A) \[2\alpha {{v}^{3}}\]
B) \[2\beta {{v}^{3}}\]
C) \[2\alpha \beta {{v}^{3}}\]
D) \[2{{\beta }^{2}}{{v}^{3}}\]
Correct Answer: A
Solution :
\[\frac{dt}{dx}=2\alpha x+\beta \Rightarrow v=\frac{1}{2\alpha x+\beta }\] \[\because \] \[a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}\] \[a=v\frac{dv}{dx}=\frac{-v.2\alpha }{{{(2\alpha x+\beta )}^{2}}}=-2\alpha .v.{{v}^{2}}=-2\alpha {{v}^{3}}\] \[\therefore \] Retardation \[=2\alpha {{v}^{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
