A cyclist is riding with a speed of `27 km h^(-1)`. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate `0.5 ms^(-1)`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Explanation

To solve the problem step by step, we will break down the components of the cyclist's motion as he approaches the circular turn.

### Step 1: Convert Speed from km/h to m/s
The cyclist's speed is given as \(27 \, \text{km/h}\). We need to convert this speed into meters per second (m/s) using the conversion factor \(1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\).

\[ \text{Speed in m/s} = 27 \, \text{km/h} \times \frac{1}{3.6} \approx 7.5 \, \text{m/s} \]
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