A) 24 m
B) 40 m
C) 56 m
D) 16 m
Correct Answer: C
Solution :
Distance travelled by the particle is \[x=40+12t-{{t}^{3}}\] \[v=\frac{dx}{dt}=12-3{{t}^{2}}\] But final velocity \[v=0\] \[12-3{{t}^{2}}=0\] \[{{t}^{2}}=\frac{12}{3}=4\] \[t=2S\] \[x=40+12(2)-8\] \[=56\,m\]You need to login to perform this action.
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