A) \[{{v}_{0}}t+\frac{1}{3}b{{t}^{2}}\]
B) \[{{v}_{0}}t+\frac{1}{3}b{{t}^{3}}\]
C) \[{{v}_{0}}t+\frac{1}{6}b{{t}^{3}}\]
D) \[{{v}_{0}}t+\frac{1}{2}b{{t}^{2}}\]
Correct Answer: C
Solution :
\[\frac{dv}{dt}=bt\Rightarrow dv=bt\ dt\Rightarrow v=\frac{b{{t}^{2}}}{2}+{{K}_{1}}\] At \[t=0,\ v={{v}_{0}}\Rightarrow {{K}_{1}}={{v}_{0}}\] We get \[v=\frac{1}{2}b{{t}^{2}}+{{v}_{0}}\] Again \[\frac{dx}{dt}=\frac{1}{2}b{{t}^{2}}+{{v}_{0}}\Rightarrowx=\frac{1}{2}\frac{b{{t}^{2}}}{3}+{{v}_{0}}t+{{K}_{2}}\] At \[t=0,\ x=0\Rightarrow {{K}_{2}}=0\] \[\therefore \]\[x=\frac{1}{6}b{{t}^{3}}+{{v}_{0}}t\]
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