A student running at 9m/s is 40m behind Eddy when Eddy starts from rest on his motor bike with an acceleration

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A student running at 9m/s is 40m behind Eddy when Eddy starts from rest on his motor bike with an acceleration of 0.9 m/s^2. (a) How long does it take for the student to catch Eddy? (b) What is the time interval during w/c the student remains ahead of Eddy? (c) Represent the motion of Eddy and the Student by graph.

A student running at 9m/s is 40m behind Eddy when Eddy starts from rest on his motor bike with an acceleration
let it be t sec's when eddy and student meet.





in t sec's eddy travels 0+.5*.9*t*t and student travels 9t metres.


since they meet, .45*t*t+ 40= 9t


solve the above quadratic to get two answers for when they meet.. 6.6666 and 13.3333.





which means they meet first after 6.6666 secs when student overtakes eddy and next after 13.3333 secs when eddy (who is accelerating) overtakes student.





there fore answer a) 6.666secs


b)13.3333-6.6666=6.6666 secs apprx


as for the graph.. the motion of student is a straight line on a distance time graph. that of eddy is a parabola.