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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