Thanks in advance for any help. I need help with this whole problem.
The distance s(t) between an object and its starting point is given by the antiderivative of the 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.
Please show your attempt.
To find the distance between the object and its starting point after 15 seconds, we first need to find the antiderivative of the velocity function v(t).
The antiderivative is the reverse of differentiation. Since the velocity function v(t) is a polynomial function, we can integrate it term by term.
The antiderivative of 0.4t^2 is (0.4/3)t^3, and the antiderivative of 6 is 6t.
So, the antiderivative of v(t) = 0.4t^2 + 6 is s(t) = (0.4/3)t^3 + 6t + C, where C is the constant of integration.
To find the value of C, we can use the initial condition that s(0) = 0 (since the object's starting point is the reference point). Substituting t = 0 into the equation, we get:
0 = (0.4/3)(0)^3 + 6(0)+ C
0 = 0 + 0 + C
C = 0
So, the equation for the distance function becomes: s(t) = (0.4/3)t^3 + 6t.
To find the distance between the object and its starting point after 15 seconds, substitute t = 15 into the equation:
s(15) = (0.4/3)(15)^3 + 6(15)
s(15) = (0.4/3)(3375) + 90
s(15) = 150 + 90
s(15) = 240 meters
Therefore, the distance between the object and its starting point after 15 seconds is 240 meters.