$A$ cyclist is riding with a speed of $27 \; km/h$. 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 $0.50 \; m/s^2$. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

(N/A) Speed of the cyclist,$v = 27 \; km/h = 7.5 \; m/s$.
Radius of the circular turn,$r = 80 \; m$.
Centripetal acceleration is given as:
$a_c = \frac{v^2}{r} = \frac{(7.5)^2}{80} = 0.703 \; m/s^2 \approx 0.7 \; m/s^2$.
Tangential acceleration is given as $a_T = 0.5 \; m/s^2$.
Since the angle between $a_c$ and $a_T$ is $90^{\circ}$,the resultant acceleration $a$ is given by:
$a = \sqrt{a_c^2 + a_T^2} = \sqrt{(0.7)^2 + (0.5)^2} = \sqrt{0.49 + 0.25} = \sqrt{0.74} \approx 0.86 \; m/s^2$.
Let $\theta$ be the angle of the resultant acceleration with the direction of centripetal acceleration.
$\tan \theta = \frac{a_T}{a_c} = \frac{0.5}{0.7} = 0.714$.
$\theta = \tan^{-1}(0.714) \approx 35.5^{\circ}$ with the direction of centripetal acceleration.