Question

A cyclist moving on a level road at $4$ m/s stops pedalling and lets the wheel come to rest. The retardation of the cycle has two components: a constant $0.08{\text{m/s}}^2$ due to friction in the working parts and a resistance of $0.02v^2$/m where $v$ is speed in meters per second. The distance traveled by cycle before it comes to rest is: (Consider $\ln 5=1.61$).

• A. $\frac{161}{2}$
• B. $161$
• C. $\frac{161}{4}$
• D. $\frac{161}{8}$

Answer 1

The correct option is C $\frac{161}{4}$

Let the cyclist starting to move from the point $O$ and moving along $OX,$ attain a velocity $v$ at point $P$ in time $t$ such that $OP=x.$ Let the acceleration of the moving cycle at $P$ be $a.$Then we know that
$v=\frac{dx}{dt}$ and $a=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}=v\frac{dv}{dx}\cdots \left(1\right)$
By hypothseis, retardation $=0.08+0.02v^2=0.02(4+v^2)$
$\Rightarrow v\frac{dv}{dx}=-0.02(4+v^2)$
$\Rightarrow dx=-\frac{1}{0.02}\frac{vdv}{4+v^2}\cdots \left(2\right)$
Integrating equation $\left(2\right)$ between the limits $x=0;v=4$ m/s and $x=x^{\prime }$ meters, $v=0,$ we get
$\underset{0}{\overset{x^{\prime }}{\int }}dx=-\frac{1}{0.04}\underset{4}{\overset{0}{\int }}\frac{2vdv}{4+v^2}$
$\Rightarrow x^{\prime }=-\frac{1}{0.04}{\left[\ln (4+v^2)\right]}_4^0$
$\Rightarrow x^{\prime }=-\frac{1}{0.04}[\ln 4-\ln 20]$
$\Rightarrow x^{\prime }=\frac{\ln 5}{0.04}=\frac{1.61}{0.04}=\frac{161}{4}$ m