A lifeguard at the beach notices a drowning swimmer. The lifeguard immediately l

Question

A lifeguard at the beach notices a drowning swimmer. The lifeguard immediately leaves his station and runs 50.0m to the water, and then begins swimming towards the drowning person. It takes a total of 30.0 seconds for the lifeguard to reach the swimmer. The lifeguard can run 7.0m/s on sand, and can swim 3.0m/s through water.

Questions: How far did the lifeguard have to swim to reach the drowning swimmer? What was the lifeguard's average velocity for the whole process?
Using this formula: 1)V=delta x/delta y
average velocity=change of the distance/change of time
2) X(sub 1)=X(sub 0) + V(average)*[t(sub 1)-t(sub 0)]

Explanation / Answer

We know the following to start: Total time: 30s Land Distance: 50m Land Speed: 7 m/s Water Speed: 3 m/s We need the following: Water Distance: Land Time: Water Time: Average velocity: The Land time is the first step. Since Distance = velocity * time, 50 m = 7 m/s * t t = 50/7 s = 7.14 seconds Since we know the total amount of time, the time he spent in the water is: 30s - 7.14s = 22.86 seconds The distance he travelled can be calculated, since Distance = velocity * time: Distance = 3 m/s * 22.86 s Distance = 68.57 m His average velocity is just the weighted average of his two different velocities: V avg = [ 7 m/s * 7.714 s + 3 m/s * 22.86 s]/30 s v avg = 3.95 m/s

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