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

To solve the problem of the cyclist coming to rest, we can break down the solution into several steps:

### Step 1: Understand the Problem
The cyclist is initially moving at a speed of \( v_0 = 4 \, \text{m/s} \) and experiences two types of retardation:
1. A constant retardation due to friction: \( a_1 = 0.08 \, \text{m/s}^2 \)
2. A velocity-dependent retardation: \( a_2 = 0.02v^2 \)

The total retardation \( a \) can be expressed as:
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