Dog sliding on frictionless pond with wind force

[macgirl06]

Don't know what I am doing wrong

An adventurous dog 7.50 kg dog runs onto a frozen, frictionless pond with an initial velocity of 4.31 m/s along the positive x-axis. The dog slides across the ice while being pushed by a steady wind with a force of 16.1 N in the positive y-direction. What is the magnitude of the dog's displacement (relative to where he came onto the ice) after 2.55s?

Can't I just use, Xf=Xi + Vixt + 0.5a(t^2) and Yf=Yi + Viyt + 0.5a(t^2)

 

[denverdoc]

post a little more work, maybe.

x(t) is simple and uses one term.
y(t) is same, but diff terms.

displacement is not x(t)+y(t), however, in case that's what was assumed.

 

[Doc Al]

Can't I just use, Xf=Xi + Vixt + 0.5a(t^2) and Yf=Yi + Viyt + 0.5a(t^2)

Sure. That will give you the components of the displacement, if you use the correct initial values and accelerations. What did use? Hint: Each component has a different acceleration and initial speed.