Answer to Question #146034 in Mechanical Engineering for Yashwanth

Question

What is the value of the crosstrack error, governed by the ODE e'(t) = -ke(t), at t=2 given that e(0)=4 and k = 1?

Please give your answer with the precision of 2 decimal places.

Answer 1

Let us rewrite the differential equation as $\frac{d e(t)}{d t} = - k e(t)$, or $\frac{d e(t)}{e(t)} = - k dt$.

Integrating from both sides, obtain $\ln \frac{e(t)}{C} = - k t$, from where $e(t) = C e^{-k t}$.

Using initial condition, $e(0) = 4 = C$, hence $e(t) = 4 e^{- k t}$.

For $k=1$, $e(2) = 4 e^{-2} \approx 0.54$.