Question

The relation between time $t$ and distance $x$ for a moving body is given as $t=m{x}^2+nx$, where $m$ and $n$ are constants. The retardation of the motion is :
$\left(\text{Where}v\text{stands for velocity}\right)$

• A. $2m{v}^3$
• B. $2mn{v}^3$
• C. $2n{v}^3$
• D. $2n^2{v}^3$

Answer 1

Answer: (A)

Step 1: Given data

Let t, x, v, and a are time, distance, velocity and acceleration.

m and n are constant.
Relation between t and x is $t=m{x}^2+nx$

Step 2: Formula used

$\frac{dx}{dt}=v$

$\frac{dv}{dt}=a$

Step 3: Finding the velocity

Differentiating w.r.t. $x$, we get
$\frac{dt}{dx}=\frac{1}{v}=2mx+n$
$v=\frac{1}{2mx+n}$ .......(1)

Step 4: Finding the acceleration
Again differentiating w.r.t. $t$, we get
$a=\frac{dv}{dt}=-\frac{2m}{(2mx+n{)}^2}\left(\frac{dx}{dt}\right)$
Substituting $\left(1\right)$ in above equation,
$a=-\left(2m\right)v^3$
So, the retardation will be,
$-a=2m{v}^3$
Hence, the acceleration of the body is $2m{v}^3$.