Question

Jeff recently drove to visit his parents who live 780 miles away. On his way there his average speed was 11 miles per hour faster than on his way home (he ran into some bad weather). If Jeff spent a total of 26 hours driving, find the two rates.

Answer 1

The problem requires setting up two equations based on the distance Jeff traveled to his parents' place and back, then solving these equations to find his average speeds for both trips. Jeff's average speed going to his parents was denoted as 's' mph and on the way back 's - 11' mph. The total time spent driving was 26 hours.

Step-by-step explanation:

Jeff drove to visit his parents who live 780 miles away. Let's denote his average speed on the way there as s mph, which implies that his average speed on the way home was s - 11 mph due to bad weather. The time taken to travel to his parents' place is t hours and for the return trip is 26 - t hours, since he spent 26 hours driving in total.

Given that Distance = Speed × Time, we can set up the following two equations representing the two trips:

1. 780 = s × t
2. 780 = (s - 11) × (26 - t)

We can solve for t from the first equation:

t = 780 / s

Then, substituting t into the second equation, we get:

780 = (s - 11) × (26 - 780/s)

Solving the above equation involves simplifying the expression and finding the roots of the resulting quadratic equation.

Upon finding s and t, we'll have determined Jeff's average speeds for both the trip to his parents' house and the return trip.

Answer 2

66 mph there, 55 mph back

Step-by-step explanation:

Each leg of the trip he drove 780 miles. The rate going there was faster that the rate coming home, so we have two equations based on distance...This would also effect the times going there and back, so we have two times as well...

Distance = (rate)(time) so

Going there...

780 = (r+ 11)(t1)

and coming back...

780 = (r)(t2)

We know that t1 + t2 = 26, so solve both equations for t, then add them together....

780/(r + 11) = t1 and 780/r = t2

780/(r + 11) + 780/r = t1 + t2

780/(r + 11) + 780/r = 26

now solve for r

780r + 780(r + 11) = 26[r(r + 11)] (multiply both sides by r(r + 11) to get rid of the fractions)

Simplify...

780r + 780r + 8580 = 26r² + 286r

We now have a quadratic, so get everything to one side, and solve...

26r² - 1274r - 8580 = 0

r² - 49r - 330 = 0 (divide both sides by 26)

(r - 55)(r + 6) = 0 (factor)

So

r - 55 = 0, then r = 55

and

r + 6 = 0, then r = -6 ***This answer is ignored because we are talking about rate, and rate in regards to speed is always positive, so

r = 55 is the only acceptable answer.

His rate going there was 66 mph, and his rate coming back was 55 mph