Solution 18.4c
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(New page: From part a) <math>a=\frac{dv}{dt}=36-6t</math> We see <math>a</math> decreases as the time <math>t</math> increases. Thus the maximum acxceleration is at <math>t=0</math> giving <math...)
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Revision as of 17:52, 6 October 2010
From part a)
\displaystyle a=\frac{dv}{dt}=36-6t
We see \displaystyle a decreases as the time \displaystyle t increases. Thus the maximum acxceleration is at \displaystyle t=0 giving
\displaystyle \text{a}\ \text{=}\ \text{36 m}{{\text{s}}^{-\text{2}}}
