Question

In 1976, Gerald Hoagland drove a car over 8.0 ´ 102 km in reverse. Fortunately for Hoagland and motorists in general, the event took place on a special track. During this drive, Hoagland’s average velocity was about -15.0 m/s. Suppose Hoagland decides during his drive to go forward. He applies the brakes, stops, and then accelerates until he moves forward at same speed he had when he was moving backward. How long would the entire reversal process take if the average acceleration during this process is +2.5 m/s2?

Answer 1

when he applies the breaks and the avg acceleration is +2.5m/s^2 then it will take him 8*10^2/2.5=320s then he speeds up taking another 320 seconds so it takes him 640 seconds which is 10 minutes and 40 seconds