Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functions17.v (mi/h)v4075 mi20100 mi00 5t (h)The distance traveled is the area under the curve. This can be divided into two simplegeometric shapes–a triangle and a rectangle. The area isd = area of triangle + area of rectangle( )1=2 × base × height + (length × width)=( ) (12 × 5 h × 30mi + 5 h × 20 mi )hh= (75 mi) + (100 mi)= 175 mi19. It means that the velocity of the car increases at a rate of 7.5 feet per secondevery second.21. It means that the velocity of the car is decreasing at a rate of 18 feet per secondevery second.23. We are trying to solve for v, so v = v 0 + at is the formula we want to use. Butnotice that the acceleration a is given in m/s 2 , while time t is given as 3 min. We mustfirst make the units match, so convertt = 3 min= 3 min × 60 s1 min= 180 sNow plug into the formula. Note that the rocket “accelerates from rest,” which meansthat the initial velocity is v 0 = 0 m/s.Version: Fall 2007