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