Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
