Question #214610
An automobile accelerates from rest at 1 ft/s
1
Expert's answer
2021-07-19T03:15:50-0400

If we were to calculate the distance covered, the following will be the steps.

First distancev=u+atv=0+301v=30ft/sDistance,s=ut+12at2=0.5900=450ftSecond distanceDistance,s=vt=30120=3600ftThird distancev=u+at0=30+a15a=2ft/s2Distance,v2=u2+2ast0=90022s    s=225ftTotal=450+3600+225=4275ft=1303.02mFirst \space distance \\ v=u+at \\ v=0+30*1\\ v=30ft/s\\ Distance, s= ut+\frac{1}{2}at^2=0.5*900= 450 ft\\ \\ Second \space distance \\ Distance, s= v*t=30*120= 3600 ft\\ Third \space distance \\ v=u+at \\ 0=30+a*15\\ a=2ft/s^2\\ Distance, v^2=u^2+2ast\\ 0=900-2*2*s \implies s= 225 ft\\ Total = 450+3600+225=4275ft=1303.02 m


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