A particle moving at a velocity of 8 m/s in the positive x direction is given an acceleration of 5.2 m/s 2 in the positive y direction for 7 s. What is the final speed of the particle?

Answer 1

Calculate separately the x and y components of the velocity, and then get the total speed and direction.

v_x = constant = 8 m/s

v_y(t) = v_0y + a_yt

v_0y = 0 (no initial speed in the y direction)

a_y = 5.2 m/s²

t = 7s

Plug the numbers for v_0y, a_y, and t into

v_y(t) = v_0y + a_yt

and get v_y in m/s. Then get the total speed v from

v = √(v_x² + v_y²)

Get the direction from

θ = arctan(v_y/v_x)

Plug in the numbers and finish from here. Note that in this problem, both v_x and v_y are positive, so θ is in quadrant I.