Question

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a= (4.50 m/s²) + (7.00 m/s²). At time t = 0, the velocity is (3.6 m/s)i. What are magnitude and angle of its velocity when it has been displaced by 12.7 m parallel to the x axis?

Answer 1

To find the magnitude and angle of the pebble's velocity, we need to first calculate its final velocity using the given acceleration and displacement. The displacement in the x-direction is 12.7 m, so we can use the equation vf = vi + at to find the final velocity in the x-direction. The angle of the velocity can be found using the equation tanθ = vy / vx.

Step-by-step explanation:

To find the magnitude and angle of the pebble's velocity, we need to first calculate its final velocity using the given acceleration and displacement. The displacement in the x-direction is 12.7 m, so we can use the equation vf = vi + at to find the final velocity in the x-direction. Since the acceleration is constant and in the x-direction, the initial velocity in the x-direction (3.6 m/s) remains constant. So, vf = 3.6 m/s + (4.50 m/s²)(12.7 m) = 3.6 m/s + 57.15 m/s = 60.75 m/s.

Next, we can find the angle of the velocity by using the equation tanθ = vy / vx. Since the acceleration is only in the y-direction, there is no change in the y-velocity. Therefore, vy = 0 m/s. Substituting these values into the equation, we get tanθ = 0 m/s / 60.75 m/s = 0. So, the angle θ is 0° (or 180°, depending on the convention used).