A particle P moves in a straight line starting from point O with velocity #2#m/s the acceleration of P at time t after leaving O is #2*t^(2/3)# m/s^2 Show that #t^(5/3)=5/6# When velocity of P is #3#m/s?

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Answer 1 · 2018-05-14T07:10:46

#"See explanation"#

#a = {dv}/{dt}#

#=> dv = a dt#

#=> v - v_0 = 2 int t^(2/3) dt#

#=> v = v_0 + 2 (3/5) t^(5/3) + C#

#t = 0 => v = v_0 => C = 0#

#=> 3 = 2 + (6/5) t^(5/3)#

#=> 1 = (6/5) t^(5/3)#

#=> 5/6 = t^(5/3)#