# Distance travelled on a curvelinear path and the coordinate of points

A race car travels in a curvilinear path at points A, B, and C. The following data is given:

At point A

• time 0 seconds
• Speed is 195.1696800 $\large{\frac{m}{s}}$
• tangential acceleration is 0.22 $\large{\frac{m}{s^2}}$
• normal acceleration is -0.0709320 $\large{\frac{m}{s^2}}$.

At point B

• time 0.02 seconds
• Speed is 195.3692400 $\large{\frac{m}{s}}$
• tangential acceleration is 0.22 $\large{\frac{m}{s^2}}$
• normal acceleration is -0.0659368 $\large{\frac{m}{s^2}}$.

At point C

• time 0.04 seconds
• Speed is 195.4690200 $\large{\frac{m}{s}}$
• tangential acceleration is 0.21 $\large{\frac{m}{s^2}}$
• normal acceleration is -0.0619406 $\large{\frac{m}{s^2}}$.

If A has coordinate $A(0,0)$, how can I calculate the distance travelled between A and B, and the distance travelled between B and C? How can I calculate the coordinates of point B and C?

Here is what I have tried: I am not really sure how to calculate the distance travelleed between each points; this is what I have done and it may be wrong.

In curvilinear motion, the tangential acceleration is the derivative of velocity with respect to time, so I integrated tangential acceleartion twice (using trapezoidal rule) to get the distance travelled between each points. The distance travelled between each points is exactly the arc length between those pair of points so I have $\ell=r\theta$ where $\ell$ is the distance travelled between each pair of points (arc length) and $r$ is radius of curvature. $r$ can be calculated from $a_n=v^2/r$ where $v$ is speed and $a_n$ is normal acceleration. So now I can calculate $\theta$ and once I caculate $\theta$, then I can calculate the coordinate of the next point (I have A, so I first compute B and if I have the coordinate of B, then I can compute the coordinate of C):

\begin{align} x_2&=x_0+(x_1-x_0)\cos\theta-(y_1-y_0)\sin\theta\\ y_2&=y_0+(x_1-x_0)\sin\theta+(y_1-y_0)\cos\theta \end{align}

For each arc, I start with $x_1$ and $y_1$ and I wish to compute $x_2$ and $y_2$. $C(x_0,y_0)$ is the coordinate of the center of curvature for each arc. Since any arbitrary point has equations $$x=x_0+r\cos\phi,\quad\text{and }\quad y=y_0+r\sin\phi$$

Then $(x_1,y_1)$ should satisfy this, so this gives me: $$x_0=x_1-r,\quad\text{and }\quad y_0=y_1$$

• Set up the equations of equilibrium in polar coordinates with position vector $r e^{ i \theta}$ so that there is a common origin for $A,B,C$. An unknown force may be assumed as a function of $r$ or $\theta$. Differentiation involves two components of velocity and two for acceleration. Positions obtained by integration. Study of planetary motion may be helpful. – Narasimham May 13 '15 at 2:38
• @Narasimham: Sadly, I can not understand your answer. Can you give a detailed answer? – user77791 May 13 '15 at 2:46

## 1 Answer

Please read up more under Curvilinear motion in the plane. There may be no sweet shortcut. Suggesting an ODE or differential equations approach when the postions are obtained by integration of $\ddot r , \ddot \theta$.

When $r$ and $\theta$ vary with respect to time, components of

$$V: ( r \dot \theta, \dot r )$$

$$Acceleration : ( r \ddot \theta, r {\dot \theta}^2 , 2 \dot \theta \dot r, \ddot r )$$

at each of the three points $A,B,C$.

• This does not answer my question. Sorry. – user77791 May 13 '15 at 9:59