Question

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.

Answer 1

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 \(x(6)\)