A) 32 m
B) 54 m
C) 81 m
D) 24 m
Correct Answer: B
Solution :
Key Idea: At the instant when speed is maximum, its acceleration is zero. |
Given, die position \[x\] of a particle with respect to time \[t\] along x-axis |
\[x=9{{t}^{2}}-{{t}^{3}}\] ...(i) |
Differentiating Eq. (i), with respect to time, we get speed, i.e., |
\[v=\frac{dx}{dt}=\frac{d}{dt}(9\,{{t}^{2}}-{{t}^{3}})\] |
or \[v=18\,t-3{{t}^{2}}\] ...(ii) |
Again differentiating Eq. (ii), with respect to time, we get acceleration, i.e., |
\[a=\frac{dv}{dt}=\frac{d}{dt}(18\,t-3{{t}^{2}})\] |
or \[a=18-6t\] ...(iii) |
Now, when speed of particle is maximum, its acceleration is zero, i.e., |
\[a=0\] |
i.e., \[186t=0\] or \[t=3s\] |
Putting in Eq. (i), we obtain position of particle at that time |
\[x=9{{(3)}^{3}}-{{(3)}^{3}}=9(9)-27\] |
\[=81-27=54\,m\] |
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