Question

An accepted relationship between stopping distance (d) feet, and the speed of a car, (v) in mph is d(v)=1.1v+0.05v^2 on dry level concrete.If an accident occurs 150 feet ahead, what is the maximum speed at which one can travel to avoid being involved in the accident?​

Answer 1

To find the maximum speed at which one can travel to avoid being involved in the accident, we need to determine the speed at which the stopping distance is equal to or less than 150 feet.

Step-by-step explanation:

To find the maximum speed at which one can travel to avoid being involved in the accident, we need to determine the speed at which the stopping distance is equal to or less than 150 feet. We can do this by setting the equation d(v) = 1.1v + 0.05v^2 equal to 150 and solving for v.

1.1v + 0.05v^2 = 150

0.05v^2 + 1.1v - 150 = 0

This is a quadratic equation that can be solved using the quadratic formula. Plug in the values of a (0.05), b (1.1), and c (-150) into the formula.

v = (-1.1 ± sqrt(1.1^2 - 4(0.05)(-150))) / (2(0.05))

v = (-1.1 ± sqrt(1.21 + 30)) / 0.1

v = (-1.1 ± sqrt(31.21)) / 0.1

Simplifying further gives:

v ≈ 6.96 or v ≈ -40.96

Since speed cannot be negative, the maximum speed at which one can travel to avoid being involved in the accident is approximately 6.96 mph.