A) \[x(t)={{x}_{0}}+{{v}_{0}}t+\frac{1}{6}\alpha {{t}^{3}}+\frac{1}{12}\beta {{t}^{4}}\]
B) \[x(t)={{x}_{0}}+{{v}_{0}}t+\frac{1}{6}\alpha {{t}^{2}}+\frac{1}{24}\beta {{t}^{3}}\]
C) \[x(t)={{x}_{0}}+{{v}_{0}}t+\frac{1}{12}\alpha {{t}^{2}}+\frac{1}{6}\beta {{t}^{3}}\]
D) \[x(t)={{x}_{0}}+{{v}_{0}}t+\frac{1}{6}\alpha {{t}^{2}}+\frac{1}{12}\beta {{t}^{3}}\]
Correct Answer: A
Solution :
Rate of change of velocity gives acceleration\[(\overrightarrow{a})\]of the particle. That is \[\overrightarrow{a}=\frac{dv}{dt}\] Given, \[a=\alpha t+\beta {{t}^{2}}\] \[\Rightarrow \] \[\frac{dv}{dt}=(\alpha t+\beta {{t}^{2}})\] \[\Rightarrow \] \[dv=(\alpha t+\beta {{t}^{2}})dt\] Integrating it within the condition of motion \[\int_{{{v}_{0}}}^{v}{dv}=\int_{0}^{t}{(\alpha t+\beta {{t}^{2}})}dt\] \[\Rightarrow \] \[v-{{v}_{0}}=\frac{1}{2}\alpha {{t}^{2}}+\frac{\beta {{t}^{2}}}{3}\] \[\Rightarrow \] \[v={{v}_{0}}+\frac{1}{2}\alpha {{t}^{2}}+\frac{1}{3}\beta {{t}^{3}}\] \[\Rightarrow \] \[\frac{dx}{dt}={{v}_{0}}+\frac{1}{2}\alpha {{t}^{2}}+\frac{1}{3}\beta {{t}^{3}}\] Integrating it, using\[\int{{{x}^{n}}}dx=\frac{{{x}^{n+1}}}{n+1},\]we have \[\int_{{{x}_{0}}}^{x}{dx}=\int_{0}^{t}{({{v}_{0}}+\frac{1}{2}\alpha {{t}^{2}}+\frac{1}{3}\beta {{t}^{3}})}dt\] \[\Rightarrow \] \[x-{{x}_{0}}={{v}_{0}}t+\frac{1}{6}\alpha {{t}^{3}}+\frac{1}{12}\beta {{t}^{4}}\] \[\Rightarrow \] \[x={{x}_{0}}+{{v}_{0}}t+\frac{1}{6}\alpha {{t}^{3}}+\frac{1}{12}\beta {{t}^{4}}\] Or \[x(t)={{x}_{0}}+{{v}_{0}}t+\frac{1}{6}\alpha {{t}^{3}}+\frac{1}{12}\beta {{t}^{4}}\]
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