The relation between time t and distance x for a moving body is given as t = mx² + nx, where m and n are constants. The retardation of the motion is : (When v stands for velocity)

<p>The relationship between time $ t $ and distance $ x $ for a moving body is given by $ t = mx^2 + nx $, where $ m $ and $ n $ are constants. To determine the retardation (negative acceleration) of the motion, let's follow the steps to derive it:</p> <p>Given:</p> <p>$ t = mx^2 + nx $</p> <p>First, differentiate $ t $ with respect to $ x $:</p> <p>$ \frac{dt}{dx} = 2mx + n $</p> <p>Since velocity $ v $ is defined as $ \frac{dx}{dt} $, we can write:</p> <p>$ \frac{1}{v} = \frac{dt}{dx} = 2mx + n $</p> <p>Thus,</p> <p>$ v = \frac{1}{2mx + n} $</p> <p>Next, to find the acceleration $ a $ (which is the derivative of velocity with respect to time), we start with the chain rule:</p> <p>$ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $</p> <p>Since $ v = \frac{dx}{dt} $, substituting $ x $ gives:</p> <p>$ \frac{dx}{dt} = v $</p> <p>Rewrite it:</p> <p>$ \frac{dv}{dt} = \frac{dv}{dx} \cdot v $</p> <p>Differentiate $ v $ with respect to $ x $:</p> <p>$ v = (2mx + n)^{-1} $</p> <p>$ \frac{dv}{dx} = -\frac{2m}{(2mx + n)^2} $</p> <p>Then,</p> <p>$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot v $</p> <p>Substitute $ v = \frac{1}{2mx + n} $ into the expression:</p> <p>$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot \frac{1}{2mx + n} $</p> <p>Simplify the expression:</p> <p>$ \frac{dv}{dt} = -2m \left( \frac{1}{2mx + n} \right)^3 $</p> <p>$ a = -2m v^3 $</p> <p>So, the retardation (negative acceleration) is:</p> <p>$ a = -2m v^3 $</p> <style> <p>table {</p> <p>width: 100%;</p> <p>border-collapse: collapse;</p> <p>margin: 20px 0;</p> <p>}</p> <p>th {</p> <p>background-color: blue;</p> <p>color: white;</p> <p>border: 1px solid lightgrey;</p> <p>padding: 10px;</p> <p>}</p> <p>td {</p> <p>border: 1px solid lightgrey;</p> <p>padding: 10px;</p> <p>text-align: left;</p> <p>}</p> </style>