benny93xp Posted August 26, 2017 Share Posted August 26, 2017 i attach here the question and my attempts .. quite confident with Question x, Question y but not Question Z . am not sure if the concept still applies. Link to comment Share on other sites More sharing options...
Country Boy Posted September 5, 2017 Share Posted September 5, 2017 (edited) There is information left out of "problem z". Are we to assume that each leg, A to B and B to C, starts initial speed 0 or does A to B start with initial speed 0 and B to C with the speed with which we arrive at B? If we start a leg with speed [tex]v_0[/tex] and have constant (or average) acceleration a then in time t we will have gone a distance [tex](a/2)t^2+ v_0t[/tex]. Assuming that we started from A to B at initial speed 0 and it takes 1.5 hours at constant (average) acceleration [tex]11 m/s^2[/tex] then the distance from A to B is (5.5)(5400)+ 0(5400)= 29700 meters or 29.7 kilometers. The final speed would be 11(5400)= 59400 m/s. Now, do we start the leg from B to C at that speed or do we stop and start over again? Edited September 5, 2017 by Country Boy Link to comment Share on other sites More sharing options...
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