A cyclist 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 $m{s}^{-1}$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Speed of the cyclist $\left(v\right)=27km{h}^{-1}=27\times \frac{5}{18}(\because 1km{h}^{-1}=\frac{5}{18}m{s}^{-1})=\frac{15}{2}m{s}^{-1}=7.5m{s}^{-1}$

Radius of the circular turn® = 80 m

$\therefore$Centripetal acceleration acting on the cyclist $a_c=\frac{v^2}{r}=\frac{{(15/2)}^2}{80}=\frac{225}{4\times 80}m{s}^{-2}$

Tangential acceleration applied by brakes $a_T=0.5m{s}^{-2}$

Centripetal acceleration and tangential acceleration act perpendicular to each other

$\therefore$Resultant acceleration $a=\sqrt{a_c^2+a_T^2}=\sqrt{{\left(0.7\right)}^2+{\left(0.5\right)}^2}=\sqrt{0.49+0.25}=\sqrt{0.74}=0.86m{s}^{-2}$

If resultant acceleration makes an angle $\theta$ with the direction of velocity, then 

$\tan \theta =\frac{a_c}{a_T}=\frac{0.7}{0.5}=1.4=\tan {54}^{0}28\text{'}$

$\theta ={54}^{0}28\text{'}$