Question

Question: (II) Uphill escape ramps are sometimes provided at the side of steep downhill highways for trucks with overheated brakes. For a simple 11° upward ramp, what minimum length would be needed for a runaway truck traveling at 140 km/h. Note the large size of your calculated length. (If sand is used for the bed of the ramp, its length can be reduced by a factor of about 2.)

Short answer

The minimum length of the ramp should be 404.41 m.

Step-by-step solution

Step 1. Given data and assumption

Escape ramps use gravitational acceleration to stop the car in an emergency.

On these ramps, the component of gravitational acceleration acts against the motion.

Given data:

The angle of incline is $\theta =11°$.

The initial speed of the truck along the x-axis is

$\begin{array}{c}{v}_{\mathrm{i}}=140\frac{\mathrm{km}}{\mathrm{h}}\\ =140\frac{\mathrm{km}}{\mathrm{h}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{h}}{3600\mathrm{s}}\\ =38.89\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

The final speed of the truck is $v_{\mathrm{f}}=0$.

Assumption:

Let m be the mass of the truck.

Let the length of the ramp be $\Delta x$.

Step 2. Calculation of the length of the ramp

The normal force acting on the truck is $N=mg\cos \theta$.

The net force on the truck along the x-direction is

$F=-mg\sin \theta$.

Then, the acceleration of the truck is

$\begin{array}{c}a=\frac{-mg\sin \theta }{m}\\ =-g\sin \theta \end{array}$.

Now,

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2a\Delta x\\ v_{\mathrm{f}}^2=v_{\mathrm{i}}^2-2g\sin \theta \Delta x\\ 0^2={\left(38.89\frac{\mathrm{m}}{\mathrm{s}}\right)}^2-\left(2\times 9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\times \sin {11}^o\right)\times \Delta x\\ \Delta x=404.41\mathrm{m}\end{array}$

Hence, the minimum length of the ramp should be 404.41 m.