Answer: 49.92 m
Explanation:
In this situation the following equation will be useful:
[tex]V^{2}=V_{o}^{2} +2 a d [/tex]
Where:
[tex]V=0 m/s[/tex] is the final velocity of the car, when it finally stops
[tex]V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s[/tex] is the initial velocity of the car
[tex]a=-6.4 m/s^{2}[/tex] is the constant acceleration of the car after the driver slams on the brakes
[tex]d[/tex] is the stopping distance
Isolating [tex]d[/tex]:
[tex]d=\frac{-V_{o}^{2}}{2a}[/tex]
[tex]d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}[/tex]
[tex]d=41.919 m \approx 41.92 m[/tex]
