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Solution 18.4c

From Mechanics

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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)

a=dtdv=366t

We see a decreases as the time t increases. Thus the maximum acxceleration is at t=0 giving

a = 36 ms2