# Ok, a simple question...

by Matty T
Tags: simple
 P: 1,004 Are we to assume that the acceleration from 0 to 60 mph is constant, and that the acceleration from 60 mph to 100 mph is constant (but different)? Anyway, you can solve this problem with one formula: $$V_f^2 = V_0^2 + 2ax$$ $$a = \frac{V_f - V_0}{t}$$ $$x = \frac{V_f^2 - V_0^2}{2\frac{V_f - V_0}{t}} = \frac{t}{2}(V_f + V_0)$$ Part 1: When the cars hit 60mph, car A will be at 57.6 meters and car B will be at 64.4 meters. Part 2: Car A's acceleration from 60 to 100 takes 7.3 seconds, whereas car B's takes only 6.8 seconds. The distance car A will pass between 60mph and 100mph is 261 meters, whereas car B will only pass 243.2 meters. The total distance of car A is 318.6 meters, and the total distance of car B is 307.6. So car A will be in front of car B by 11 meters when they both hit 100 mph. Of course, I'm not sure I'm right and I'm probably not.