Question

Uphill escape ramps are sometimes provided to 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 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.)

Answer 1

404.4 m

Step-by-step explanation:

Converting the initial speed from km/h to m/s then

`140* \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s</p><p>`

The acceleration is resolved as shown in the figure hence

deceleration of the truck along the inclined plane will be

`a=-g sin \theta` where g is acceleration due to gravity

Substituting g with
`9.81 m/s^(2)` then

`a=-9.81 m/s^(2) sin 11^(\circ)=-1.871836245\approx -1.87 m/s^(2)</p><p>`

Using kinematic equation

`v^(2)=u^(2)+2as` and making s the subject then

`s=\frac {v^(2)-u^(2)}{2a}` where v and u are final and initial velocities respectively

Substituting 0 for v, 38.89 m/s for u and
`-1.87 m/s^(2)` then

`s=\frac {0^(2)-38.89^(2)}{2* -1.87}=404.3936096 m\approx 404.4 m`