A) \[\sqrt{15}\]
B) \[\sqrt{14}\]
C) 7
D) \[\sqrt{7}\]
Correct Answer: B
Solution :
\[=-\frac{2}{3}\] (say) \[(V-20)=-\frac{2}{3}\,(S-0)\] \[\Rightarrow \] \[v=20-\frac{2}{3}\,S\] \[S=15\] \[{{\left. v=\frac{ds}{dt} \right|}_{S=15\,m}}{{\left. =-\frac{2}{3}\,\frac{dS}{dt} \right|}_{S=15\,m}}=-\frac{20}{3}\,m{{s}^{-2}}\] \[=\frac{dv}{dt}=-\frac{2}{3}\,\frac{dS}{dt}\] \[\therefore \] \[{{\left. \frac{dV}{dt} \right|}_{S=15\,m}}{{\left. =-\frac{2}{3}\frac{dS}{dt} \right|}_{S=15\,m}}=-\frac{20}{3}\,m{{s}^{-2}}\] So, the point on the place is (1, 2, 3) \[=(2mv)\] The distance of P(1, 2, 3) from origin is \[=(2mv)\,(nt)\]
You need to login to perform this action.
You will be redirected in
3 sec
