- #1

Sean1218

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## 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?

## Homework Equations

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?