To find the magnitude and direction of the net acceleration of the cyclist on the circular turn, we'll need to consider both the centripetal acceleration due to the circular motion and the tangential acceleration due to the braking.
1. **Convert Speed from km/hr to m/s:**
The cyclist's initial speed is \(27 \text{ km/hr}\). To convert this to meters per second:
\[
27 \text{ km/hr} = \frac{27 \times 1000 \text{ meters}}{3600 \text{ seconds}} = 7.5 \text{ m/s}
\]
2. **Calculate Centripetal Acceleration:**
The formula for centripetal acceleration (\(a_c\)) is:
\[
a_c = \frac{v^2}{r}
\]
where \(v\) is the speed and \(r\) is the radius of the turn.
Given:
- \(v = 7.5 \text{ m/s}\)
- \(r = 80 \text{ m}\)
\[
a_c = \frac{(7.5)^2}{80} = \frac{56.25}{80} = 0.703 \text{ m/s}^2
\]
3. **Calculate Tangential Acceleration:**
The tangential acceleration (\(a_t\)) due to braking is given as \(0.5 \text{ m/s}^2\) (deceleration).
4. **Find the Magnitude of the Net Acceleration:**
The net acceleration (\(a_{net}\)) is the vector sum of the centripetal acceleration and the tangential acceleration. Since these accelerations are perpendicular to each other, we use the Pythagorean theorem to find the net acceleration:
\[
a_{net} = \sqrt{a_c^2 + a_t^2}
\]
Substituting the values:
\[
a_{net} = \sqrt{(0.703)^2 + (0.5)^2}
\]
\[
a_{net} = \sqrt{0.493 + 0.25} = \sqrt{0.743} \approx 0.862 \text{ m/s}^2
\]
5. **Determine the Direction of the Net Acceleration:**
The direction of the net acceleration is given by the angle \(\theta\) between the net acceleration vector and the centripetal acceleration vector. This angle can be found using the tangent function:
\[
\tan \theta = \frac{a_t}{a_c}
\]
Substituting the values:
\[
\tan \theta = \frac{0.5}{0.703} \approx 0.711
\]
Therefore:
\[
\theta = \tan^{-1}(0.711) \approx 35.3^\circ
\]
The net acceleration makes an angle of approximately \(35.3^\circ\) with the direction of the centripetal acceleration (which points toward the center of the circle).
In summary, the magnitude of the net acceleration is approximately \(0.862 \text{ m/s}^2\), and the direction is \(35.3^\circ\) relative to the centripetal acceleration direction.
