Kinematics question (uniformly accelerated linear motion)

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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?
 

Answers and Replies

  • #2
jhae2.718
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You need to take the distance between the two into account.
 
  • #3
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Yes, but I'm not sure how.
 
  • #4
jhae2.718
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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?
 
  • #5
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at x = 50 or d2i = d1i + 50, i remember this question from last year, but can't remember where I went with this
 
  • #6
jhae2.718
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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.
 
  • #7
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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
 
  • #8
jhae2.718
Gold Member
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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.
 

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