# 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.

What was calculated was the average acceleration over the 10 second period. The acceleration changes direction during the 10 seconds, so the average is not the same as the instantneous acceleration.

I agree with your result, though you seem to have chosen a rather complicated way to arrive at it. I would simply draw the diagram: 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.

• Are you sure this is a uniform circular motion question? If the velocity 20m/s is perpendicular to the radius vector, then it is and your answer is correct. But what if the velocity points radially outward? Then the displacement will be different. The radius vector will change with time. – Mephistopheles Feb 26 '16 at 18:41
• I watched the video. I'm sure. – John Rennie Feb 26 '16 at 18:43
• @Mephistophales I'm not sure what you mean by 'velocity points radially outward', and how would that change the answer. – john Feb 26 '16 at 20:27
• @john Sorry I wasn't clearer about what I meant by that. What I meant was the velocity points outward away from the instantaneous center of rotation during the entire 10 seconds and has the constant magnitude 20 m/s during that time. So its angle changes. In that case there would be different x and y accelerations for those times. the x acceleration would be +28.28 m/s^2 and the y acceleration would be +1.65 m/s^2 for time 0 to 5 sec and -1.65 m/s^2 for time 5 to 10 sec. I haven't worked out the details but the displacement in that case might be different. – Mephistopheles Feb 26 '16 at 21:37
• @john It's a moot point anyway if the path is circular as you say. The original problem wording wasn't very clear to me. – Mephistopheles Feb 26 '16 at 21:39