A) \[24\,\,m\]
B) \[40\,\,m\]
C) \[56\,\,m\]
D) \[16\,\,m\]
Correct Answer: C
Solution :
Key Idea: Speed is rate of change of distance. Distance travelled by the particle is \[x=40+12t-{{t}^{3}}\] We know that, speed is rate of change of distance\[i.e.,\,\,v=\frac{dx}{dt}\] \[\therefore \] \[v=\frac{d}{dt}(40+12t-{{t}^{3}})\] \[=0+12-3{{t}^{2}}\] But final velocity\[v=0\] \[\therefore \] \[12-3{{t}^{2}}=0\] or \[{{t}^{2}}=\frac{12}{3}=4\] or \[t=2\,\,s\] Hence, distance travelled by the particle before coming to rest is given by \[x=40+12(2)-{{(2)}^{3}}\] \[=40+24-8=64-8\] \[=56\,\,m\]
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