Amoderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration = (4.80 m/s2)ã® + (7.00 m/s2)äµ. at time t = 0, the velocity is (4.4 m/s)ã®. what are magnitude and angle of its velocity when it has been displaced by 12.4 m parallel to the x axis?

Acceleration, a = ( 4.80 i+7.00 j )

We have acceleration a = Horizontal component i + Vertical component j

                               So  Horizontal component =4.80

                                      Vertical component =7.00

We have horizontal displacement ,

Here s = 12.4 m, u = 4.4 m/s and a = 4.80

Substituting

             Negative time is not possible.

Now we need to find velocity after 1.53 seconds

We have v = u+at

Horizontal case

        v = 4.4 + 4.80 * 1.53 = 11.744 i

Vertical case

        v = 0 + 7.00 * 1.53 = 10.71 j

So Velocity when it has been displaced by 12.4 m parallel to the x axis = (11.744 i+10.71 j) m/s

Magnitude =

Angle , tan θ =

   θ = 42.36⁰

c

explanation:

3.2x10^7