Solve Uniform Acceleration: Find a = -12.4872 cm/s^2

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hsma
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1. An object moving with uniform acceleration has a velocity of 17.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.60 s later is -5.00 cm, what is its acceleration?




2.
[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]




3.
X = 5
Xo = 3
V = 17
Vf = 17
a = ?
t = 2.6

Does these values seem correct? I plug them into the equation above but I am not getting the right answer. i got -12.4872 cm/s. what am i doing wrong?
 
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hsma said:
1. An object moving with uniform acceleration has a velocity of 17.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.60 s later is -5.00 cm, what is its acceleration?




2.
[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]




3.
X = 5
Xo = 3
V = 17
Vf = 17
a = ?
t = 2.6

Does these values seem correct? I plug them into the equation above but I am not getting the right answer. i got -12.4872 cm/s. what am i doing wrong?


The value you're using for X is wrong—note that the object starts to the right of the origin and ends at the left for this period of time.
 
oh man -5!, i thought this whole time it was 5. well this clears up a lot of headache