The deceleration experienced by a moving moter boat, after its engine is cut-off is given (d v/d t)=-k v3 where k is constant and v0 is the magnitude of the velocity at time t after the cut off is

Question

The deceleration experienced by a moving moter boat, after its engine is cut-off is given (d v/d t)=-k v3 where k is constant and v0 is the magnitude of the velocity at time t after the cut off is (A) v0 c-k t (B) (v0/√2 v02 k t+1) (C) v0 (D) (v0/2)

Tardigrade Admin · Accepted Answer

Here, (d v/d t)=-k v3 or (d v/v3)=-k d t or ∫ limitsv0v (d v/v3)=∫ limits0t-k d t or [(-1/2 v2)]v0v=-k d t or (1/2 v02)-(1/2 v2)=-k t or (1/2 v2)=(1/2 v2)+k t or (1/2 v2)=(1+2 v02 k t/2 v02) or v2 =(v02/2 v02 k t+1) or v =(v0/√(2 v02 k t+1))

Q. The deceleration experienced by a moving moter boat, after its engine is cut-off is given $\frac{d v}{d t} = - k v^{3}$ where $k$ is constant and $v_{0}$ is the magnitude of the velocity at time $t$ after the cut off is

$v_{0} c^{- k t}$

$\frac{v _{0}}{2 v _{0}^{2} k t + 1}$

$v_{0}$

$\frac{v _{0}}{2}$

Q. The deceleration experienced by a moving moter boat, after its engine is cut-off is given dtdv​=−kv3 where k is constant and v0​ is the magnitude of the velocity at time t after the cut off is

v0​c−kt

2v02​kt+1​v0​​

v0​

2v0​​

Here, dtdv​=−kv3 or v3dv​=−kdt or v0​∫v​v3dv​=0∫t​−kdt or [2v2−1​]v0​v​=−kdt or 2v02​1​−2v21​=−kt or 2v21​=2v21​+kt or 2v21​=2v02​1+2v02​kt​ or v2=2v02​kt+1v02​​ or v=(2v02​kt+1)​v0​​

Q. The deceleration experienced by a moving moter boat, after its engine is cut-off is given $\frac{d v}{d t} = - k v^{3}$ where $k$ is constant and $v_{0}$ is the magnitude of the velocity at time $t$ after the cut off is

$v_{0} c^{- k t}$

$\frac{v _{0}}{2 v _{0}^{2} k t + 1}$

$v_{0}$

$\frac{v _{0}}{2}$