A road makes a ${90}^{\circ }$ bend with a radius of $190 m$. A car enters the bend moving at $20 m/s$. Finding this too fast, the driver decelerates at $0.92 m/s^2$. Determine the angle(in degrees) made by acceleration of the car  relative to the tangent line to the circle when its speed rounding the bend has dropped to $15 m/s$.

Since it is rounding a curve, the car has a radial acceleration associated with its changing direction, in addition to the tangential deceleration that changes its speed. We are given that $a_T=-0.92 m/s^2$ since the car is slowing down, the tangential acceleration is directed opposite the velocity.

The radial acceleration is $a_r=\frac{v^2}{r}=\frac{{\left(15\right)}^2}{190}=1.2 m/s^2$

Magnitude of net acceleration. $a=\sqrt{a_r^2+a_T^2}={\left[{\left(1.2\right)}^2+{\left(0.92\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.5 m/s^2$

and points at angle $\theta = {\tan }^{-1}\left(\frac{a_r}{a_T}\right)= {\tan }^{-1}\left(\frac{1.2}{0.92}\right)={53}^{\circ }$

relative to the tangent line to the circle.