Kinematics Problem and final velocity

  • Thread starter Brett
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  • #1
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Problem: The acceleration of a particle is given by Ax(t) = -2.00m/s^2 + (3.00m/s^2)t. A) Find the initial velocity such that the particle will have the same x-coordinate at t= 4.00s as it had at t= 0. B) What will the velocity be at t= 4.00s?

Work so far:

Integrated to get these:

Vx(t) = -2t + (3/2)t^2
x(t) = -t^2 + (1/2)t^3

I am just stuck on what to do. I dont need it worked out as much as I just need a push in the right direction. Thanks.
 

Answers and Replies

  • #2
15
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Both your equations are wrong. When you integrated the acceleration, you forgot about the constant of integration or the initial velocity.
 
  • #3
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PureEnergy said:
Both your equations are wrong. When you integrated the acceleration, you forgot about the constant of integration or the initial velocity.

So then they are:

Vx(t) = Vi + [-2t + (3/2)t^2]
x(t) = Xi + Vi(t) + [-t^2 + (1/2)t^3]

With Vi meaning initial velocity and Xi meaning initial position.

I just solved it, thanks for the correction.
 
Last edited:

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