# Kinematics question (uniformly accelerated linear motion)

## Homework Statement

Two friends see each other in a store and are initially 50m apart. The first friend starts walking towards the second at a constant speed of 0.50 m/s. At the same instant, the second friend accelerates uniformly from rest at a rate of 1.0 m/s^2 toward the first friend. How long before the two friends can shake hands?

v = d/t
d = vt + 1/2at^2

## The Attempt at a Solution

Distance between them = 50m
Person A (on left side):
v = 0.50 m/s (is this just v or v1 or v2? not sure if i can even use the uniform acceleration equations so it's just v?)
d = vt
d = 0.50t (1)

Person B (right side):
a = -1.0 m/s^2
v1 = 0 m/s
d = 1/2(-1)t^2 (2)

(1) = (2)
0.50t = 1/2(-1)t^2
0 = -1/2t^2 - 0.50t
0 = -1/2t(t + 1)

t = 0 or t = -1

I did something wrong since t being 0 or -1 wouldn't make sense.

Not sure what else I can do or how to deal with the distance especially since they don't necessarily meet at halfway point. Any help?

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jhae2.718
Gold Member
You need to take the distance between the two into account.

Yes, but I'm not sure how.

jhae2.718
Gold Member
If we say that the first person is at x=0, and the distance between the two is 50 m, how can we write person 2's initial position?

at x = 50 or d2i = d1i + 50, i remember this question from last year, but can't remember where I went with this

jhae2.718
Gold Member
Right, so basically for person 2, the initial position is 50 m, so you need to add that to your equation for person 2, and then equate both and solve for t.

How would I phrase the problem if I had to answer it on a test? just d = v1(t) + 1/2a(t^2) + 50 as the equation? Just asking in terms of variables and such

jhae2.718
Gold Member
You could do it that way, or you could call it d0, or some other name, and just say it is the initial position at t=0.