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 $0.50\; m/s$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

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:

$a_{e}=\frac{v^{2}}{r}$

$=\frac{(7.5)^{2}}{80}=0.7\, m / s ^{2}$

The situation is shown in the given figure

Suppose the cyclist begins cycling from point $P$ and moves toward point $Q$. At point $Q$ he applies the breaks and decelerates the speed of the bicycle by $0.5\, m / s ^{2}$

This acceleration is along the tangent at $Q$ and opposite to the direction of motion of the cyclist.

since the angle between $a_{\varepsilon}$ and $a_{ r }$ is $90^{\circ},$ the resultant acceleration $a$ is given by:

$a=\sqrt{a_{c}^{2}+a_{1}^{2}}$

$=\sqrt{(0.7)^{2}+(0.5)^{2}}$

$=\sqrt{0.74}=0.86 \,m / s ^{2}$

$\tan \theta=\frac{a_{c}}{a_{T}}$

Where $\theta$ is the angle of the resultant with the direction of velocity

$\tan \theta=\frac{0.7}{0.5}=1.4$

$\theta=\tan ^{-1}(1.4)$

$=54.46^{\circ}$