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Answer 1 · 2018-01-16T02:23:20

The initial velocity of the particle will simply be the derivative of the displacement function evaluated when $t = 0$ $t=0$

$s (t) = e^{2 t} - 5 t$ $s(t)=e^(2t)-5t$

$\frac{d s (t)}{d t} = 2 e^{2 t} - 5$ $(ds(t))/(dt)=2e^(2t)-5$

$\frac{d s (0)}{d t} = 2 e^{2 \times 0} - 5$ $(ds(0))/(dt)=2e^(2xx0)-5$

$\frac{d s (t)}{d t} = 2 e^{0} - 5$ $(ds(t))/(dt)=2e^(0)-5$

$\frac{d s (t)}{d t} = 2 (1) - 5$ $(ds(t))/(dt)=2(1)-5$

$v (0) = - 3$ $v(0)=-3$

I hope that helps :)