- #1

harujina

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## Homework Statement

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?

## Homework Equations

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 ]

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... 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?

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