Centrapetal acceleration, \(a _{ c }=\frac{ v ^2}{ r }=\frac{20^2}{80}\)

\(=5\,m / s ^2\)

If the brake is applied in the movement of body at \(P\) of circular turn , acceleration along the target \(=5 \,m / sec\)

Angle between both the acceleration \(=90^{\circ}\)

the magnitude of resultant acceleration

\(a =\sqrt{ a _{ c }^2+ a _5^2}=\sqrt{5^2+5^2}=5 \sqrt{2}\, m / sec\)

List the resultant acceleration make an angle \(\theta\) with tangent

\(\tan \theta=\frac{ a _{ c }}{ a _5}=\frac{5}{5}=1\)

\(\theta=45^{\circ}\)

Therefore required angle is \(90^{\circ}+45^{\circ}\)

\(=135^{\circ}\)