During the first part of a trip,a canoeist travels 38miles at a certain speed.The canoeist travles 14miles on?

Question

second part of the trip at a speed of 5mph slower.The total time for the trip is 3hours. What was the speed on each part of the trip?????????

thank you for the help

Answer 1

Let

V = speed for the first 38 miles

(V - 5) = speed for the last 14 miles

T1 = time for the first 38 miles

T2 = time for the last 14 miles

T1 + T2 = 3

For the first 38 miles,

T1 = 38/V

and for the last 14 miles,

T2 = 14/(V - 5)

and since T1 + T2 = 3, then

38/V + 14/(V - 5) = 3

Cross multiplying the above,

38(V - 5) + 14(V) = 3(V)(V - 5)

38V - 190 + 14V = 3V^2 - 15V

52V - 190 = 3V^2 - 15V

Rearranging the above,

3V^2 - 15V - 52V + 190 = 0

3V^2 - 67V + 190 = 0

Using the quadratic formula,

V = 19 mph and V = 3.33 mph

IGNORE the root V = 3.33 mph (since this is not applicable as the speed in the second part of the trip is 5 mph slower) and therefore the speed in the first 38 miles,

V = 19

and the speed in the last 14 miles,

(V - 5) = 14 mph.

Answer 2

19 miles per hour for the first 2 hours then 14 miles per hour for the last hour