
# Integration in the Context of Motion


Above, we defined velocity as the derivative of position and acceleration as the derivative of velocity. Integration allows us to go the other way! As you learned in first semester calculus, integration allows us to generate an object's velocity as a function of time given its acceleration and an initial velocity. We can also find the object's position function by integrating the velocity and using an initial position.

Let's consider an example.

Example $$\PageIndex{8}$$: Finding an object's velocity from its acceleration

An object has an acceleration over time given by $$\vecs a(t) = \sin t \,\hat{\mathbf i} - 2t \,\hat{\mathbf j}$$, and it's initial velocity was $$\vecs v(0) = 2 \,\hat{\mathbf i} + \,\hat{\mathbf j}$$.

a. Find the object's velocity as a function of time.

b. Assuming the object was located at the point (2, 3, -1) when time $$t = 0$$, determine the object's position function and find its location at time $$t = 3$$ sec.

Solution

a. First, we find the velocity as the antiderivative of the acceleration.

\begin{align*} \vecs v(t) = \int \vecs a(t) \, \,dt &= \int \left( \sin t \,\hat{\mathbf i} - 2t \,\hat{\mathbf j}\right)\,dt \\ &= -\cos t \,\hat{\mathbf i} - t^2\,\hat{\mathbf j} + \vecs C_1 \end{align*}

Now we use the initial velocity to determine $$\vecs C_1$$.

\begin{align*} \vecs v(0) = -\cos 0 \,\hat{\mathbf i} - (0)^2\,\hat{\mathbf j} + \vecs C_1 &= 2 \,\hat{\mathbf i} + \,\hat{\mathbf j} \\ -\hat{\mathbf i} + \vecs C_1 &= 2 \,\hat{\mathbf i} + \,\hat{\mathbf j} \\ \\ \text{And so} \quad \vecs C_1 &= 3 \,\hat{\mathbf i} + \,\hat{\mathbf j} \end{align*}

Incorporating this constant vector into our velocity function from above, we obtain the velocity describing this object's motion over time:

$\vecs v(t) = \left( 3 -\cos t \right) \,\hat{\mathbf i} + \left(1 - t^2\right) \,\hat{\mathbf j}$

b. Since the object's position is at the point (2, 3, -1) when time $$t = 0$$, we know $$\vecs r(0) = 2 \,\hat{\mathbf i} + 3 \,\hat{\mathbf j} - \,\hat{\mathbf k}$$.

Now, to determine the object's position function, we integrate it's velocity.

\begin{align*} \vecs r(t) = \int \vecs v(t) \, \,dt &= \int \bigg[\left( 3 -\cos t \right) \,\hat{\mathbf i} + \left(1 - t^2\right) \,\hat{\mathbf j}\bigg]\,dt \\ &= \left( 3t -\sin t \right) \,\hat{\mathbf i} + \left(t - \frac{t^3}{3}\right) \,\hat{\mathbf j} + \vecs C_2 \end{align*}

Now we use the initial position to determine $$\vecs C_2$$.

\begin{align*} \vecs r(0) = \left( 3(0) -\sin 0 \right) \,\hat{\mathbf i} + \left( 0 - \frac{(0)^3}{3}\right) \,\hat{\mathbf j} + \vecs C_2 &= 2 \,\hat{\mathbf i} + 3 \,\hat{\mathbf j} - \,\hat{\mathbf k}\\ \\ \text{And so} \quad \vecs C_2 &= 2 \,\hat{\mathbf i} + 3 \,\hat{\mathbf j} - \,\hat{\mathbf k} \end{align*}

Incorporating this constant vector into our position function from above, we obtain:

$\vecs r(t) = \left( 2 + 3t -\sin t \right) \,\hat{\mathbf i} + \left(3 + t - \frac{t^3}{3}\right) \,\hat{\mathbf j} - \,\hat{\mathbf k}$

To find the object's position at time $$t = 3$$ seconds, we just evaluate this position function at $$t = 3$$.

\begin{align*} \vecs r(3) &= \left( 2 + 3(3) -\sin 3 \right) \,\hat{\mathbf i} + \left(3 + 3 - \frac{(3)^3}{3}\right) \,\hat{\mathbf j} - \,\hat{\mathbf k} \\ &= \left( 11 -\sin 3 \right) \,\hat{\mathbf i} - 3 \,\hat{\mathbf j} - \,\hat{\mathbf k} \end{align*}

This position vector indicates that the object will be located at the point $$(11 -\sin 3, -3, -1)$$ at time $$t = 3$$ seconds.