Solved

The Initial Value Problem (Lx)d2xdt2(dxdt)2=Lg,x(0)=0,x(0)=0( L - x ) \frac { d ^ { 2 } x } { d t ^ { 2 } } - \left( \frac { d x } { d t } \right) ^ { 2 } = L g , x ( 0 ) = 0 , x ^ { \prime } ( 0 ) = 0

Question 26

Multiple Choice

The initial value problem (Lx) d2xdt2(dxdt) 2=Lg,x(0) =0,x(0) =0( L - x ) \frac { d ^ { 2 } x } { d t ^ { 2 } } - \left( \frac { d x } { d t } \right) ^ { 2 } = L g , x ( 0 ) = 0 , x ^ { \prime } ( 0 ) = 0 is a model of a chain of length LL falling to the ground, where x(t) x ( t ) represents the length of chain on the ground at time tt . The solution for v=dxdtv = \frac { d x } { d t } in terms of xx is


A) v=Lg(L2(Lx) 2) v = \sqrt { L g \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) }
B) v=(Lg(L2+(Lx) 2) ) /(Lx) v = \left( \operatorname { Lg } \left( L ^ { 2 } + ( L - x ) ^ { 2 } \right) \right) / ( L - x )
C) v=(Lg(L2(Lx) 2) ) /(Lx) v = \left( \operatorname { Lg } \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) \right) / ( L - x )
D) v=Lg(L2+(Lx) 2) /(Lx) v = \sqrt { L g \left( L ^ { 2 } + ( L - x ) ^ { 2 } \right) } / ( L - x )
E) v=Lg(L2(Lx) 2) /(Lx) v = \sqrt { L g \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) } / ( L - x )

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions