Two objects A and B are moving in opposite directions toward each other in one d

Question

Two objects A and B are moving in opposite directions toward each other in one dimensional motion along the East-West (-X, +X) line. Initially they have velocities of equal magnitude, 3.00m/sec, with the first particle subject to a uniform acceleration of 2.00m/sec2 directed in the positive +X direction and B, which is 20.00 meters away in the +X direction from A ,is subject to a uniform acceleration of 1.00m/sec2 directed in the -X direction.
(a): Obtain the distance from the initial position A of the first object and the direction of the point where they meet and the time t at which the meeting occurs.
(b): Also obtain the magnitudes and directions of the velocities Va and the Vb respectively of the two particles at the meeting point O.

[Hints: (1): You can use the four basic equations relating velocity, uniform acceleration, time and displacement for one dimensional motion derived in class.
(2) For any quadratic equation that may occur use the fact that for Ax2+BX+C=0, the roots are: (the quadratic equation)

Explanation / Answer

This may not give you the answer, but it will at least help you with what's going on. A velocity = 3.0 acceleration = +2.0 x (position) = -20 B v = 3.0 a = -1.0 x = 0 I'll write it out iteratively as A(3,2,-20); B(3,-1,0) t=0 A(3,2,-20); B(3,-1,0) t=1 A(5,2,-17); B(2,-1,3) t=2 A(7,2,-12); B(1,-1,5) t=3 A(9,2,-5); B(0,-1,5) t=4 A(11,2,4); B(-1,-1,5) t=5 A(13,2,15); B(-2,-1,4) So, A overtakes B somewhere shortly after 4 seconds. You could start with the values I've provided at 4 seconds, which could give an easy problem, or you could start from the beginning. Relevant equation would be the equation of motion that states: x (position) = xi (initial position) + v0 (initial velocity) * t (time) + 1/2 * a (acceleration) * t^2 Given that we want positions to be equal, this should just be plug and chug.

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