Question

Two particles move along an x axis. the position of particle 1 is given by x = 9.00t2 + 6.00t + 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 23.0 m/s. when the velocities of the particles match, what is their velocity?

Answer 1

For particle 1, we can find for velocity by taking the 1st derivative of the equation.

x = 9 t^2 + 6 t + 2

dx / dt = 18 t + 6

v1 = 18 t + 6

For particle 2, we are given the acceleration a, and initial velocity v0, from this, we can get the velocity v using the formula:

v = v0 + a t

v2 = 23 – 8 t^2

So when v1 = v2, we get t:

18 t + 6 = 23 – 8t^2

8t^2 + 18t = 17

t^2 + 2.25 t = 2.125

Completing the square:

(t + 1.125)^2 = 3.390625

t + 1.125 = ± 1.84

t = -2.965, 0.715

Since time cannot be negative,

t = 0.715 seconds

v1 = 18 t + 6 = 18.87 m/s = v2

Answer:

18.87 m/s