Homework SourseFor several weeks the highway department has been recording
For several weeks, the highway department has been recording the speed of freeway traffic flowing past a certain downtown exit. The data suggest that between the hours of 1:00 and 6:00 P.M. on a normal weekday, the speed of traffic at the exit is approximately
S(t) = t^3 - 10.5t^2 + 30t - 26 miles per hour, where t is the number of hours past noon. a
At what time between 1:00 and 6:00 is the traffic moving the slowest? What is the speed?
S(t) = t^3 - 10.5t^2 + 30t - 26 miles per hour, where t is the number of hours past noon. a
At what time between 1:00 and 6:00 is the traffic moving the slowest? What is the speed?
Solution
S(t) = t^3 - 10.5t^2 + 30t - 26
S\'(t) = 3*t^2 - 21*t +30 = 3(t^2-7*t+10)
this will be 0 at t=2,5
now calculate S\'\'(t) = 3(2*t-7)
at t=2 , S\'\'(t) = 4-7=-3<0 so this is maxima
at t=5 , S\'\'(t) = 10-7 =3>0 so this is minima
so trafic is moving at time 5 hours past noon and speed will be =| 125 - 10.5*25 +30*5 -26 |= |125-262.5+150-26 |= 13.5
so sloest speed is 13.5 miles/hour
