A cylclist 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 of `0.5 ms^-2`. 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 calculate the tangential acceleration, centripetal acceleration, and then find the net acceleration of the cyclist.

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

\[ \text{Speed in m/s} = \frac{27 \times 1000 \, \text{m}}{3600 \, \text{s}} = 7.5 \, \text{m/s} \]

### Step 2: Calculate the Tangential Acceleration
The cyclist is applying brakes, which means he is decelerating at a constant rate of \(0.5 \, \text{m/s}^2\). This is the tangential acceleration (\(a_t\)).

\[ a_t = -0.5 \, \text{m/s}^2 \quad (\text{negative because it's deceleration}) \]

### Step 3: Calculate the Centripetal Acceleration
Centripetal acceleration (\(a_c\)) is given by the formula:

\[ a_c = \frac{v^2}{r} \]

Where:
- \(v = 7.5 \, \text{m/s}\) (speed of the cyclist)
- \(r = 80 \, \text{m}\) (radius of the circular turn)

Now substituting the values:

\[ a_c = \frac{(7.5)^2}{80} = \frac{56.25}{80} = 0.703125 \, \text{m/s}^2 \]

### Step 4: Calculate the Net Acceleration
The net acceleration (\(a_{net}\)) is the vector sum of the tangential acceleration and centripetal acceleration. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:

\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]

Substituting the values:

\[ a_{net} = \sqrt{(-0.5)^2 + (0.703125)^2} \]

Calculating each term:

\[ a_{net} = \sqrt{0.25 + 0.495\ldots} = \sqrt{0.745\ldots} \approx 0.864 \, \text{m/s}^2 \]

### Step 5: Determine the Direction of the Net Acceleration
The direction of the net acceleration can be found using the angle \(\theta\) with respect to the radial direction (centripetal acceleration). The angle can be found using:

\[ \tan(\theta) = \frac{a_t}{a_c} \]

Substituting the values:

\[ \tan(\theta) = \frac{-0.5}{0.703125} \]

Calculating \(\theta\):

\[ \theta = \tan^{-1}\left(\frac{-0.5}{0.703125}\right) \approx -36.87^\circ \]

This indicates that the net acceleration is directed inward (towards the center of the circular path) and downward (due to the braking).

### Final Result
The magnitude of the net acceleration is approximately \(0.864 \, \text{m/s}^2\) and it is directed at an angle of approximately \(-36.87^\circ\) from the radial direction towards the center of the circular path.

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