A jogger maintains a speed of 2.5 m/s for 216 m until heencounters a stoplight,

Question

A jogger maintains a speed of 2.5 m/s for 216 m until heencounters a stoplight, and he abruptly stops and waits 28s for thelight to change. He then resumes his excercise and maintains aspeed of 3.4 m/s for the remaining 52 m to his home.What is theaverage velocity for this entire interval? I am not sure which equation to use or what goes where in theequation. I used X-Xo= Vot +1/2at2 But when I entered in my final answer it said myanswer was wrong. So I'm not sure if I'm even on the right track atall. A jogger maintains a speed of 2.5 m/s for 216 m until heencounters a stoplight, and he abruptly stops and waits 28s for thelight to change. He then resumes his excercise and maintains aspeed of 3.4 m/s for the remaining 52 m to his home.What is theaverage velocity for this entire interval? I am not sure which equation to use or what goes where in theequation. I used X-Xo= Vot +1/2at2 But when I entered in my final answer it said myanswer was wrong. So I'm not sure if I'm even on the right track atall.

Explanation / Answer

   Totaldistance   d   =   216 +   52 =   268 m    totaltime   t   =   216/2.5 +   28 +   52/3.4 =   129.69 s    averagevelocity   vavg   =   d/ t   =   268 / 129.69 =   2.066 m/s

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