My Math Forum One Dimensional Motion

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September 8th, 2016, 07:28 PM   #1
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Joined: Nov 2015
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Math Focus: Calculus and Physics
One Dimensional Motion

Two particles move along an x axis. The position of particle 1 is given by x = 7.05t2 + 3.00t + 2.00 (in meters and seconds); the acceleration of particle 2 is given by a = −8.00t (in meters per seconds squared and seconds) and, at t = 0, its velocity is 16 m/s. When the velocities of the particles match, what is their velocity?

(my work is attached)

So my first plan of attack was to take the derivative of my position function and then integrate my acceleration function. When I do this, I should get two velocity functions. I did not no where to go from here, but then I sketched the graphs and saw that they should intersect when their velocity and time are the same (logically). So I set both the functions equal to each other and solved for t. I then plugged that value in for t to the functions and I got the same velocity back, 13.13 m/s. Could someone please tell me why I am totally wrong?

Thank you so much
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 September 8th, 2016, 07:48 PM #2 Senior Member     Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Okay so I forgot to carry my power after I integrated... How does one learn to stop making careless mistakes in math? Seriously... Algebraically, how do I go about solving equations when I have a t^2 on one side and a t^1 on the other such as: 14.1t+3.00 = -4.00t^2 + 16 I would need to take the square root of both sides, I am just not sure when to implement that step. I am sorry I have holes in my algebra knowledge.
 September 8th, 2016, 08:29 PM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,881 Thanks: 1503 Move all terms to one side, setting the equation equal to zero ... $4t^2+14.1t-13=0$ ... use the quadratic formula. Thanks from SenatorArmstrong

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