Question 53: Motion in 1D

Question

The relation between time t and distance x for a moving body is given as t = mx
2
+ nx, where m and n are constants. The retardation of the motion is : (When v stands for velocity)

Accepted Answer

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:
Given:
$ t = mx^2 + nx $
First, differentiate $ t $ with respect to $ x $:
$ \frac{dt}{dx} = 2mx + n $
Since velocity $ v $ is defined as $ \frac{dx}{dt} $, we can write:
$ \frac{1}{v} = \frac{dt}{dx} = 2mx + n $
Thus,
$ v = \frac{1}{2mx + n} $
Next, to find the acceleration $ a $ (which is the derivative of velocity with respect to time), we start with the chain rule:
$ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $
Since $ v = \frac{dx}{dt} $, substituting $ x $ gives:
$ \frac{dx}{dt} = v $
Rewrite it:
$ \frac{dv}{dt} = \frac{dv}{dx} \cdot v $
Differentiate $ v $ with respect to $ x $:
$ v = (2mx + n)^{-1} $
$ \frac{dv}{dx} = -\frac{2m}{(2mx + n)^2} $
Then,
$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot v $
Substitute $ v = \frac{1}{2mx + n} $ into the expression:
$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot \frac{1}{2mx + n} $
Simplify the expression:
$ \frac{dv}{dt} = -2m \left( \frac{1}{2mx + n} \right)^3 $
$ a = -2m v^3 $
So, the retardation (negative acceleration) is:
$ a = -2m v^3 $

Answer

2 mv
3

Answer

2 mnv
3

Answer

2nv
3

Answer

2n
2
v
3

Motion in 1D

The relation between time t and distance x for a moving body is given as t = mx 2 + nx, where m and n are constants. The retardation of the motion is : (When v stands for velocity)

Choose the correct answer: