Solution 18.4c
From Mechanics
(Difference between revisions)
(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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<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 | + | 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> | ||
Current revision
From part a)
\displaystyle a=\frac{dv}{dt}=36-6t
We see \displaystyle a decreases as the time \displaystyle t increases. Thus the maximum acceleration is at \displaystyle t=0 giving
\displaystyle \text{a}\ \text{=}\ \text{36 m}{{\text{s}}^{-\text{2}}}
