Solution 18.4c - Revision history

Ian at 17:12, 27 March 2011

Ian — Sun, 27 Mar 2011 17:12:08 GMT

←Older revision		Revision as of 17:12, 27 March 2011
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	<math>a=\frac{dv}{dt}=36-6t</math>		<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	+	We see <math>a</math> decreases as the time <math>t</math> increases. Thus the maximum acceleration is at <math>t=0</math> giving

	<math>\text{a}\ \text{=}\ \text{36 m}{{\text{s}}^{-\text{2}}}</math>		<math>\text{a}\ \text{=}\ \text{36 m}{{\text{s}}^{-\text{2}}}</math>