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

Answer:

404.4 m

Explanation:

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

$140\times \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s$

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

$a=-9.81 m/s^{2} sin 11^{\circ}=-1.871836245\approx -1.87 m/s^{2}$

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

$s=\frac {0^{2}-38.89^{2}}{2\times -1.87}=404.3936096 m\approx 404.4 m$