Three forces act on a moving object. one force has a magnitude of 80.0 n and is directed due north. another force has a magnitude of 60.0 n and is directed due west. what must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?
force 1 = 80.0 n due north.
force 2 = 60.0 n due west.

In order for the object to continue with constant velocity (no acceleration), the forces on it must be balanced.  That means they all have to add up to zero.

-- The 3rd force has to cancel 80N north, so part of it has to be 80N south.

-- It also has to cancel 60N west, so another part of it has to be 60N east.

The combined magnitude of these two 'components' is

Mag = √(80² + 60²)

Mag = √(6,400 + 3,600)

Mag = √(10,000)

Magnitude = 100 Newtons

For the direction of the new force . . .

We know it'll be pushing partly toward South and partly toward East. So when you put these together, it's pushing somewhere in the neighborhood of Southeast.

If you draw yourself a picture of the two components, and then draw the resultant (the combination of both), you'll have a right triangle.  You'll see that the components are the legs of the triangle, and the angle 'below' east has a tangent that's (the 80) divided by (the 60) ... which is 1.333 .

Now you go to your calculator and find the angle that has a tangent of 1.333 .  That's 53.13° .

So the new third force, with a magnitude of 100 Newtons, has to point in the direction of 53.13° south of east, in order to exactly cancel out the other two forces.  There's no other way to do it with just one single new force.

good for her? ? where is the question?

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer