An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 12.0 m from time t = 1.00 s to time t = 2.00 s. Calculate the acceleration of the object.
x(t)=v0t+at22x(t) = v_0t +\frac{at^2}{2}x(t)=v0t+2at2
v0=0v_0 =0v0=0
x(t)=at22x(t) = \frac{at^2}{2}x(t)=2at2
x(1)=a2x(1) = \frac{a}{2}x(1)=2a
x(2)=2ax(2) = 2ax(2)=2a
x(2)−x(1)=12x(2)-x(1)=12x(2)−x(1)=12
x(2)−x(1)=2a−a2=3a2x(2)-x(1)=2a-\frac{a}{2}=\frac{3a}{2}x(2)−x(1)=2a−2a=23a
3a2=12\frac{3a}{2}=1223a=12
a=8a=8a=8
Answer: 8 ms2\text{Answer: }8 \ \frac{m}{s^2}Answer: 8 s2m
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