The question can be found in the URL httppostimgorgimage4hf4

The question can be found in the URL

http://postimg.org/image/4hf4fp7yp/

If the triathlete\'s running speed is 8.95 m/s and swimming speed is 1.879 m/s, calculate the value of x so that the triathlete reaches her bike in the least amount of time.

Calculate the minimum time required to reach the bicycle

Solution

a = 130 m

b = 90 m

x+y = 179.9

sin theta1 = x/sqrt(a^2+x^2)

sin thetat2 = y/sqrt(b^2+y^2)

sintheta1 / v1 = sintheta2/v2

x/v1*sqrt(a^2+x^2) = y/v2*sqrt(b^2+y^2)

y = 179.9 - x

x/(v1*sqrt(a^2+x^2) = (179.9-x)/(v2*sqrt(b^2+(179.9-x)^2)

x/(1.879*sqrt(130^2^2+x^2)) = (179.9-x)/(8.95*sqrt
(90^2+(179.9-x)^2)

8.95*x*sqrt(90^2+(179.9-x)^2) =

(1.879*(179.9-x)*sqrt(130^2+x^2))

x = 24.03 m    <------------answer

y = 155.87 m

v1 = .879

v2 = 8.95

t1 = sqrt(x^2+a^2)/v1 = 70.4 s

t2 = sqrt(y^2+b^2) / v2 = 20.11 s

T = t1+t2 = 90.51 s <,,------------answer