Here, $v=27\ \mathrm{km\,h}^{-1}=7.5\ \mathrm{m\,s}^{-1}$ ; $r=80\ \mathrm{m}$ Centripetal acceleration, $a_c = \frac{v^2}{r} = \frac{(7.5)^2}{80} = 0.7\ \mathrm{m\,s}^{-2}$ Suppose that the cyclist applies brakes at the point A of the circular turn. Then, retardation produced dueto the brakes, say $a_T$ will act opposite to the velocity, v figure.Thus, $a_T = -0.5\ \mathrm{m\,s}^{-2}$Therefore, total acceleration is given by$a = \sqrt{a_c^2 + a_T^2}$ $= \sqrt{(0.7)^2 + (-0.5)^2}$ $= \sqrt{0.49 + 0.25} = \sqrt{0.74} = 0.86\ \mathrm{m\,s}^{-2}$Ifθ is the angle between the total acceleration and the velocity of the cyclist, then $\tan \theta = \frac{a_c}{a_T} = \frac{0.7}{0.5} = 1.4$ or $\theta = 54^{\circ} 28'$