Sunday

February 1, 2015

February 1, 2015

Posted by **ben** on Sunday, September 17, 2006 at 2:29pm.

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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