Question

Two particles move along an x axis. The position of particle 1 is given by x ! 6.00t 2 # 3.00t # 2.00 (in meters and seconds); the acceleration of particle 2 is given by a ! "8.00t (in meters per second squared and seconds) and, at t ! 0, its velocity is 20 m/s. When the velocities of the particles match, what is their velocity?

Answer 1

Answer:

15.6m/s

Explanation:

V1=$\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2)$ because the derivate of the position is the velocity

V1=12t+3

V2=20+$\int\limits^_ {} \,$-8t because the integral of the acceleration is the velocity

V2=$20-4t^{2}$

V1=V2 to see when the velocities of particles match

12t+3=20-4t^2

4t^2+12t-17=0 we resolve this with $\frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}$

and we take the positif root

t=1.05 sec

if we evaluate the velocity (V1 or V2) the result is 15.6m/s