PJ+Gaudio

1.) I. Deltax=190m Vmax= 305m/min a=1.22m/s^2 V= integral(1.22) so 1.22t+C=Vmax c=0

5.083=1.22t t=4.162s

x=integral from 0 to 4.167(1.22tdt)= 1.22t^2/2 +D= 10.592 meters

II. 4.167*2=8.334 a and -a occur at the same rate, doubling time gives us our answer.

2.) 50m t=1 second delay a=-9.8m/s

I. V=-9.8t+C X=-9.8/2t^2

9.8t^2/2=50 so t=3.194

II. 50=1/2(V0+V)t=3.19 seconds times 2 = about 6.4 seconds

Professor's Note : very Good effort - Give more explanation !! and try to avoid numerical values in the final answer -- check my page

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Visited the page, no work is done for this week :(