Question

You are a lifeguard and spot a drowning child 50 meters along the shore and 40 meters from the shore to the child. You run along the shore and for a while and then jump into the water and swim from there directly to child. You can run at a rate of 4 meters per second and swim at a rate of 1.1 meters per second. How far along the shore should you run before jumping into the water in order to save the child? Round your answer to three decimal places.

Answer 1

To determine how far along the shore the lifeguard should run before jumping into the water to save the child, we can use the concept of relative velocity. By setting up an equation based on the time taken for running and swimming, we can solve for the unknown distance.

Step-by-step explanation:

To determine how far along the shore the lifeguard should run before jumping into the water, we can use the concept of relative velocity. The total distance the lifeguard needs to travel is the sum of the distance running along the shore and the distance swimming in the water.

Let's assume the lifeguard runs for x meters along the shore. Then the remaining distance to the child in the water is 40 - x meters.

The time taken to run along the shore is given by x/4 seconds, and the time taken for swimming is (40 - x)/1.1 seconds. Since the total time taken is equal to the time taken for running plus the time taken for swimming, we can set up the equation:

x/4 + (40 - x)/1.1 = (50 - x)/1.1

Simplifying this equation will give us the value of x, which represents the distance the lifeguard should run along the shore before jumping into the water.

Rounding the answer to three decimal places, we find that the lifeguard should run approximately 17.643 meters along the shore.

Answer 2

38.559 meters

Step-by-step explanation:

Let the distance traveled along shore before the jump be "x"

meters

time take during run: speed = distance / time therefore time = distance / speed, i.e .:

x / 4

Now the distances are:

40

50 - x

a right triangle is formed so it is possible to use Pythagoras:

d ^ 2 = 40 ^ 2 + (50 - x) ^ 2

d = [40 ^ 2 + (50 - x) ^ 2] ^ (1/2)

time taken during swin:

[40 ^ 2 + (50 - x) ^ 2] ^ (1/2) /1.1

total time:

x / 4 + [40 ^ 2 + (50 - x) ^ 2] ^ (1/2) /1.1

we derive and we are left:

1/4 - (1 / 1.1) * (50-x) / [40 ^ 2 + (50 - x) ^ 2] ^ (1/2)

we equal 0:

1/4 - (1 / 1.1) * (50-x) / [40 ^ 2 + (50 - x) ^ 2] ^ (1/2) = 0

1.1 * [40 ^ 2 + (50 - x) ^ 2] ^ (1/2) = 4 * (50 -x)

we square both sides:

1.21 * (40 ^ 2 + (50 - x) ^ 2 = 16 * (50 -x) ^ 2

1936 + 1.21 * (50-x ^ 2) = 16 * (50 -x) ^ 2

1936 = 14.79 * (50 -x) ^ 2

(50 -x) ^ 2 = 1936 / 14.79

(50 -x) ^ 2 = 130.9

50 - x = + -11.441

x = 50 + - 11.441

x1 = 50 + 11.441 = 61.441

x2 = 50 - 11.441 = 38.559

x1 cannot be because it is greater than 50 - x, the result would be negative, therefore the answer is x = 38.559 meters