The distance traveled by car
"d=v_it+\\frac{at^2}{2}"
"3000=6\\times 60+\\frac{a\\times 60^2}{2}"So, the acceleration of the car
"a=\\frac{22}{15}\\:\\rm{m\/s^2}=1.47\\:\\rm{m\/s^2}"The final velocity
"v_f=v_i+at=6+\\frac{22}{15}\\times 60=94\\:\\rm{m\/s}"
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