$2.$ Antilock Braking Systems $($ABS$)$ allow vehicles to come to an emergency stop under conditions of the static coefficient of friction, $\backslash (\backslash$mu$\_\{\backslash$mathrm$\{$s$\}\}\backslash ),$ between the tire and the road. Vehicles without ABS will lock up $($skid$)$ the tires under conditions of the dynamic coefficient of friction, $\backslash (\backslash$mu$\_\{$k$\}\backslash ),$ between the tire and the road.

Experimental data from a modern vehicle with mass of mvehicle was collected with and without ABS activated. Data with and without ABS activated was collected from several low speed stops $($starting speed before braking was viow$\_$speed$\_$test. The stopping distance with ABS activated was measured to be LaBs, and the stopping distance without ABS was measured to be $\backslash (\backslash$mathrm$\{$L$\}\_\{\backslash$text $\{$skid $\}\}\backslash ).$

Calculate:

Using the data provided for the low speed test you are going to calculate predicted stopping distances for the same vehicle, but from a higher starting speed, vhigh$-$speed.

For your exam question you will be asked to calculate the difference in stopping distances with and without $\backslash ($ A B S $\backslash )$ activated $(\backslash (\backslash$Delta $\backslash$mathrm$\{$L$\}=\backslash$mathrm$\{$L$\}\_\{\backslash$text $\{$skid $\}\}-\backslash$mathrm$\{$L$\}\_\{\backslash$mathrm$\{$ABS$\}\}\backslash )).$

Tip: Neglect weight transfer $($assume the vehicle acts like a block with a single normal force and friction force$)$