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.

A tour bus normally leaves for its destination at 500 pm for a 364 mile trip This week however the bus leaves at 530 pm To arrive on time the driver drives 4 mi class=

Respuesta :

52mph

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

[tex]\frac{364}{x}-\frac{364}{x+4}=0.5[/tex]

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:

[tex]\begin{gathered} Speed=\frac{d}{t}\Rightarrow st=d\Rightarrow t=\frac{d}{s} \\ \\ \frac{d}{s}-\frac{d}{s+4}=t \end{gathered}[/tex]

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

[tex]\begin{gathered} \frac{364}{x}-\frac{364}{x+4}=0.5\:\:\:(From\:the\:text\:t=0.5\:hour) \\ \\ \frac{364}{x}x\left(x+4\right)-\frac{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)\:\:\:\:\times10\:\: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_=\frac{-20\pm\sqrt{20^2-4\cdot\:5\left(-14560\right)}}{2\cdot\:5} \\ \\ x_1=52,\:x_2=-56 \end{gathered}[/tex]

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