = [tex] \frac{2}{3} * 45 = 30 miles [/tex]
