The velocity of a body is vec v(t) = A vec i + Bt vec j +…

The velocity of a body is $\vec v(t) = A\,\vec i + Bt\,\vec j + Ct^2\,\vec k$, where $A$, $B$, $C$ are constants. Find the change of its position vector between the times $t_1$ and $t_2$.

Explanation of Method

Displacement from a velocity vector.

Step 1: Displacement from velocity

The displacement is the integral of the velocity over time (the area under the velocity–time graph), taken separately for each component: $\Delta\vec r = \int_{t_1}^{t_2} \vec v(t)\,dt$.

$v_x = A$ $\Delta x$ $t_1$ $t_2$ $v_y = Bt$ $\Delta y$ $t_1$ $t_2$ $v_z = Ct^2$ $\Delta z$ $t_1$ $t_2$

Each component of the displacement is the area under that component's velocity–time graph between $t_1$ and $t_2$: a rectangle, a trapezoid, and the area under a parabola.

Step 2: Component by component

\[ \Delta x = \int_{t_1}^{t_2} A\,dt = A\,(t_2 - t_1), \]

\[ \Delta y = \int_{t_1}^{t_2} Bt\,dt = \frac{B}{2}\left(t_2^2 - t_1^2\right), \]

\[ \Delta z = \int_{t_1}^{t_2} Ct^2\,dt = \frac{C}{3}\left(t_2^3 - t_1^3\right). \]

Step 3: Result

\[\begin{aligned} \Delta\vec r = {}& A(t_2 - t_1)\,\vec i + \frac{B}{2}(t_2^2 - t_1^2)\,\vec j \\ &+ \frac{C}{3}(t_2^3 - t_1^3)\,\vec k \end{aligned}\]

Check

Differentiating a position that gives this displacement, $\vec r(t) = At\,\vec i + \tfrac{B}{2}t^2\,\vec j + \tfrac{C}{3}t^3\,\vec k$, returns $A\,\vec i + Bt\,\vec j + Ct^2\,\vec k = \vec v(t)\ \checkmark$

Result

\[\boxed{\,\begin{aligned} \Delta\vec r = {}& A(t_2 - t_1)\,\vec i + \frac{B}{2}(t_2^2 - t_1^2)\,\vec j \\ &+ \frac{C}{3}(t_2^3 - t_1^3)\,\vec k \end{aligned}\,}\]