"F=ma=m\\frac{dv}{dt}\\to v=\\frac{1}{m}\\int F(t)dt"
"v=\\frac{1}{m}\\int (t^2+6-t^3+5)dt=\\frac{1}{m}(t^3\/3+6t-t^4\/4+5t)+C="
"=\\frac{1}{m}(t^3\/3+11t-t^4\/4)+C"
if "t=0" "v=0 \\to" "C=0"
We have
"v=\\frac{1}{m}(t^3\/3+11t-t^4\/4)=\\frac{1}{120\/9.81}(3^3\/3+11\\cdot 3-3^4\/4)=1.78 m\/s"
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