An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.94 cm. If its x-coordinate 2.35 s later is −5.00 cm,what is its acceleration?

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u=11 cm/s

s=-5cm

t=2.35

we can apply s=ut+1/2at^2 to solve equation

-5=11*2.35+1/2*a*2.35^2

a=-11.17 cm/s

so the acceleration is

-11.17 cm/s

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