Example 4 The velocity, \( v \) in feet per second, of a car that slammed on its brakes can be determined based on the length of skid marks that the tires left on the ground. This relationship is given by \[ v(d)=\sqrt{2 g f d} \] In this formula, \( g \) represents acceleration due to gravity ( \( 32 \frac{\mathrm{ft}}{\sec ^{2}} \) ), d is the length of the skid marks in feet, and \( f \) is a constant representing the friction of the surface. A car lost control on wet asphalt, with a friction coefficient of 0.5 , leaving 200 foot skid marks. How fast was the car travelling when it lost control? Show solution ? Exercise 3.8.4 Suppose a car is travelling 40 feet/sec and it loses control on wet asphalt. How long will its skid marks be? \( d= \) \( \square \) feet

We begin with the formula for the car’s velocity when it skids:   v(d) = √(2 · g · f · d) where   g = 32 ft/sec² (acceleration due to gravity),   f = 0.5 (friction coefficient for wet asphalt), and   d = length of the skid marks in feet. Example 4: A car left 200-foot skid marks. Substituting the values into the formula gives:   v(200) = √(2 · 32 · 0.5 · 200)        = √(2 · 32 · 0.5 · 200)        = √(32 · 200)     (since 2 · 0.5 = 1)        = √6400        = 80 ft/sec Thus, the car was traveling at 80 feet per second when it lost control. Exercise 3.8.4: Now suppose the car is traveling at 40 ft/sec. To find the length of its skid marks, we start with the same formula but solve for d. Begin with:   v = √(2 · g · f · d) Square both sides to get:   v² = 2 · g · f · d Now, solve for d:   d = v² / (2 · g · f) Substitute the values (v = 40 ft/sec, g = 32 ft/sec², f = 0.5):   d = 40² / (2 · 32 · 0.5)     = 1600 / (2 · 32 · 0.5)     = 1600 / (32)       (again, because 2 · 0.5 = 1)     = 50 ft Thus, if the car is traveling at 40 feet per second on wet asphalt, its skid marks will be 50 feet long.