Electric Vehicle Acceleration & Trajectory Recovery Solution

In electric and autonomous vehicle engineering, onboard inertial measurement units (IMUs) log high-frequency 3-axis accelerometer data. When GPS satellite signals are occluded (such as inside tunnels or subterranean parking garages), vehicle stability control systems reconstruct the vehicle's instantaneous velocity and position trajectory by performing successive vector integrations over time (dead reckoning).

An autonomous electric delivery vehicle enters an underground facility at $t = 0\text{ seconds}$. The onboard IMU records a 3-axis acceleration vector (in $\text{m/s}^2$) modeled by:

1. Use indefinite vector integration and the initial velocity condition to derive the vehicle's 3D velocity vector function $\vec{v}(t)$.
2. Use indefinite vector integration and the initial position condition to derive the vehicle's 3D position trajectory function $\vec{r}(t)$.
3. Calculate the net displacement vector $\Delta \vec{r}$ of the vehicle during its first 3 seconds inside the facility ($0 \le t \le 3$) using a definite integral.

Step-by-Step Solution:

1. Derive Velocity Vector Function $\vec{v}(t)$: Integrate acceleration component-by-component:

   $$\vec{v}(t) = \int \vec{a}(t) \, dt = \left\langle \int 6t \, dt, \int 12t^2 \, dt, \int 4 \, dt \right\rangle = \langle 3t^2, 4t^3, 4t \rangle + \vec{C}_1$$
   Apply initial velocity $\vec{v}(0) = \langle 2, -1, 5 \rangle$:
   $$\vec{v}(0) = \langle 0, 0, 0 \rangle + \vec{C}_1 = \langle 2, -1, 5 \rangle \implies \vec{C}_1 = \langle 2, -1, 5 \rangle$$
   Therefore:
   $$\vec{v}(t) = \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle \text{ m/s}$$

2. Derive Position Vector Function $\vec{r}(t)$: Integrate velocity component-by-component:

   $$\vec{r}(t) = \int \vec{v}(t) \, dt = \left\langle \int (3t^2 + 2) \, dt, \int (4t^3 - 1) \, dt, \int (4t + 5) \, dt \right\rangle = \langle t^3 + 2t, t^4 - t, 2t^2 + 5t \rangle + \vec{C}_2$$
   Apply initial position $\vec{r}(0) = \langle 0, 10, -3 \rangle$:
   $$\vec{r}(0) = \langle 0, 0, 0 \rangle + \vec{C}_2 = \langle 0, 10, -3 \rangle \implies \vec{C}_2 = \langle 0, 10, -3 \rangle$$
   Therefore:
   $$\vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ meters}$$

3. Calculate Net Displacement $\Delta \vec{r}$ over $[0, 3]$: Evaluate the definite integral of velocity:

   $$\Delta \vec{r} = \int_{0}^{3} \vec{v}(t) \, dt = \vec{r}(3) - \vec{r}(0)$$
   Evaluating $\vec{r}(3)$:
   • $x(3) = (3)^3 + 2(3) = 27 + 6 = 33$
   • $y(3) = (3)^4 - (3) + 10 = 81 - 3 + 10 = 88$
   • $z(3) = 2(3)^2 + 5(3) - 3 = 18 + 15 - 3 = 30$
     $$\vec{r}(3) = \langle 33, 88, 30 \rangle \text{ m}$$
     Subtracting initial position $\vec{r}(0) = \langle 0, 10, -3 \rangle$:
     $$\Delta \vec{r} = \langle 33 - 0, 88 - 10, 30 - (-3) \rangle = \langle 33, 78, 33 \rangle \text{ meters}$$

Meaning of the Answer & Real-Life Application:

Over the 3-second interval, the vehicle undergoes a 3D spatial translation of $\langle 33, 78, 33 \rangle\text{ meters}$, achieving a direct linear displacement magnitude of $\|\Delta \vec{r}\| = \sqrt{33^2 + 78^2 + 33^2} = \sqrt{1089 + 6084 + 1089} = \sqrt{8262} \approx 90.9\text{ meters}$. Autonomous vehicle systems use these continuous vector integral calculations to update local obstacle maps and prevent collisions when GPS signals are unavailable.