Gadget said:
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 can't 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
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 x
0 to when it's at current position "x", and v
0 is the car's (unknown) velocity when it was at initial position x
0.
let x
0 be the car position at time t
0=8 sec. Then let x
1 be the car position at time t
1=10 sec, and let x
2 be the car position at time t
2=12 sec. Then you can form 2 equations in 2 unknowns (v
0 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:
(t
1 - t
0) = 10 - 8 = 2 sec
(x
1 - x
0) =
Read From Graph
(t
2 - t
0) = 12 - 8 = 4 sec
(x
2 - x
0) =
Read From Graph (= 64 meters which u already know)
SOLVE for (a) by placing these values in the above 2 equations and eliminating v
0.
.
