## anonymous one year ago A particle moves with constant acceleration of 3.2 m/s2. At t = 4 s, it is at x = 110 m; at t = 6 s, it has a velocity v = 19 m/s. Find its position at t = 6 s.

use the equations of motion: $$x = ut + \frac{1}{2}at^2$$ $$v = u + at$$ plug in given information: $$110 = u (4) + \frac{1}{2}a (4)^2$$ $$19 = u + a(6)$$ you now have 2 unknowns and 2 equations, you can solve for u and a then find $$x(6)$$