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., 18 ? 6t = 0 or t = 3s Putting in Eq. (i), we obtain position of particle at that time \[x=9{{(3)}^{3}}-{{(3)}^{3}}=9(9)-27\] \[=81-27=54\,m\]You need to login to perform this action.
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