Question

A driver traveling at 65 mi/h rounds a curve on a level grade to see a truck overturned across the roadway at a distance of 350 ft. If the driver is able to decelerate at a rate of 10ft/s2, at what speed will the vehicle hit the truck Plot the result for reaction times ranging from 0.50 to 5.00 s in increments of 0.5 s. Comment on the results.

Answer 1

Answer:

When the driver first sees the overturned truck, he will continue to move at 65mi/h during this reaction time. During this time, the vehicle will travel 1.47*65*t feet, or 95.5f ft.

Distance available for braking is 350-95.5t feet. This is the distance that is available for deceleration before the vehicle hits the overturned truck. The formula for braking distance is

d=(si²-sf²)/30(F +/- 0.01G)

In this case, the initial speed (Si) is 65mi/h. The friction factor, F, is related to the deceleration rate, and is computed by dividing the deceleration rate by the deceleration rate due to gravity, or 10/32.2 =0.31. The grade is level, i.e., G = 0. The braking distance is 350 – 95.5t.

Therefore,

350-95t=(65²-sf²)/(30*0.31)

This equation is solved for various values of t from 0.50 to 5.00 s. Note that at the point where the reaction distance becomes more than 350ft, the final speed is a constant 65mi/h, and the braking distance is essentially “0.”

Check the attachment below for the table and graph.

*COMMENT*

The vehicle is going to hit the overturned truck in any event. For reaction times over approximately 3.75s, the vehicle will hit the truck at full speed, 65mi/h.