A) 20 m
B) 18 m
C) 16 m
D) 25 m
Correct Answer: B
Solution :
[b] \[\frac{dv}{dt}=6t+5\] \[\Rightarrow \]\[dv=(6t+5)dt\] \[v=6\frac{{{t}^{2}}}{2}+5t\] (on integration) \[\frac{ds}{dt}=3{{t}^{2}}+5t\] \[\Rightarrow \]\[ds=(3{{t}^{2}}+5t)dt\] Integrating, \[s=3\frac{{{t}^{3}}}{3}+5\frac{{{t}^{2}}}{2}={{t}^{3}}+5\frac{{{t}^{2}}}{2}\] \[\therefore \] At \[t=2\,s,\] distance covered, \[s={{(2)}^{3}}+\frac{5{{(2)}^{2}}}{2}=18\,m\]You need to login to perform this action.
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