A cyclist is moving with a speed of 6 m/s. As he approaches a circular turn on the road of radius 120 m, he applies brakes and reduces his speed at a constant rate of 0.4 m/s². The magnitude of the net acceleration of the cyclist on the circular turn is

See fig. (12). As the cyclist approaches the circular turn at B, he decelerates at the rate of a_t shown by vector BE. At B, the centripetal acceleration directed towards centre C is acting on the cyclist. This is given by

${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{(6{)}^2}{120}=0\cdot 3\mathrm{m}/{\mathrm{s}}^2$

The resultant acceleration a is given by vector BD. The magnitude of a (i.e., a) is given by 

$\begin{array}{l}\mathrm{a}=\sqrt{\left({\mathrm{a}}_{\mathrm{t}}^2+{\mathrm{a}}_{\mathrm{c}}^2\right)}={\left\{(0\cdot 4{)}^2+(0\cdot 3{)}^2\right\}}^{1/2}\\ =0\cdot 5\mathrm{m}/{\mathrm{s}}^2\end{array}$