Uniform Acceleration Motion (UAM)

AI Thread Summary
The discussion focuses on solving a problem involving Uniform Acceleration Motion (UAM) where an automobile and a race car start from rest, with the race car initially 121 meters behind. The automobile accelerates at 2.00 m/s², while the race car accelerates at 3.20 m/s². The user successfully calculates the time it takes for the race car to overtake the automobile as 14.2 seconds, the distance at which this occurs as approximately 322.67 meters, and both vehicles' speeds at that point as 45.40 m/s. The user expresses confusion about when to use specific equations for UAM, and the community provides guidance on formula selection and corrections. Overall, the user gains confidence in their solutions after receiving feedback.
xelrae
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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 haven't found any decent ones that answers my questions.

Homework Statement


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?

Homework Equations



http://en.wikipedia.org/wiki/Equations_of_motion
im sorry for this, i don't 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 don't know how to use the itex thing :( I am sorry

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 canceled 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 canceled 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 don't know WHEN to use WHAT equation.
It really gives me a hard time that's why i resorted to this website in high hopes of finding great understanding. Hope you guys can help me :\ THANKS!
 
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Welcome to PF, xlrae.
Part (a) looks great, though the V=V0t + 1/2(a(t^2)) was confusing. Should be d = Vot + 1/2at² of course. You can copy the ² and other symbols from this page: https://www.physicsforums.com/blog.php?b=346
The trick to getting the right formulas is to write down your list of formulas and beside each, write what it is for. There are several formulas for constant accelerated motion, but you need the one giving distance as a function of time.

I think you have a mistake in (b). Use either
d = 1/2at² with the a=2 or d = 1/2at² - 121 with a = 3.2.

I used V = Vi + a*t for (c) and got the same answer as you.
 
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wow thanks for the reply!
actually, our teacher presented the answers and solutions to all the questions and i am right! thanks for the help!
 
Most welcome. Good luck with the next set of problems.
 
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