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.)

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opudodennis opudodennis · Answer 1 · 2019-12-29T11:55:36+01:00

Answer:

404.4 m

Explanation:

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

[tex]140\times \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s

[/tex]

The acceleration is resolved as shown in the figure hence

deceleration of the truck along the inclined plane will be

[tex]a=-g sin \theta[/tex] where g is acceleration due to gravity

Substituting g with [tex]9.81 m/s^{2}[/tex] then

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

[/tex]

Using kinematic equation

[tex]v^{2}=u^{2}+2as[/tex] and making s the subject then

[tex]s=\frac {v^{2}-u^{2}}{2a}[/tex] where v and u are final and initial velocities respectively

Substituting 0 for v, 38.89 m/s for u and [tex]-1.87 m/s^{2}[/tex] then

[tex]s=\frac {0^{2}-38.89^{2}}{2\times -1.87}=404.3936096 m\approx 404.4 m[/tex]