A) 24 \[\text{m}{{\text{s}}^{\text{-2}}}\]
B) zero
C) 6 \[\text{m}{{\text{s}}^{\text{-2}}}\]
D) 12 \[\text{m}{{\text{s}}^{\text{-2}}}\]
Correct Answer: D
Solution :
Given, \[x=8+12\,t-{{t}^{3}}\] We know \[v=\frac{dx}{dt}\]and\[a=\frac{dv}{dt}\]So,\[v=12-3\,{{t}^{2}}\]and\[a=-6t\] \[At\,t=2s\]\[v=0\]and\[a=-6t\]At \[t=2s\] \[a=-12\,m/{{s}^{2}}\] So, retardation of the particle \[=12\,\,m/{{s}^{2}}\].
You need to login to perform this action.
You will be redirected in
3 sec
