Answer:
23.98 m
14.55058 m/s
Step-by-step explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s²
![v^2-u^2=2g\mu s\\\Rightarrow s=(v^2-u^2)/(2g\mu)\\\Rightarrow s=(0^2-20^2)/(2* 9.81* -0.85)\\\Rightarrow s=23.98\ m](https://img.qammunity.org/2020/formulas/physics/college/lhl16z4a40j8590vvvjuc27e2dmmgz78gl.png)
Minimum distance the driver would need to start braking in order to stop before the intersection is 23.98 m
![v^2-u^2=2\mu gs\\\Rightarrow -u^2=2\mu gs-v^2\\\Rightarrow u=√(v^2-2\mu gs)\\\Rightarrow u=√(0^2-2* 9.81* -0.45* 23.98)\\\Rightarrow u=14.55058\ m/s](https://img.qammunity.org/2020/formulas/physics/college/i6gfkngpae8215men43yb4o4sufuoxhur1.png)
The speed would be 14.55058 m/s