Posted by ben on Sunday, September 17, 2006 at 2:29pm.
What are the magnitude and direction of the particle’s acceleration? If a particle is moving along an x axis with a constant acceleration. Some points on the graph are (0, -2), (1,0), (2,6).
I got 6 m/s to the right. is this correct?
<< got 6 m/s to the right. is this correct? >>
No. The answer must have dimensions of m/s^2, not m/s.
Is the first variable in parentheses time (t)? If so, the particle moves 2 units to the right in the first second and 6 units to the right (for a total of 8 units) in the second second.
Use the equation:
change in X = (1/2) a t^2
When t = 1 s, X =2m, so
2 = (1/2) a
a = 4 m/s^2 during the first second.
Try convincing yourself that the particle accelerates at the same rate during the time from t=1 to t=2 s
I am assuming that distances are measured in meters.
I'm supposing that the first variable is t and the second is x. If this is correct then
t x
0 -2
1 0
2 6
If the accleration is constant, then we have a second degree equation that looks like
x = ((1/2)a*t^2 + b*t + c
When t=0, x=c=-2 so we have c figured out.
When t=1, (1/2)a*1+b*1-2=0 and
when t=2, (1/2)a*4+b*2-2=6
This is two equations in two unknowns, so if you add -4 times the first to the second you get
-2b+6=6 so b=0
I'll let you figure the constant for a, which happens to be the acceleration too.
You have the direction right, it's positive, but I don't think the magnitude is right.
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