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N april 1974, steve prefontaine completed a 10-km race in a time of 27 min, 43.6 s. suppose \"pre\" was at the 7.15-km mark at a time of 25.0 min. if he accelerates for 60 s and maintains the increased speed for the duration of the race, calculate the acceleration that he had. assume his instantaneous speed at the 7.15-km mark was the same as his overall average speed up to that time.

Respuesta :

solution:[tex]speed is 7.99km ,u=\frac{7990}{(25\times60)}=5.33m/s\\ now let the final velocity after 60s from 7.99km is v.\\ v-u =at\\v=5.33+a\times60\\  =60s+5.33\\now distance travelled during this time will be\\ s_{2}=vt\\ v\times103.6\\ =(60a+5.33)\times103.6\\ =551.84+6216am\\ now total distance travelled  is \\  also\\ s_{1}=ut+(\frac{1}{2})at^{2}      =5.33\times60+(0.5)9\times3600\\ =1800a+319.6m\\ Now the remaining time to complete rest race was\\ t=(27\times60+43.6)60-60(25\times60)\\ =103.6s\\

Now, distance travelled during this time will be\\ s_{2}=vt\\ v\times103.6\\ =(60a+5.33)\times103.6\\ =551.84+6216am\\ now total distance travelled  is \\ s_{1}+s_{2}+7990=10000\\ 1800a+319.6+551.84+6216a=2010\\ 8016a=2010-871.44\\ 8016a=1138.56\\ a=0.142m/s^2[/tex]

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