Answer:
Approximately [tex]\rm 1.73 \; \rm m \cdot s^{-2}[/tex].
Explanation:
The acceleration on the car comes in two parts:
These two accelerations are perpendicular to each other.
Convert the current velocity to meters-per-second ([tex]\rm m \cdot s^{-1}[/tex].)
[tex]\begin{aligned} v &= 58.0\; \rm km \cdot h^{-1} \\ &= \left(58.0\; \rm km \cdot h^{-1}\left/\frac{3600\; \rm s}{1\; \rm h}\right.\right) \times \frac{1000\; \rm m}{1 \; \rm km} \\ &= \frac{58.0}{3600} \times 1000 \; \rm m \cdot s^{-1} \\ & \approx 16.1\; \rm m \cdot s^{-1}\end{aligned}[/tex].
Convert the acceleration due to braking to meters-per-second-squared ([tex]\rm m \cdot s^{-2}[/tex].)
[tex]\begin{aligned}a_{\text{tangential}} &= 4.0\; \rm km \cdot h^{-1} \cdot s^{-1} \\ &= \left(4.0\; \rm km \cdot h^{-1} \cdot s^{-1}\left/\frac{3600\; \rm s}{1\; \rm h}\right.\right) \times \frac{1000\; \rm m}{1 \; \rm km} \\ &= \frac{4.0}{3600} \times 1000 \; \rm m \cdot s^{-1} \cdot s^{-1} \\ & \approx 1.111111\; \rm m \cdot s^{-2} \end{aligned}[/tex].
Calculate the acceleration due to the circular motion:
[tex]\begin{aligned} a_\text{centripetal} &= \frac{v^2}{r} \\ &= \frac{\left(16.1\; \rm m \cdot s^{-1}\right)^2}{195.0\; \rm m} \\ & \approx 1.33112\; \rm m \cdot s^{-2}\end{aligned}[/tex].
Since the two accelerations are perpendicular to each other, the resultant acceleration can be found using the Pythagorean Theorem
[tex]\begin{aligned}a &= \sqrt{\left(a_\text{tangential}\right)^2 + \left(a_\text{centripetal}\right)^2} \\ &\approx \sqrt{1.111111^2 + 1.33112^2} \\ &\approx 1.73\; \rm m \cdot s^{-2}\end{aligned}[/tex].
