A) 0
B) 4 unit
C) 8 unit
D) 12 unit
Correct Answer: C
Solution :
[c] Given rule is: Distance, \[s=2-3t+4{{t}^{3}}\] \[\Rightarrow \] Velocity \[=\frac{ds}{dt}=-3+12{{t}^{2}}\] \[\Rightarrow \] Acceleration \[=\frac{{{d}^{2}}s}{d{{t}^{2}}}=16t\] Since, velocity is zero \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{ds}{dt}=0;\,\,\,\Rightarrow \,\,\,\,0=-3+12{{t}^{2}}\] \[\Rightarrow t=\frac{\sqrt{3}}{12}=\frac{1}{2}\Rightarrow \frac{{{d}^{2}}s}{d{{t}^{2}}}=16t\] Acceleration (when velocity is zero) \[=16\times \frac{1}{2}=8\] unit.
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