My Math Forum Distance, speed and time problem

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 January 20th, 2012, 11:24 PM #1 Member   Joined: Jan 2012 Posts: 33 Thanks: 0 Distance, speed and time problem Can someone solve this ? A plane flying at 300 mph has a 3-hr head start on a “chase plane,” which has a speed of 800 mph. How far from the airport will the chase plane overtake the first plane?
 January 20th, 2012, 11:40 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Text exercise It is preferable to begin a new topic for a new question, so that topics are easy to follow and add to. Thus, I split this topic from the original into its own. The relationship between distance d, constant speed v and time t is: $d=v\cdot t$ When the chase plane overtakes the slower plane, their distance form the airport will be equal. Let t be the time in hours the slower plane is in the air, and so t - 3 will be the time the chase plane is in the air. So we have: $$$300\text{ mph}$$t\text{ hr}=$$800\text{ mph}$$(t-3)\text{ hr}$ Now we must solve for t: $$$300\text{ mph}$$t\text{ hr}=$$800\text{ mph}$$t\text{ hr}-$$800\text{ mph}$$3\text{ hr}$ $$$800\text{ mph}$$t\text{ hr}-$$300\text{ mph}$$t\text{ hr}=2400\text{ mi}$ $$$500\text{ mph}$$t\text{ hr}=2400\text{ mi}$ $t\text{ hr}=\frac{2400\text{ mi}}{500\text{ mph}}=\frac{24}{5}\:\text{hr}=4.8\text{ hr}$ Now to find d, we use: $d\text{ mi}=$$300\text{ mph}$$$$4.8$$\text{ hr}=$$800\text{ mph}$$$$4.8-3$$\text{ hr}=1440\text{ mi}$
 January 21st, 2012, 01:19 AM #3 Member   Joined: Jan 2012 Posts: 33 Thanks: 0 Re: Distance, speed and time problem Thank you for your answer. I can solve long equations but when it comes to solve text ecercises i dont understand them..I think i dont have logic thinking like you

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