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This question is about the content presented in section 14.4 of Stewart Calculus 5th edition (Motion in Space: Velocity and Acceleration). It is particularly about Tangential and Normal Components of Acceleration.

Suppose that the vector function of the motion of the particle is given by $\mathbf{r}(t)=(r_1,r_2,r_3)$. Then the speed of the particle is calculated by $$ |\mathbf{r}'(t)|=\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2} $$ The tangential component of the acceleration can be calculated by $$ a_T=\frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{|\mathbf{r'}(t)|} $$ and the normal component of the acceleration can be calculated from

$$a_N=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r'}(t)|}$$

Now, if we only have the numerical values of $|\mathbf{r'}(t)|$, $a_N$, $a_T$ for $t=0,0.1,0.2,0.3,0.4,0.5,\ldots,10$ seconds, then how can we calculate the position at each $t$; that is what is $\mathbf{r}(0)$, $\mathbf{r}(0.1)$, $\mathbf{r}(0.2)$, ..., and $\mathbf{r}(10)$?

EDIT: This is not a book exercise; I am just using the theory from the textbook. For a race car, I have various $t$s from $t=0$ to $t=120$ seconds (in the increments of $0.02$ seconds). For each of these times, I also have the $a_T$, $a_N$, and speed of the car. Now I want to find out at each $t$, what is the position of the car? (assuming the car is at $x=0$ and $y=0$ at $t=0$).

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  • $\begingroup$ You can only approximate it, since the function is doing god knows what in between. $\endgroup$ – Archaick May 10 '15 at 4:09
  • $\begingroup$ Fair enough; how would you approximate it? $\endgroup$ – user77791 May 10 '15 at 4:10
  • $\begingroup$ You need to have $r(0)$ among other things. $\endgroup$ – copper.hat May 10 '15 at 4:13
  • $\begingroup$ @copper.hat: Assume that $\mathbf{r}(0)=\mathbf{0}$. $\endgroup$ – user77791 May 10 '15 at 4:14
  • $\begingroup$ You need the normal & tangent vectors, so you really don't have enough info. For example, if you rotate the whole system you will get the same measured values above, but clearly the solution will be a rotated one. There is not enough information here. If you were given $r'(t)$ then you might be able to approximate the normal. tangent vectors. $\endgroup$ – copper.hat May 10 '15 at 4:18
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You can't. All you have is the speed at certain times. You have no information about the direction and you don't know the speed between the tenths of seconds. Maybe between $0$ seconds and $0.1$ seconds the speed was multiplied by $1000$. Maybe between each tenth second there is a sharp turn.

You can do a numerical integration. You can then update the velocity using the acceleration, update the orientation of the car using the velocities, and update the position using the velocities. This could be done in a program or easily in an Excel sheet. The columns I would make are
time
longitudinal acceleration (LA)
tangential acceleration (TA)
pointing angle =atan2(VX,vY)
x velocity (VX(t+delta t)=VX(t)+x component of acceleration * delta t)
y velocity (VY(t+delta t)=VY(t)+y component of acceleration * delta t)
x position (X(t+delta t)=X(t)+x component of velocity * delta t)
y position (Y(t+delta t)=Y(t)+y component of velocity * delta t)

All this assumes the acceleration doesn't change much in each interval. You might want to average the values at the start and end of the interval to get the value you use in the update.

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  • $\begingroup$ This is not a book exercise. For a race car, I have various $t$s from $t=0$ to $t=120$ seconds (in the increments of $0.02$ seconds). For each of these times, I also have the $a_T$, $a_N$, and speed of the car. Now I want to find out at each $t$, what is the position of the car? (assuming the car is at $x=0$ and $y=0$ at $t=0$). $\endgroup$ – user77791 May 10 '15 at 4:22
  • $\begingroup$ Is this a 1D problem because the car is going down a track? This is important because now we know the direction of velocity and don't care about normal acceleration at all. Now you have some choices. You can just take the speed at each timestep and assume it is constant over the timestep. Then the distance traveled in the timestep is the speed times the timestep. You could fit a higher order polynomial through the speeds and integrate that polynomial to get position, which assumes the speed is smooth. $\endgroup$ – Ross Millikan May 10 '15 at 4:33
  • $\begingroup$ How about if, we first calculate the distance travelled between each timestep? we can use trapezoidal rule for two t values and two speeds at each time, to calculate the distance travelled. The distance travelled is exactly the arc length which is radius * angle. From this we can calculate the coordinate of the point considering that when we are moving a curved path from A to B, it is actually a rotation and therefore we can use rotation formula to find the coordinate of the point. $\endgroup$ – user77791 May 12 '15 at 1:38
  • $\begingroup$ If you start using velocity to compute change in position, the ending velocity will be the same as the starting because you have not applied the acceleration. The method will still work. If you average the acceleration over the interval, essentially using the trapezoidal method, you will get an integration that is second order accurate in velocity. You can then integrate the position the same way. If you start with the velocity you will only be first order accurate. That may not be a problem if the acceleration changes slowly. $\endgroup$ – Ross Millikan May 12 '15 at 4:34
  • $\begingroup$ Do you mean integrating longitudinal and latitudinal accelerations? In curvilinear motion, we only have tangential velocity. $\endgroup$ – user77791 May 12 '15 at 8:33

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