since v(t) = s'(t), then
s(t) = (.4t^3)/3 + 6t + c
at the start (t=0) s(0) = c
at t=15, s(15) = .4(3375)/3 + 6(15) + c
so the distance traveled
= .4(3375)/3 + 6(15) + c - c
= 540
I got this problem today in school, and I don't know how to go about soving it. Can someone show me. Thanks. Here it is.
The distance s(t) between an object and its starting point is given by the antiderivative of velocity function v(t). Find the distance between the object and its starting point after 15 seconds if v(t)=0.4t^2+6 meters per second.
My teacher said that I wouldn't be able to solve it, so I really want some help with this thanks.
1 answer
1 answer