A particle passes through the origin with an initial speed of 5.6 m/s in the positive y direction. The particle accelerates in the negative x direction at 5.2 m/s2. (a) What are its x and y positions after 4.0 s? m (x position) m (y position) (b) What are vx and vy at this time? m/s (vx) m/s (vy) This question seems pretty hard to me, I have tried about 4 times to get the answer to no avail, heres what I have done. To try and find out where its x and y positions were after 4.0 seconds, I plugged it into the Position Versus Time Equation, xf=xi+vixt+.5axt^2 Since I have the givens time = 4.0 Vi=5.6 a=-5.2^2 I plug it into the equation. xf=xi+vixt+.5axt^2 xf=0+5.6(4)+.5(-5.2)^2(4.0)^2 xf=238.72 however, this answer for part A. was not correct. Can anyone tell me what I am doing wrong here?