Answer
$v = 15.6~m/s$
Work Step by Step
Particle 1:
$x = 6.00t^2+3.00t+2.00$
$v = 12.00t+3.00$
Particle 2:
$a = -8.00t$
$v = -4.00t^2+v_0$
$v = -4.00t^2+20$
We can equate the two expressions for $v$ to find the time $t$ when the velocities are equal:
$12.00t+3.00 = -4.00t^2+20$
$4.00t^2+12.00t-17 = 0$
We can use the quadratic formula:
$t = \frac{-12.00\pm \sqrt{(12.00)^2-(4)(4.00)(-17)}}{2(4.00)}$
$t = \frac{-12.00\pm \sqrt{416}}{8.00}$
$t = -4.05, 1.05$
Since time $t$ is positive, the solution is $t = 1.05~s$
We can find the velocity:
$v = 12.00t+3.00$
$v = (12.00)(1.05)+3.00$
$v = 15.6~m/s$
