During the first part of a trip, a bicyclist travels 56 miles at a certain speed.The return trip is made at a speed that is 6 mph slower. Total time for the round trip is 16hr. Find the bicyclist's speed on each part of the trip.
certain speed ---- x mph
slower speed ---- x-6 mph
time at certain speed = 56/x
time at slower speed = 56/(x-6)
56/x + 56/(x-6) = 16
multiply each of the 3 terms by x(x-6)
you will get a quadratic, according to my definition, x > 6 , so reject any invalid answers after you solve for x.
To solve this problem, let's assign variables to the unknown quantities.
Let's say the speed during the first part of the trip is "s" mph.
Since the return trip is made at a speed that is 6 mph slower, the speed on the return trip would be "s - 6" mph.
We know that distance = speed * time, so we can write the following equations:
For the first part of the trip:
Distance = 56 miles
Time = 56/s
For the return trip:
Distance = 56 miles
Time = 56/(s - 6)
Since the total time for the round trip is 16 hours, we can add the times of both parts of the trip and set it equal to 16:
56/s + 56/(s - 6) = 16
To solve this equation, we can multiply through by s(s - 6) to remove the denominators:
56(s - 6) + 56s = 16s(s - 6)
Distributing the terms:
56s - 336 + 56s = 16s^2 - 96s
Combining like terms:
112s - 336 = 16s^2 - 96s
Rearranging the terms and setting the equation equal to zero:
16s^2 - 208s + 336 = 0
Now we have a quadratic equation, and we can solve it by factoring or using the quadratic formula. In this case, factoring is not straightforward, so let's use the quadratic formula:
s = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 16, b = -208, and c = 336.
Plugging in these values:
s = (-(-208) ± √((-208)^2 - 4 * 16 * 336)) / (2 * 16)
Simplifying:
s = (208 ± √(43264 - 21504)) / 32
s = (208 ± √21760) / 32
s = (208 ± 147.47) / 32
s = (208 + 147.47) / 32 or s = (208 - 147.47) / 32
Simplifying further:
s = 355.47 / 32 or s = 60.53 / 32
s ≈ 11.11 or s ≈ 1.89
Since the speed of the bicyclist cannot be negative, we discard the solution s ≈ 1.89.
Therefore, the bicyclist's speed during the first part of the trip is approximately 11.11 mph, and the speed on the return trip is approximately 11.11 - 6 = 5.11 mph.