Thursday1 9th April2 2006

Apuntes

Thursday, 9th April, 2006 4. The study --- ``` v (m/s) Speed time graph, v-t 5 4 3 2 1 4 8 12 16 20 t/s S/m 150 100 50 4 8 12 16 20 t/s Position time graph, s-t In a uniform motion, the speed is constant so, the acceleration is zero as the average speed is 10/10/18 (Time) V = rac{d}{t} => d = v t => s - s_0 = v t - b => s - s_0 + v t Movement equation page 138 21) Draw graphs of a motion showing a constant speed of 20 m/s. Determine the distance traveled between t = 10s and t = 20s. Check that you get the same figure for both V (m/s) Area: 20 m/s x 10s = 200m t = 10s d = 2 p (d) = v t 10s - 0 s = 20 m/s x 10s = 200 m 200s - 0 s = 20 m/s x 20s = 400 m d2 - s - s_0 = 400 - 200m = 200m 20 0 10 10 20 t/s t = 10s ``` --- ```svg <svg width="200" height="200" xmlns="http://www.w3.org/2000/svg"> <line x1="50" y1="150" x2="150" y2="50" stroke="black" /> <line x1="50" y1="150" x2="50" y2="50" stroke="black" /> <line x1="50" y1="150" x2="150" y2="150" stroke="black" /> <line x1="100" y1="150" x2="100" y2="100" stroke="black" stroke-dasharray="4" /> <line x1="150" y1="150" x2="150" y2="50" stroke="black" stroke-dasharray="4" /> <line x1="50" y1="100" x2="150" y2="100" stroke...