During the first part of a trip a canoeist travels 93 miles at a certain speed. The conoiest travels 5 miles on the second trip at 5mph slower. The total time for the trip is 2 hrs. What was the speed on each part of the trip?
Thank you for all your help.
To solve this problem, we can break it down into two parts: the first part of the trip and the second part of the trip. Let's assume that the speed during the first part of the trip is "x" miles per hour.
During the first part of the trip, the canoeist travels 93 miles. We can use the formula "distance = speed × time" to find the time taken for this part of the trip. So, for the first part:
93 = x × time1
Next, we are told that during the second part of the trip, the canoeist travels 5 miles at a speed that is 5 mph slower than the first part of the trip. Therefore, the speed during the second part of the trip is (x - 5) miles per hour.
Using the same formula, the equation for the second part of the trip can be written as:
5 = (x - 5) × time2
The total time for the entire trip is given as 2 hours. Therefore, time1 + time2 = 2.
Now, we have a system of equations:
93 = x × time1
5 = (x - 5) × time2
time1 + time2 = 2
To solve this system of equations, we can use substitution or elimination method.
Let's solve it using substitution. From the first equation, we can isolate time1:
time1 = 93 / x
Now substitute the value of time1 in the third equation:
(93 / x) + time2 = 2
Rearrange the equation:
time2 = 2 - (93 / x)
Substitute the values of time1 and time2 in the second equation:
5 = (x - 5) × (2 - (93 / x))
Simplify the equation:
5 = 2x - 10 - (93 * (x - 5)) / x
Multiply throughout by x to remove the fraction:
5x = 2x^2 - 10x - 93(x - 5)
Expand and simplify:
5x = 2x^2 - 10x - 93x + 465
Combine like terms:
2x^2 - 98x + 465 = 0
Now we have a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula. In this case, the quadratic formula would be the most efficient method.
Using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -98, and c = 465.
x = (-(-98) ± sqrt((-98)^2 - 4 * 2 * 465)) / (2 * 2)
Simplify:
x = (98 ± sqrt(9604 - 3720)) / 4
x = (98 ± sqrt(5884)) / 4
x = (98 ± 76.66) / 4
Now, we have two possible values for x:
x1 = (98 + 76.66) / 4 = 44.665
x2 = (98 - 76.66) / 4 = 5.84
However, we can eliminate x2 = 5.84 because we know that the speed during the second part of the trip is (x - 5). Therefore, (5.84 - 5) does not satisfy the condition of being 5 mph slower.
Therefore, the speed during the first part of the trip is approximately 44.665 mph.
To find the speed during the second part, we subtract 5 from the speed of the first part:
Speed during the second part = 44.665 - 5 = 39.665 mph.
So, the speed during the first part of the trip is approximately 44.665 mph, and the speed during the second part is approximately 39.665 mph.