# Constant acceleration physics problem

1. Jul 23, 2005

Ok I have a question for everyone. Please help!!!! I was given a graph with the problem, but I am going to try to put it in words. In the X-axis there is seconds. In the Y-axis there is meters. From 8 sec - 12 sec the car traveled 64 meters. The distance before the 64 meters is not given. It also states that the car is traveling under constant acceleration. Question is, what is the acceleration. I tried to solve it by first getting a velocity, 64 meters / 4 sec = 16 m/s, I used v= v_o + a t to get acceleration, a = 4 m/s^2, then I tried to use x = x_o + v_o t + 1/2 a t^2, but I realized that I cant use the v_o as zero because I am only taking the 4 sec not the entire 12 sec. What is it that I need to do. Find the velocity at 8 sec? How do you do that if a distance wasn't given to me?
Thank You

2. Jul 23, 2005

### geosonel

you're correct that under constant acceleration (a), the x position is given by:

$$x \ = \ x_{0} + v_{0} \Delta t + \frac{a ( \Delta t )^{2}}{2}$$

or:

$$\left ( x - x_{0} \right ) \ = \ v_{0} \Delta t + \frac{a \left ( \Delta t \right )^{2}}{2}$$

where Δt is the elapsed time from when the car was at initial position x0 to when it's at current position "x", and v0 is the car's (unknown) velocity when it was at initial position x0.

let x0 be the car position at time t0=8 sec. Then let x1 be the car position at time t1=10 sec, and let x2 be the car position at time t2=12 sec. Then you can form 2 equations in 2 unknowns (v0 and a) and solve for (a):

$$( x_{1} - x_{0} ) \ = \ v_{0} ( t_{1} - t_{0} ) + \frac{a ( t_{1} - t_{0} )^{2}}{2}$$

$$( x_{2} - x_{0} ) \ = \ v_{0} ( t_{2} - t_{0} ) + \frac{a ( t_{2} - t_{0} )^{2}}{2}$$

where:

(t1 - t0) = 10 - 8 = 2 sec
(x1 - x0) = Read From Graph
(t2 - t0) = 12 - 8 = 4 sec
(x2 - x0) = Read From Graph (= 64 meters which u already know)

SOLVE for (a) by placing these values in the above 2 equations and eliminating v0.

Last edited: Jul 23, 2005
3. Jul 24, 2005

### VietDao29

I wonder if you could read this from graph...
Another hint if you like : You should assume at the time t_0 = 0, the car starts to go. That means at t_0 = 0s, v_0 = 0 m/s.
$$v = v_0 + at = at$$ as $$v_0 = 0 \mbox{m / s}$$
so at t = 8 seconds $$v_8 = 8a$$.
Can you find the distance the car goes between t = 8 and t = 12 with respect to a? From there, you can easily solve the problem.
Hope you get it.
Viet Dao,

Last edited: Jul 24, 2005
4. Jul 24, 2005

### mukundpa

If you can get slop of the tangent at t = 8 second that will be velocity at that time.

5. Jul 24, 2005

### marlon

Gadget, i know your question has been answered but nevertheless i would like to point your attention to this intro on Newtonian Physics. Perhaps it might be of some interest to you in the near future .
Newtonian Mechanics Intro

regards
marlon