Sam is running at 3.8 m/s and is 75m behind John who is running at a constant velocity of 4.2m/s. If Sam accelerates at 0.15m/s^2, how long will it take him to catch John?
d = (vf+vi/2)t
vf = vi + at
d = vit + 1/2at^2
vf^2 = vi^2 + 2ad
d = vft - 1/2at^2
The Attempt at a Solution
I have asked this before, but have now come back with a little more understanding than before.
I've done the following so far:
Xs (displacement of Sam) = viΔt + 1/2aΔt^2
= (3.8m/s)Δt + (0.075m/s^2)Δt^2
Xj (displacement of John) = 75 + 4.2m/sΔt
[-> from: (xf-xi) = VavΔt
xf = xi + VavΔt ]
... Ok so now I do (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt
to find the time when they are in the same position/displacement, correct?
But I'm so confused as to what I would do after that?
EDIT; Also, I got 1005.33s for the time, and I don't think that's right?