Question

2–3. Displacement, distance, and position Consider an object moving along a line with the following velocities and initial positions. Assume time t is measured in seconds and velocities have units of m/s.

d. Determine the position function s(t) using the Fundamental Theorem of Calculus (Theorem 6.1). Check your answer by finding the position function using the antiderivative method.

v(t) = 12t²-30t+12, for 0 ≤ t ≤ 3; s(0)=1

Accepted Answer

Recall that the position function $$s(t)$$ can be found from the velocity function $$v(t) by $$integrating $$v(t)$$ over time and adding the initial position $$s(0). $$According to the Fundamental Theorem of Calculus, we have:

$$s(t) = s(0) + \int_0^t v(x) \, dx$$ Substitute the given velocity function $$v(t) = 12t^2$ - 30t + 12$$ into the integral:

$$s(t) = 1 + \int_0^t (12x^2 - 30x + 12) \, dx$$ Compute the integral by integrating each term separately:

- The integral of $$12x^2 is$$ $$12 \cdot \frac{x^3$}{3} = 4x^3$  
- The integral of -30x$ is -30$$ \cdot \frac{$$x^2$$}{2} = $$-15x^2  
- $$The integral of $$12 is$$ $$12x$$ Combine the results of the integrals and evaluate the definite integral from 0 to $$t$$:

$$\int_0^t (12x^2 - 30x + 12) \, dx = \left[4x^3 - 15x^2 + 12x\right]_0^t = 4t^3 - 15t^2 + 12t$$ Add the initial position $$s(0) = 1 to $$the integral result to write the position function:

$$s(t) = 1 + 4t^3 - 15t^2 + 12t

To $$check your answer using the antiderivative method, differentiate $$s(t)$$ and verify that $$s'(t) = v(t).$$

Verified video answer for a similar problem:

Key Concepts

Fundamental Theorem of Calculus

Antiderivative Method

Initial Conditions in Differential Equations