A) 24 m
B) 40 m
C) 56 m
D) 16 m
Correct Answer: C
Solution :
Distance travelled by the particle is \[x=40+12t-{{t}^{3}}\] We know that velocity is rate of change of distance \[\therefore \,\,\,\,\,\,\,\,\,\,\nu =\frac{d}{dt}(40+12t-{{t}^{3}})=0+12-3{{t}^{2}}\] but final velocity \[\nu = 0\] \[12=3{{t}^{2}}=0\,\,\,or\,\,\,{{t}^{2}}=\frac{12}{3}=4\] Or \[t=2\,s\] Hence, distance travelled by the particle before coming to rest is given by \[\operatorname{x}=40+12\left( b \right)-{{\left( b \right)}^{3}}=56\,\,m\]You need to login to perform this action.
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