Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 5. A jet comes in for a landing on solid ground with a speed of 100 m//s, and its acceleration can have a maximum magnitude of 5.00 m//s^2 as it comes to rest. (a) From the instant the jet touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this jet land at a small tropical island airport where the runway is 0.800 km long? (c) Explain your answer. Hint v = v_0 + at, v^2 = (v_0)^2 + 2a∆x Answer Show Steps Givens: v_0 = 100 m//s v_f = 0 a = -5.00 m//s (a) Knowns: v_f = v_0 + at => t = (v_f - v_0)/a t = (0-100 m//s)/(-5.00m//s^2) = color(red)(20.0 s) (b) the plane cannot land (c) Solve the distance need for a stoping: Knowns (v_f)^2 = (v_0)^2 + 2a∆x => ∆x = ((v_f)^2 - (v_0)^2 )/(2a) ∆x = (0^2-(100 m//s)^2)/(2 times -5.00 m//s^2) = 1000m color(red)(∆x = 1.00 km > 0.800km)