Question

A particle is moving such that its height h at time t is given by h(t) = 2 + 8t - 3t^2 + 1/5t^3. The average velocity of the particle on the period [0,3] is

Answer 1

• t=3

$\\ \Large\sf\longmapsto h(t)$

$\\ \Large\sf\longmapsto 2+8t-3t^2+\dfrac{1}{5}t^3$

$\\ \Large\sf\longmapsto 2+8(3)-3(3)^2+\dfrac{1}{5}(3)^3$

$\\ \Large\sf\longmapsto 2+24-3(9)+\dfrac{27}{5}$

$\\ \Large\sf\longmapsto 26-27+5.4$

$\\ \Large\sf\longmapsto -2+5.4$

$\\ \Large\sf\longmapsto h(t)=3.4m$