calcuating time and distance

Question

Shannon is a lifeguard and spots a drowning child 30 meters along the shore and 40 meters from the shore to the child. Shannon runs along the shore for a while and then jumps into the water and swims from there directly to the child. Shannon can run at a rate of 3 meters per second and swim at a rate of 1.1 meters per second. How far along the shore should Shannon run before jumping into the water in order to save the child? Round your answer to three decimal places.

Answer 1

X is the distance she runs before diving into the water..

distance = rate * time or D = rt so t = d/r

total time = time running + time swimming

T(x) = x/3 + sqrt( (30-x)^2+ 40^2)/1.1

= x/3 + sqrt( 900 - 60x + x^2 + 1600)/1.1

= x/3 + (10/11) sqrt( x^2 - 60x + 2500)

Then

0 = dT/dx = (1/3) + (10/11) * (1/2) * (2x-60) / sqrt(x^2-60x+2500)

= (1/3) + (10/11) (x-30) / sqrt(x^2-60x+2500)

Solving...

-11/30 = (x-30)/sqrt(x^2-60x+2500)

Squaring both sides:

121/900 = (x^2-60x+900)/(x^2-60x+2500)

cross multiplies:

121(x^2-60x+2500) = 900(x^2-60x+900)

121(x^2-60x+2500) = 900(x^2-60x+900+1600) - 900*1600

121(x^2-60x+2500) = 900(x^2-60x+2500) - 900*1600

Let Q = x^2-60x + 2500 be the quadratic

Then

121 Q = 900 Q - 900*1600

900*1600 = 900Q-121Q = 779 Q

Finally, upon solving for Q:

900*1600/779 = Q

completing of the square in reverse:

900*1600/779 = x^2 - 60x + 2500 = x^2 - 60x +900 + 1600

900*1600/779 = (x-30)^2+1600

solving via square root method:

x = +or- sqrt(900*1600/779 - 1600) + 30

It turns out that the positive branch results in negative measures...

x = 14.235364

Answer 2

distance = rate of speed times time

d=rt

time = distance divided by rate of speed

t = d/r

Let x = the distance along the shore line that he runs

then sqr(40^2+(30-x)^2) = distance he swims to the child

= sqr(1600 +(900-60x +x^2)) =sqr(2500-60x+x^2)

that's the hypotenuse of a right triangle with sides 40 and 30-x.

total time running and swimming =

t = x/3 + (1/1.1)sqr(2500-60x+x^2)

take the derivative, set equal to zero and solve for x

that will give the value of x that minimizes time

the calculations get a little tedious, but x should be about 14 or 15 meters, the distance he runs along the shore line

Answer 3

Let (30 −x) is the distance she runs before diving into the water..

Total time = Time running + Time swimming

T( x )= (30 −x) / 3 + ( 1600 + x² )^1/2 / 1.1

T' (x) = -1 /3 + 10x /[ 11 √(1600 + x²)

T' (x) = 0 ⇔ -1 /3 + 10x /[ 11 √(1600 + x²) ⇔ 30 x = 11 √(1600 + x²) ⇔ 900x² = 121 (1600 + x²)

779 x² = 121 1600

x = 440 / √779

x = 15.76

30 −x = 14.24