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.
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
To find the distance between the object and its starting point after 15 seconds, we need to find the antiderivative of the velocity function and then evaluate it at t = 15.
The antiderivative of v(t) = 0.4t^2 + 6 is obtained by integrating each term separately. The integral of 0.4t^2 is (0.4/3)t^3, and the integral of 6 is 6t. Adding these results together gives us the antiderivative as (0.4/3)t^3 + 6t + C, where C is the constant of integration.
Now, we want to evaluate the antiderivative at t = 15 to find the distance after 15 seconds. Plugging in t = 15, we get:
s(15) = (0.4/3)(15)^3 + 6(15) + C
Since C is a constant, it will cancel out when we subtract s(0) (the distance at the starting point) from s(15). Therefore, we don't need to know the value of C to solve this problem.
Let's calculate the distance s(15):
s(15) = (0.4/3)(15)^3 + 6(15) + C
s(15) = (0.4/3)(3375) + 90 + C
s(15) = 1125 + 90 + C
s(15) = 1215 + C
Therefore, the distance between the object and its starting point after 15 seconds is 1215 meters + C.