Question

A tour bus normally leaves for its destination at 5:00 p.m. for a 364 mile trip. This week however, the bus leaves at 5:30 p.m. To arrive on time, the driver drives 4 miles per hour faster than usual. What is the bus' usual speed?The bus' usual speed is __miles an hour.

Answer 1

52mph

1) We can solve this problem, using one equation:

`(364)/(x)-(364)/(x+4)=0.5`

Note that we have used a ratio, in which on the numerator we can write the distance and the bottom number stands for the time. So basically we are writing an expression based on this:

`\begin{gathered} Speed=(d)/(t)\Rightarrow st=d\Rightarrow t=(d)/(s) \\ \\ (d)/(s)-(d)/(s+4)=t \end{gathered}`

2) Now, let's solve it to find the speed:

`\begin{gathered} (364)/(x)-(364)/(x+4)=0.5\:\:\:(From\:the\:text\:t=0.5\:hour) \\ \\ (364)/(x)x\left(x+4\right)-(364)/(x+4)x\left(x+4\right)=0.5x\left(x+4\right) \\ \\ 64\left(x+4\right)-364x=0.5x\left(x+4\right)\:\:\:\:*10\:\:Get\:rid\:of\:the\:decimal\:point \\ \\ 640\left(x+4\right)-3640x=5x\left(x+4\right) \\ \\ 640x+2560-3640x=5x^2+20x \\ \\ 5x^2+20x-14560=0 \\ \\ x_=(-20\pm√(20^2-4\cdot\:5\left(-14560\right)))/(2\cdot\:5) \\ \\ x_1=52,\:x_2=-56 \end{gathered}`

As there are no negative velocities, then we can discard the negative root for that and tell that the usual speed is 52 mph