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, we will break it down into steps:

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

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

### Step 2: Identify the radius of the circular turn
The radius of the circular turn is given as \(R = 80 \, \text{m}\).

### Step 3: Calculate the centripetal acceleration
Centripetal acceleration (\(a_c\)) can be calculated using the formula:

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

Substituting the values:

\[ a_c = \frac{(7.5)^2}{80} = \frac{56.25}{80} = 0.703125 \, \text{m/s}^2 \approx 0.7 \, \text{m/s}^2 \]

### Step 4: Identify the tangential acceleration
The cyclist is applying brakes, which means there is a tangential acceleration (\(a_t\)) acting in the opposite direction to the motion. The rate of deceleration is given as \(0.5 \, \text{m/s}^2\). Therefore:

\[ a_t = -0.5 \, \text{m/s}^2 \]

### Step 5: Calculate the net acceleration
The net acceleration (\(a_{net}\)) is the vector sum of the centripetal acceleration and the tangential acceleration. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the net acceleration:

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

Substituting the values:

\[ a_{net} = \sqrt{(0.7)^2 + (-0.5)^2} = \sqrt{0.49 + 0.25} = \sqrt{0.74} \approx 0.86 \, \text{m/s}^2 \]

### Step 6: Determine the direction of the net acceleration
The direction of the net acceleration can be found using the tangent of the angle (\(\theta\)) formed between the net acceleration and the centripetal acceleration:

\[ \tan(\theta) = \frac{a_t}{a_c} = \frac{-0.5}{0.7} \]

Calculating the angle:

\[ \theta = \tan^{-1}\left(\frac{-0.5}{0.7}\right) \]

This angle indicates that the net acceleration is directed inward towards the center of the circular path but has a downward component due to the braking.

### Final Answer
The magnitude of the net acceleration of the cyclist is approximately \(0.86 \, \text{m/s}^2\) and it is directed towards the center of the circular turn with a downward component.

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