Answer:
15.6m/s
Explanation:
V1=[tex]\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2)[/tex] because the derivate of the position is the velocity
V1=12t+3
V2=20+[tex]\int\limits^_ {} \,[/tex]-8t because the integral of the acceleration is the velocity
V2=[tex]20-4t^{2}[/tex]
V1=V2 to see when the velocities of particles match
12t+3=20-4t^2
4t^2+12t-17=0 we resolve this with [tex]\frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}[/tex]
and we take the positif root
t=1.05 sec
if we evaluate the velocity (V1 or V2) the result is 15.6m/s
