# Uniform Acceleration Motion (UAM)

by xelrae
Tags: acceleration, motion, uniform
 P: 2 Hello I am new here. I came here because i was trying to search for a tutorial regarding Uniform acceleration motion or UAM but i am so unlucky since i havent found any decent ones that answers my questions. 1. The problem statement, all variables and given/known data An automobile and a race car start from rest at the same instant, with the race car intially behind by 121m. The automobile has a constant acceleration of 2.00m/s^2, and the race car has an acceleration of 3.20m/s^2 a) How much time does it take for the race car to overtake the automobile? b) Where does the car overtake the automobile? c) What is the speed of each when they are abreast? 2. Relevant equations http://en.wikipedia.org/wiki/Equations_of_motion im sorry for this, i dont know how to put the subscripts and other small symbols :( EDIT: v_{ave} = (v_i + v_f)/2 v = v_0 + a t x = x_0 + v_0 t + (1/2) a t^2 v$^{}$ = v$_{}$$^{}$ + 2 a \Delta x i dont know how to use the itex thing :( im sorry 3. The attempt at a solution A) my solution for a is simply equating them using the forumla V=V0t + 1/2(a(t^2)) (i cancelled out V0t since the initial velocity is zero) but adding 121m to the side of the automobile since it is ahead so that would give me 1/2(3.2)(t^2) = 1/2(2.0)(t^2)+121m 1.6(t^2) = 1(t^2)+121m 0.6(t^2) = 121m final answer is 14.2 B) x=V0t + 1/2(3.20)((14.2)^2) again, i cancelled out V0t since the initial velocity is zero so my final answer is about 322.67m C) RACE CAR V^2=V0^2 + 2ad V^2 = 2(3.2m/s^2)(322m) V=Square root of 2060.8 m^2/s^2 V=45.40 m/s AUTOMOBILE V^2=V0^2 + 2ad V^2 = 2(2.00m/s^2)(322m) V=Square root of 1288 m^2/s^2 V=45.40 m/s There. My major problem regarding this topic is i dont know WHEN to use WHAT equation. It really gives me a hard time thats why i resorted to this website in high hopes of finding great understanding. Hope you guys can help me :\ THANKS!