Respuesta :
The given question is incomplete. The complete question is as follows.
The speed limit on a particular city street is 45 miles per hour (which is 20 m/s). A driver approaching a stop light slams on their brakes, locking the wheels of their car in place so that they no longer rotate. If the coefficient of friction between the car's tires and the street is 0.9, what is the minimum distance the driver would need to start braking in order to stop before the intersection?
When it rains, the coefficient of friction drops to 0.5. If the driver still wants to be able to stop in the same distance as in part A, what is the maximum speed they can have when it rains?
Explanation:
(a) The relation between coefficient of friction and force of friction is as follows.
[tex]F_{f} = \mu \times mg[/tex]
Also, a = [tex]\mu g[/tex]
where, a = acceleration
Hence, [tex]F_{f} = m \times a = \mu \times mg[/tex]
[tex]v^{2} - x^{2} = 2ad[/tex]
d = stopping distance
Putting the given values into the above formula as follows.
[tex]v^{2} - x^{2} = 2ad[/tex]
[tex]0 - (20)^{2} = 2 \times (-0.9 \times 9.8)d[/tex]
d = [tex]\frac{400}{2 \times 0.9 \times 9.8}[/tex]
= 22.7 m
Therefore, minimum distance the driver would need to start braking in order to stop before the intersection is 22.7 m.
(b) It is given that [tex]\mu[/tex] is 0.5.
As, a = [tex]\mu g[/tex]
= [tex]0.5 \times 9.8[/tex]
= 4.9 [tex]m/s^{2}[/tex]
and, d = 22.7 m
Also, [tex]v^{2} - u^{2} = 2 \mu gd[/tex]
u = [tex]\sqrt{2 \mu gd}[/tex]
= [tex]\sqrt{2 \times 4.9 \times 22.7}[/tex]
= 15 m/s
Therefore, the maximum speed they can have when it rains is 15 m/s.
