Automotive Cornering & Tire Traction Limits Solution

In automotive and autonomous vehicle engineering, electronic stability control (ESC) systems continuously monitor vehicle kinematics to prevent skidding. By decomposing the total acceleration vector into its tangential component $a_T$ (longitudinal drive torque/braking) and normal component $a_N$ (lateral cornering grip), vehicle computers compare lateral acceleration against the friction limit $a_{\text{grip}} = \mu_s g$ of the tire contact patch.

An autonomous electric vehicle prototype is undergoing track testing at the Las Vegas Motor Speedway. Its 2D trajectory across the asphalt is tracked in meters as a function of time $t$ (in seconds) by high-precision RTK-GPS:

1. State the formulas for the tangential component $a_T$ and normal component $a_N$ of acceleration.
2. Compute the exact velocity $\vec{v}(2)$, acceleration $\vec{a}(2)$, and speed $\|\vec{v}(2)\|$ at time $t = 2\text{ seconds}$.
3. Calculate the tangential acceleration $a_T$ and normal acceleration $a_N$ at $t = 2\text{ seconds}$.
4. Determine if the vehicle maintains tire traction or if it exceeds the critical skid limit at $t = 2\text{ seconds}$.

Step-by-Step Solution:

1. State Component Formulas:

   • Tangential Component Formula:
     $$a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}$$
   • Normal Component Formula:
     $$a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}$$

2. Compute Kinematic Vectors at $t = 2\text{ s}$: Differentiate the position vector:

   $$\vec{v}(t) = \vec{r}'(t) = \langle 3t^2 - 3, 4t, 0 \rangle$$
   $$\vec{a}(t) = \vec{r}''(t) = \langle 6t, 4, 0 \rangle$$

   Evaluate at $t = 2$:

   $$\vec{v}(2) = \langle 3(2)^2 - 3, 4(2), 0 \rangle = \langle 9, 8, 0 \rangle\text{ m/s}$$
   $$\vec{a}(2) = \langle 6(2), 4, 0 \rangle = \langle 12, 4, 0 \rangle\text{ m/s}^2$$

   Compute speed $\|\vec{v}(2)\|:$

   $$\|\vec{v}(2)\| = \sqrt{9^2 + 8^2 + 0^2} = \sqrt{81 + 64} = \sqrt{145}\text{ m/s} \approx 12.04\text{ m/s}$$

3. Calculate $a_T$ and $a_N$ at $t = 2\text{ s}$:

   • Tangential Acceleration ($a_T$):

     $$\vec{v}(2) \cdot \vec{a}(2) = (9)(12) + (8)(4) + (0)(0) = 108 + 32 = 140\text{ m}^2\text{/s}^3$$
     $$a_T = \frac{140}{\sqrt{145}} = \frac{140\sqrt{145}}{145} = \frac{28\sqrt{145}}{29} \approx 11.63\text{ m/s}^2$$

   • Normal Acceleration ($a_N$):

     $$\vec{v}(2) \times \vec{a}(2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & 8 & 0 \\ 12 & 4 & 0 \end{vmatrix} = \hat{k}\Big((9)(4) - (8)(12)\Big) = \hat{k}(36 - 96) = \langle 0, 0, -60 \rangle$$
     $$\|\vec{v}(2) \times \vec{a}(2)\| = 60$$
     $$a_N = \frac{60}{\sqrt{145}} = \frac{60\sqrt{145}}{145} = \frac{12\sqrt{145}}{29} \approx 4.98\text{ m/s}^2$$

   (Pythagorean Check: $a_T^2 + a_N^2 = \frac{140^2 + 60^2}{145} = \frac{19600 + 3600}{145} = \frac{23200}{145} = 160 = 12^2 + 4^2 = \|\vec{a}(2)\|^2 \checkmark$)

4. Traction Safety Evaluation: Compare the computed normal acceleration against the maximum friction limit:

   $$a_N \approx 4.98\text{ m/s}^2 < a_{N,\max} = 12\text{ m/s}^2$$
   Since the normal acceleration demand ($4.98\text{ m/s}^2$) is well below the tire adhesion threshold ($12\text{ m/s}^2$), the vehicle will safely maintain traction without sliding out.

Meaning of the Answer & Real-Life Application:

The total acceleration of $12.65\text{ m/s}^2$ is predominantly tangential ($a_T = 11.63\text{ m/s}^2$), meaning the electric powertrain is heavily accelerating forward. The lateral turning demand ($a_N = 4.98\text{ m/s}^2$) consumes only $41.5\%$ of the tire friction envelope, confirming that the autonomous vehicle stability controller does not need to throttle back or engage torque vectoring.