A cyclist moving on a level road at 4 m/s stops pedalling and lets the wheels come to rest. The retardation of the cycle has two components: a constant 0.08 `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. What distance is traversed by the cycle before it comes to rest? (consider 1n 5=1.61).

Explanation

Let the cyclist starting to move from the point O and moving along OX, attains 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=(dx)/(dt)` and `a=(dv)/(dt) = (d^(2)x)/(dt^(2)) = v(dv)/(dt)` ............(1)
By hypothesis, retardation `=0.08 + 0.02v^(2) = 0.02 (4+v^(2))`
or `v(dv)/(dx) = -0.02(4+v^(2))`
or `dx= -1/0.02 (vdv)/(4+v^(2))` .............(2)
Integrating equation (2) between the limits `x=0`, v=4 m/x, and `x=x^(')` meters, v=0, we get
`int_(0)^(x^(')) dx=-1/0.04 int_(4)^(0)(2vdv)/(4+v^(2))`
or `x^(') = -1/0.04 ["ln"4-"ln"20]`
`=("ln"5)/(0.04) = 1.61/0.04 = u161/4m`