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.
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
2 answers
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
3 answers