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.
speed in first trip ---- x mph
time for 1st trip = 93/x hours
speed in 2nd trip ---- x-5 mph
time for 2nd trip = 5/(x-5)
93/x + 5/(x-5) = 2
93(x-5) + 5x = 2x(x-5)
93x - 465 + 5x = 2x^2 - 10x
2x^2 - 108x + 465 = 0
x = 49.3 or x = 4.7
since the speed in the second trip is reduced by 5, then the new speed would have to be 4.7-5, which is negative. So we'll reject the second answer
check: if speed = 49.3
time = 93/49.3 + 5/44.3 = 1.999 (not bad)
This answer is wrong, If I can just get the time of the first trip, I can take it from there. I don't need to know the second trip, just the first trip. I appreciate your help. Thank you
The answer I gave you is correct for the way the question is given.
The answer was "x=49.3" or the speed on the first part was 49.3 mph and on the second part it was 44.3 mph.
I then checked the answer and got 2 hours as the total time.
The question itself is totally flawed.
I have done enough canoeing in my life time to know that you cannot do a 98 mile canoe trip in 2 hours.
Reiny is totally correct Ester.
You cannot get the correct speed by considering only the first leg of the trip.
The time of the first leg is T1 = 93/V while the time of the second leg is T2 = 5/(V-5).
Adding, 93/V + 5/(V-5) = 2
Multiplying out and simplifying yields 2V^2 - 108V + 465 = 0 which produces a positive V = 49.282 mph, the speed during the 93 miles trip, the speed during the second leg being 44.282 mph.
To solve this problem, we can start by setting up equations using the given information.
Let's say the speed during the first part of the trip is represented by 'x' miles per hour.
At this speed, the canoeist travels 93 miles, so we can use the formula:
Time (T) = Distance (D) / Speed (S)
The time taken during the first part of the trip is therefore:
T1 = 93 / x
During the second part of the trip, the canoeist travels 5 miles at a speed 5mph slower than the first part. So, the speed during the second part would be (x - 5) mph.
The time taken during the second part of the trip is:
T2 = 5 / (x - 5)
The total time for the trip is given as 2 hours:
T1 + T2 = 2
Now, we can substitute the values of T1 and T2:
93 / x + 5 / (x - 5) = 2
To solve this equation, we need to combine the fractions by finding a common denominator:
[(93(x - 5)) + (5x)] / [x(x - 5)] = 2
Simplifying the equation:
(93x - 465 + 5x) / (x(x - 5)) = 2
98x - 465 = 2x^2 - 10x
0 = 2x^2 - 10x - 98x + 465
0 = 2x^2 - 108x + 465
To solve this quadratic equation, we can either factor it, complete the square, or use the quadratic formula. In this case, the quadratic formula is the most convenient:
x = (-b ± √(b^2 - 4ac))/(2a)
where a = 2, b = -108, and c = 465
Substituting the values into the quadratic formula:
x = (-(-108) ± √((-108)^2 - 4 * 2 * 465))/(2 * 2)
x = (108 ± √(11664 - 3720))/4
x = (108 ± √7944)/4
x = (108 ± 89.16)/4
x ≈ (108 + 89.16)/4 or x ≈ (108 - 89.16)/4
x ≈ 49.54 or x ≈ 4.71
Since the speed cannot be negative, the speed during the first part of the trip is approximately 49.54 miles per hour.
The speed during the second part of the trip is 5 mph slower, so it is approximately 49.54 - 5 = 44.54 miles per hour.
Therefore, the speed on each part of the trip is approximately 49.54 mph and 44.54 mph respectively.