Question

Bob wants to catch an elevator before the door closes. He is 46ft away when the door starts to close. The door has to travel 4.2ft to fully close, and is closing at a rate of 1.5ft/s. Bob needs to get to the door while it is at least 0.7ft open. How fast does he have to run in order to catch the elevator? Give your answer in feet per second, and round to the nearest hundredth. Show your work here Enter your answer​

Answer 1

Answer: Bob has to run at a speed of 21.31ft/s to catch the elevator.

Explanation:

Let's call Bob's running speed "V".We know that the door is closing at a rate of 1.5ft/s, so the time it takes to close completely is 4.2ft / 1.5ft/s = 2.8 seconds.During this time, Bob has to travel 46ft to reach the elevator. So the time it takes Bob to reach the door is 46ft / V.We also know that the door needs to be at least 0.7ft open when Bob reaches it. So, the time that the door is open when Bob reaches it is 2.8s - (2.8s * 0.7ft / 4.2ft) = 2.8s - 0.6364s = 2.1636s.So, Bob has to cover 46ft in 2.1636s. Solving for V, we get:V = 46ft / 2.1636s = 21.31ft/s (rounded to the nearest hundredth).Therefore, Bob has to run at a speed of 21.31ft/s to catch the elevator.