# Displacement on a turn with constant speed with a pitfall

A car is driving North-West on a highway. It has a speed of $20$ m/s with the cruise control on. Ten seconds later, the car is heading North-East (still going $20$ m/s). Determine the displacement of the car during the turn.

As the magnitude of velocity is constant throughout the turn, it appears we have uniform circular motion, so we cannot use the four linear acceleration kinematic equations. I tried to set up the problem using parametric equations. Since the car covered 90 degrees in 10 seconds, the angular speed $\omega$ is $\large \frac{\pi /2}{10} \frac{rad}{s}$ , which gives one such parametrization for the car.

$\large x(t) =\frac{400}{\pi}\cos( -\frac{\pi}{20} \cdot t - \frac{3 \pi}{4}) \\ \large y(t) = \frac{400}{\pi}\sin( -\frac{\pi}{20} \cdot t - \frac{3 \pi}{4})$

The displacement over the first 10 seconds is $\mathbf{r_1} - \mathbf{r_o} = ~\langle ~x(10),~y(10)~ \rangle - \langle ~x(0),~y(0) ~\rangle = \langle 0, \large \frac{400 \sqrt 2 }{\pi} \rangle$

Which implies that the displacement is $\large \frac{400 \sqrt 2 }{\pi} \approx \normalsize 180$ meters North.

But according to this physics Youtube uploader the displacement is $144$ meters North. https://youtu.be/CQp9vSkDeyY?list=PL4B64FB6A8FE2DC5D

The equation the uploader used is $\Delta \vec{x} =\large ( \frac{\vec{v_1}+ \vec{v_2}}{2} )~ \normalsize \Delta t$.
I believe this is only valid for linear acceleration problems. Here we have centripetal acceleration.

Any critique would be most appreciated. This is not a homework question. The question sprung out viewing a Youtube on physics. The uploaders look like they are well experienced in solving physics problems.

This should make it obvious that circumference of the circle is $800$m, the radius is $400/\pi$m and therefore the vertical displacement is $400/\pi\sqrt{2}$m.