Answer:
The usually speed of the bus is 50 miles/h.
Step-by-step explanation:
Let the usual speed of the bus be x mile/hour.
We know that
[tex]speed=\frac{distance}{time}[/tex]
[tex]\Rightarrow time = \frac{distance}{speed}[/tex]
The bus travels 200 miles.
To reach its destination it takes time [tex]=\frac{200}{x}[/tex] h
This week however the bus leaves at 5:40.
The bus late 40 minutes [tex]=\frac{40}{60} h[/tex] [tex]= \frac{2}{3}h[/tex]
Now the speed of the bus is = (x+10) miles/h
The new time to reach the destination is [tex]=\frac{200}{x+10}[/tex] h
According to the problem,
[tex]\frac{200}{x}-\frac{200}{x+10}=\frac{2}{3}[/tex]
[tex]\Rightarrow 200[\frac{x+10-x}{x(x+10)}]=\frac{2}{3}[/tex]
[tex]\Rightarrow 200[\frac{10}{x^2+10x}]=\frac{2}{3}[/tex]
⇒2(x²+10x)=200×10×3
⇒x²+10x = 3000
⇒x²+10x -3000=0
⇒x²+60x-50x-3000=0
⇒x(x+60)-50(x+60)=0
⇒(x+60)(x-50)=0
⇒x= -60,50
∴x=50 [since speed does not negative]
The usually speed of the bus is 50 miles/h.
