speed and velocity

Question

A particle travels to the right at a constant rate of 7.1 m/s. It suddenly is given a vertical acceleration of 2.8 m/s 2 for 4.6 s. What is its direction of travel after the acceleration with respect to the horizontal? Answer between −180◦ and +180◦ .

a) Answer in units of ◦ .

What is the speed at this time?

b) Answer in units of m/s.

Answer 1

The velocity in the "x direction" (to the right doesn't change. It is maintained before and after the acceleration in the y direction (vertical). Initially, there is no vertical velocity (v_yi = 0) After the vertical acceleration, the velocity is: v_y = v_yi + a_yt = 0 + (2.8)(4.6) = 12.88 m/s. This is the y component (vertical component) of the velocity vector. The x component (horizontal component) of the velocity vector is still 7.1 m/s. Drawing the two vectors you create a triangle with a horizontal leg o 7.1 and a vertical leg of 12.88. To find the direction of travel, take the inverse tangent of (12.88/7.1) to get 61.13º = 61º (assuming going right is 0º)

The speed is the magnitude of the vector = sqrt(v_x² + v_y²) = sqrt(7.1² + 12.88²) = 14.7 m/s = 15 m/s