Question

You get on your bicycle and ride it with a constant acceleration of 0.5 m/s2 for 20s. After that, you continue riding at a constant velocity for a distance of 200 m. Finally, you slow to a stop, with a constant acceleration over a distance of 20 m 350 300 250 x (m) 200 150 100 50 10 30 t (s) 10 v (m/s) 10 10 a (m/s2) (a) How far did you travel while you were accelerating at 0.5 m/s2, and what was your velocity at the end of that interval? (b) After that, how long did it take you to cover the next 200 m? (c) What was your acceleration while you were slowing down to a stop, and how long did it take you to come to a stop? (d) Considering the whole trip, what was your average velocity? (e) Plot the position versus time, velocity versus time, and acceleration versus time graphs for the whole trip, in the grids provided above. Values at the beginning and end of each interval must be exact. Slopes and curvatures must be represented accurately. Do not draw any of the curves beyond the time the rider stops (or, if you do, make sure what you draw makes sense!)

Answer 1

@ xs Voxt to axt? - (O)(20) + x (0.5)(2012 at t2osce 1x = 100m "x= voxtant = 0+ (0.5)(20) x = lo mis. at t = 20 see © to 2000 - 200 = 20 seconds 7 t=2oscel 4²_urazas using this equaliers {t= 205ce] Vxf_vý = 2axx ax= x xf²_xx² = 0² (1032 100 -2.5 m/s2 ax -2.5 mise Average velocity - Total displacement - Total time taken time taken for deceleration motion Vai Vytant 0-10.4 see = -25

Total time taken T= 20 sect 2osect 4 see = 44see felel Xtola Ttolat - Total displacement total = 100m + 200m + 20m - 320 m ". Average velocity Vaug = lauga 320 (Yang= 7.2 mis x(m) t(s) v (m/s) t(s) -2.5 m/s^2 a (m/s)