A particle starts from the origin at t = 0 with a velocity of 6.0 m/s and moves in the xy plane with a constant acceleration of (-2.0 + 4.0) m/s2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin

9sqrt4

Explanation:

Assuming you mean 6.0i m/s and (-2.0i + 4.0j) m/s^2:

The maximum positive x coordinate is when the time is 3 seconds, because the horizontal velocity will be zero, and therefore will be a turning point, where after that point the x coordinate will only be more negative.

After 3 seconds the x coordinate is 6*3+1/2*-2*3^2=9, and the y coordinate will be 1/2*4*3^2=18.

The distance from the origin is sqrt(9^2+18^2) = 9*sqrt(1+4)

no. dot product   =0 when vecs at right angles

attach the statements so i can answer : )