The would have to have travelled the same distance in the same lapse of time.
Let time=t, then
Brown roadrunner would have travelled
S1=ut+(1/2)at^2
=0(t)+(1/2)2.75t^2
=1.375t^2
Black roadrunner would have travelled
S2=9.25t
Both distances S1 and S2 are in metres.
For them to meet, S1=S2, =>
1.375t^2=9.25t =>
t(9.25-1.375t)=0
Solve for t.
t=0 (when black roadrunner past brown), or
(9.25-1.375t)=0
=> t=6.727 s.
S1=S2=9.25(6.727)=62.23 m
A brown road runner is standing still pecking at soil looking for worms. Suddenly, a black road runner travelling at a constant speed of 9.25 m/s, overtakes the brown roadrunner. The brown road runner notices that the other one has a worm in its mouth and begins to accelerate at 2.75 m/s2 to catch up to and steal the worm. How far will the brown road runner have to travel to catch up to the black road runner? Assume the stationary brown road runner begins to move the moment the black road runner passes it.
2 answers
9.25 t = 1/2 * 2.75 * t^2
1.375 t^2 - 9.25 t = 0
t (1.375 t - 9.25) = 0
t = 0 ... when the black one passes
t = 9.25 / 1.375 ... brown catches up
1.375 t^2 - 9.25 t = 0
t (1.375 t - 9.25) = 0
t = 0 ... when the black one passes
t = 9.25 / 1.375 ... brown catches up
2 answers