SPRING 2002 QUIZ 2

PHYSICS 220A #85077

1. At t =0, a particle starts from rest and moves in the xy-plane with an acceleration a = (4.0i + 3.0j)m/s². At t = 3.0s, determine (a) the magnitude and direction of the particle's velocity, and (b) the position of the particle.

Solution.

(a) We find the velocity from

             v = ò₀^3.0sadt = ò₀^3.0s(4.0i + 3.0j)dt

               = [(4.0t)|₀^3.0s i+ (3.0t)|₀^3.0s j]m/s

              = 12.0m/si + 9.0m/sj.

We find the magnitude of the velocity from

             v = (v_x² + v_y²)^1/2 = [(12.0m/s)² + (9.0m/s)²]^1/2  = 15.0m/s

We find the direction of  the velocity from

            tanq = v_y/v_x = (9.0m/s)/(12.0m/s) = 0.75, which gives q  = 36.7°.

(b) We find the position from

           r = ò₀^3.0s vdt = ò₀^3.0s [(4.0tm)i + (3.0tm)j]dt

             = 1/2[(4.0t²m)i + (3.0t²m)j]|₀^3.0s

             = (18.0m)i + (13.5m)j.

2. What average force is needed to accelerate a 6.0-kg block from rest to 15m/s over a distance of 1.2m?

Solution.

We find the acceleration of the block over the distance from

              v² = v₀² + 2a(x - x₀);

              (15m/s)² = 0 + 2a(1.2m - 0), which gives a = 93.75m/s².

We find the average force from

             F_av = ma = (6.0kg)(-93.75m/s²) = 5.63 x 10²N.