Asked by Matt
                got an odd one
a person is being chased by a wombat is running in a straight line towards a tree at a speed of 5.04m/s where the tree is (d) distance away from the person
The wombat is 29.3 m behind the person running at a speed of 9.21m/s
If the person is safe upon reaching the tree whats the maximum value (d) can be for the person to be safe from the wombat
            
        a person is being chased by a wombat is running in a straight line towards a tree at a speed of 5.04m/s where the tree is (d) distance away from the person
The wombat is 29.3 m behind the person running at a speed of 9.21m/s
If the person is safe upon reaching the tree whats the maximum value (d) can be for the person to be safe from the wombat
Answers
                    Answered by
            oobleck
            
    so let's set the d so that the person just reaches the tree at the same time as the wombat. That means that they take the same time (distance/speed) to reach it.
d/5.04 = (d+29.3)/9.21
find d.
    
d/5.04 = (d+29.3)/9.21
find d.
                    Answered by
            Damon
            
    person runs distance d
wombat flies distance (d + 23.9)
both go for time t
so
d = 5.04 t so t = d/5.04
(d + 23.9) = 9.21 t
d + 23.9 = 9.21 (d/5.04) = 1.83 d
23.9 = 0.83 d
    
wombat flies distance (d + 23.9)
both go for time t
so
d = 5.04 t so t = d/5.04
(d + 23.9) = 9.21 t
d + 23.9 = 9.21 (d/5.04) = 1.83 d
23.9 = 0.83 d
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