A) 0.3 ms-2
B) 0.03 ms-2
C) 3 ms-2
D) 0.09 ms-2
Correct Answer: A
Solution :
Given, \[\text{o }\!\!|\!\!\text{ }\,=A+Bt+C{{t}^{2}}+D{{t}^{3}}\] Angular velocity \[\omega =d\text{o}|/dt=\frac{d}{dt}(A+Bt+C{{t}^{2}}+D{{t}^{3}})\] \[=B+2Ct+3D{{t}^{2}}\] Acceleration \[\alpha =\frac{d\omega }{dt}=\frac{d}{dt}(B+2Ct+3D{{t}^{2}})\] \[\frac{d\alpha }{dt}=6D=6\times 1=6m/{{s}^{2}}\] Rate of change of tangential acceleration \[\left( \frac{d{{\alpha }_{t}}}{dt} \right)=\frac{d}{dt}(r\alpha )=r\frac{d\alpha }{dt}=5\times {{10}^{-2}}\times 6\] \[=0.3\,m/{{s}^{2}}\]
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