FALL 2003 PHYSICS 100A

#85014 QUIZ 1

1. A man runs 100m to the east, and then walks 50m back to the west. His average speed during the process is 5m/s. What is his average velocity at the same time interval? Take eastward as positive direction.

Solution.

The total distance covered during the process is d = (100m + 50m) = 150m.

We find the time elapsed during the journey from

            average speed = d/t;

            5m/s = (100m + 50m)/t, which give t = 30s.

We find the average velocity from

            v_av = Dx/Dt = (x₂ - x₁)/(t₂ - t₁) = (50m - 0)/(30s) = +16.67m/s.

2. A car traveling 45km/h slows down at a constant deceleration of 0.50m/s². Calculate (a) the distance the car coasts before it stops, and (b) the time it takes to stop.

Solution.

(a) We find the distance from

            v² = v₀² + 2aDx;

            0 = [(45km/h)(1000m/km)/(3.6ks)/h)]² + 2(-0.50m/s²)Dx, which gives Dx = 1.6 x 10²m.

(b) we find the time it takes to stop from

            v = v₀ + at;

            0 = [(45km/h)(1000m/km)/(3.6ks)/h)] + (-0.50m/s²)t, which gives t = 25s.