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

$"See explanation"$

$a = {dv}/{dt}$

$=> dv = a dt$

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

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

$t = 0 => v = v_0 => C = 0$

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

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

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

A particle P moves in a straight line starting from point O with velocity 22m/s the acceleration of P at time t after leaving O is 2*t^(2/3)2*t^(2/3) m/s^2 Show that t^(5/3)=5/6t^(5/3)=5/6 When velocity of P is 33m/s?