Bill spends two days driving from point a to point b. on the first day, he drove 2 hours longer and at an average speed 5 miles per hour faster than he drove on the second day. if during the two days he drove a total of 680 miles over the course of 18 hours, what was his average speed on the second day, in miles per hour?

his average speed on the second day is 35 miles per hour

Step-by-step explanation:

Bill spends two days driving from point a to point b

Let x = his speed on the first day.

Let y = his speed on the second day

Speed = distance / time

Time = distance / speed

Total time spent on the trip is 18 hours

Let time that he spent on the first day be t hours

Time that he spent on the second day will be 18 - t

Bill drove 2 hours longer on the second day than on the first day. This means

t = 18 - t + 2

t+t = 18+2

2t = 20

t = 10

He spent 10 hours on the first day.

He spent 18-10 = 8hours on the second day.

Bill drove at an average speed of 5 miles per hour faster than he drove on the second day. This means that

x = y + 5 - - - - - - - -- 1

Distance travelled on the first day = speed on the first day × time spent. This becomes

10 × x = 10x miles

Distance travelled on the second day = speed on the second day × time spent. This becomes

8 × y = 8y miles

He drove a total of 680 miles over the course of the 18 hours. This means that

10x + 8y = 680 - - - - - - - - 2

Substituting equation 1 into equation 2, it becomes

10(y + 5) + 8y = 680

10y + 50 + 8y = 680

10y + 8y = 680 - 50

18y = 630

y = 630/18 = 35 miles per hour

x = y + 5 = 35 + 5

x = 40 miles per hour

$9.41

step-by-step explanation:

536

step-by-step explanation: