A) 36 m
B) 27 m
C) 21 m
D) 14 m
Correct Answer: D
Solution :
As given: \[\omega (t)\,=2+4{{t}^{2}}\,\left( \omega =\frac{d\theta }{dt} \right)\] So, \[\frac{d\theta }{dt}=2+4{{t}^{2}}\,\Rightarrow \,d\theta \,=(2+4{{t}^{2}})dt\] \[\Rightarrow \,d\theta \,=2dt+4{{t}^{2}}dt\] By integrating both sides, we get \[\int_{\theta }^{\theta }{d\theta =\int_{2}^{3}{2dt\,+\int_{2}^{3}{4{{t}^{2}}dt\,\Rightarrow \,\theta \,=[2t]_{2}^{3}+\left[ \frac{4{{t}^{3}}}{3} \right]_{2}^{3}}}}\] By putting the limit, \[\theta =(2\times 3-2\times 2)\,+\left[ \frac{4\times 27}{3}\,-\frac{4\times 8}{3} \right]\] \[\Rightarrow \,\theta =2+\left( \frac{108-32}{3} \right)\] \[=2+\frac{76}{3}=\frac{82}{3}\] As rotation through \[2\pi \] radian indicates that \[2\pi r\] distance moved by front wheel. So, \[\frac{82}{3}\,\,rad\,=\frac{2\pi r\times 82}{2\pi \times 3}\,=\frac{82\times 0.5}{3}\]\[=13.66\,m=14m\]
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