Doug walks at a steady pace of 5280 feet per hour. While Tommy walks at a steady pace of 3/4 of that of Doug. After one half hour, how far apart are they?

Question

anonymous · Answer

I dont know the equation to solve it. 
and I am very bad at solving fractions :(

magbak · Answer

2 miles.

anonymous · Answer

thanks I was kind of hoping to see how to solve it too

espex · Answer

The distance (d) between the two is growing at a rate of doug ($\frac{5280~feet}{1~hour}$) plus the rate of Tommy ($\frac{3}{4}*doug$). To see how far apart they get, take the distance and multiply it by the time in hours that they walk ($\frac{1}{2}hour*d$).
Substituting and expanding you get:$$\frac{1~hour}{2}*d \rightarrow \frac{1~hour}{2}*(\frac{5280~feet}{1~hour}+ (\frac{3~hour}{4}*doug))$$$$\rightarrow \frac{1~hour}{2}*(\frac{5280~feet}{1~hour}+ (\frac{3~hour}{4}*\frac{5280~feet}{1~hour}))$$$$\rightarrow\frac{1~hour}{2}*(\frac{5280~feet}{1~hour}+\frac{15840~hour~feet}{4~hour})\rightarrow \frac{1~hour}{2}*(\frac{5280~feet}{1~hour}+\frac{3960~feet}{1})$$$$\rightarrow\frac{1~hour}{2}*\frac{9240~feet}{1~hour} \rightarrow 4620~feet$$ Given a conversion factor of $\frac{1~mile}{5280~feet}$, you end up with $\frac{4620}{5280}~mile\rightarrow 0.875~miles$