Problema Solution
A tour bus leaves Sacramento every Friday evening at 5:00 pm for a 270-mile trip to Las Vegas. This week, however, the bus leaves at 5:30 pm. To arrive in Las Vegas on time, the driver drives 6 miles per hour faster than usual. What is the bus's usual speed?
Answer provided by our tutors
let
v = the bus's usual speed, v>0
d = 170 mi the distance traveled
t = the time of the travel in hours
since speed = distance/time that is v = d/t follows
v*t = d
v*t = 170 divide both sides by v
t = 170/v
This week, however, the bus leaves at 5:30 pm. To arrive in Las Vegas on time, the driver drives 6 miles per hour faster than usual.
(v + 6)*(t - 0.5) = d
(v + 6)*(t - 0.5) = 170
plug t = 170/v into the last equation
(v + 6)*(170/v - 0.5) = 170
by solving we find
v = 42.27 mph
click here to see the step by step solution of the equation
the bus's usual speed is 42.27 mph.