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)
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.
4 answers
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.
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.
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.