Question

the deceleration experienced by a moving motor boat ,after engine is cut off is given by dv/dt= -kv^3,where k is constant .if v* is the magnitude of the velocity ay cut off ,the magnitude of the velocity at a time 't' after cut off is?

Answer 1

$\mathbf{Given},$
$\bold{\frac{dv}{dt}=-kv^3}\\\\\bold{\frac{dv}{v^3}=-kdt}\\\\\bold{\int\limits^{v_0}_v{\frac{dv}{v^3}}=-k\int\limits^t_0{dt}}$
⇒[1/2v₀² - 1/2v²] = -kt
⇒1/2v₀² + kt = 1/2v²
⇒(1 + 2v₀²kt) = 2v₀²/2v² = v₀²/v²
⇒v = v₉/√(1 + 2v₉²kt)

Hence, option (4) is correct

Answer 2

Given that the deceleration of the moving boat is $\frac { dv }{ dt } =-k{ v }^{ 3 };$ where k is given as constant and v is the the velocity at the time of engine being shut down. Therefore the velocity at a time after the engine is being shut down be vt.  

vt is calculated by the formula of equation of motion which is the second equation of motion which defines velocity at any instance.

Therefore,

$vt=\frac { v }{ \left( \sqrt { 2{ v }^{ 2 }kt+1 } \right) }$.