Falldistanse

Introduksjon til integrasjon

Hans Georg Schaathun

September 2016

$$v(t) = 9{,}8t$$

$$x = h(3) - h(2)$$

$$h'(t) = v(t) = 9{,}8t$$

$$h(t) = \int v(t)dt = 4{,}9t^2 + C$$

$$x = (4{,}9t^2 + C)\big|_{t=3} - (4{,}9t^2 + C)\big|_{t=2} $$

$$x = 4{,}9\cdot3^2 - 4{,}9\cdot2^2=24{,}5$$

$$x = \int_3^2 v(t)dt$$

$$v(t) = 9{,}8t$$