Est instantaneous acceleration of the car at t=20

[karush]

A car is traveling along a straight road with values of its continuous velocity \(\displaystyle f(t)\) in ft/sec as given

\(\displaystyle t \ \ v(t)\)

\(\displaystyle 0 \ \ 90\)
\(\displaystyle 10 \ \ 75\)
\(\displaystyle 20 \ \ 80\)
\(\displaystyle 30 \ \ 100\)
\(\displaystyle 40\ \ 90\)
\(\displaystyle 50 \ \ 85\)
\(\displaystyle 60 \ \ 80\)

Est the instantaneous acceleration of the car at \(\displaystyle t=20\) sec.

there is no eq for this so using the values from table of t=10 and t=30 is elapsed time of 20sec
and from v(10)=75 and v(30)=100 we have a difference of 25 so presume we can the slope from this for instantaneous accelaeration at 20 sec

 

[Jameson]

It's been a few years since I've done this problem, but it definitely involves using a secant line to approximate the tangent line.

Using Paul's Online Math Notes as a reference, I would approximate this by:

\(\displaystyle a(20) \approx \frac{v(20)-v(10)}{20-10}\)

I think using the values for 20 and 10 is better than 30 and 10 because you don't need a spread of 20 seconds, you want to approximate the acceleration at $t=20$. The time spread between 20 and 10 is only 10 seconds and it's as small as we can get using this data.

 

[MarkFL]

I suspect the method Jameson suggests is what is expected, but since the velocity is only increasing on $10\le t<30$ going by the data points, I might be tempted to find the quadratic passing through the points:

$(10,75),\,(20,80),\,(30,100)$

and then evaluate its derivative at $t=20$.

View attachment 2454

Another approach would be to simply average the change in velocity from $t=10$ to $t=20$ with the change from $t=20$ to $t=30$. :D

Both methods give the same value...coincidence?

 

[karush]

ok, well since this is a free response question with very limited time to do it probably just taking the average is best. (however this is just a practice handout

there are some more questions on this but will deal with them tomorrow.

 

[MarkFL]

The two methods I suggested also give the same answer as the method you originally posted. I can see why the method you posted and the second method I posted are the same, but I would have to use a little algebra to demonstrate that the quadratic would always produce the same results.

 

[Ackbach]

The two methods I suggested also give the same answer as the method you originally posted. I can see why the method you posted and the second method I posted are the same, but I would have to use a little algebra to demonstrate that the quadratic would always produce the same results.

$$\frac{\frac{v(30)-v(20)}{30-20}+\frac{v(20)-v(10)}{20-10}}{2}=\frac{\frac{v(30)-v(20)+v(20)-v(10)}{10}}{2}=\frac{v(30)-v(10)}{20}.$$
Key to this working is that the two points on either side are an equal distance away from the central point, thus making the denominators the same.

Fitting a quadratic would be thus:
\begin{align*}
\ell(x)&=v(10)\frac{t-20}{10-20} \, \frac{t-30}{10-30}+
v(20)\frac{t-10}{20-10} \, \frac{t-30}{20-30}+
v(30)\frac{t-10}{30-10} \, \frac{t-20}{30-20} \\
&=v(10)\frac{(t-20)(t-30)}{200}+v(20)\frac{(t-10)(t-30)}{-100} +v(30)\frac{(t-10)(t-20)}{200}.
\end{align*}
We compute $\ell'(20)$ to obtain:
\begin{align*}
\ell'(20)&=\left[\frac{v(10)}{200}((t-30)+(t-20))-\frac{v(20)}{100}((t-10)+(t-30))+\frac{v(30)}{200}((t-10)+(t-20))\right]_{t=20} \\
&=\left[\frac{v(10)}{200}(2t-50)-\frac{v(20)}{100}(2t-40)+\frac{v(30)}{200}(2t-30)\right]_{t=20} \\
&=\frac{v(10)}{200}(-10)+\frac{v(30)}{200}(10) \\
&=\frac{v(30)-v(10)}{20},
\end{align*}
as before.

 

[MarkFL]

For the quadratic, I had in mind beginning with the function:

\(\displaystyle f(x)=ax^2+bx+c\) where $a\ne0$

And the 3 $x$-values in arithmetic progression:

\(\displaystyle x_1<x_2<x_3\)

And so, we want to demonstrate that:

\(\displaystyle f'\left(x_2\right)=\frac{f\left(x_3\right)-f\left(x_1\right)}{x_3-x_1}\)

\(\displaystyle 2ax_2+b=\frac{\left(ax_3^2+bx_3+c\right)-\left(ax_1^2+bx_1+c\right)}{x_3-x_1}\)

\(\displaystyle 2ax_2+b=\frac{a\left(x_3^2-x_1^2\right)+b\left(x_3-x_1\right)}{x_3-x_1}\)

\(\displaystyle 2ax_2+b=a\left(x_3+x_1\right)+b\)

\(\displaystyle 2x_2=x_3+x_1\)

\(\displaystyle x_2-x_1=x_3-x_2\)

Since the three $x$ values are in AP, we know this is true, and so the hypothesis is true.

 

[karush]

Wow that was pretty interesting
Another question asked was what was area underneath the graph of this.
since they only give a set of points they said to use the trapazoid method. My question is what use would knowing the area be also I don't see how a function could be derived to go thru all 7 points

 

[MarkFL]

The area under a velocity graph represents the total distance traveled. Consider the case of constant velocity:

\(\displaystyle d=vt\)

You see, here distance can be thought of as the area of a rectangle, where the base is the elapsed time and the height is the speed.

In the case of varying speed, then we use the calculus and the fact that velocity is the time rate of change of position, and so the integral of the velocity function over the time interval will represent the total distance traveled.

I would just use trapezoids as suggested...you could of course use rectangles or even parabolic, or higher order polynomials, but I think this would be overkill for the problem.

You could find a polynomial through all the points, but without a CAS, it would be extremely tedious. :D

 

[karush]

https://www.physicsforums.com/attachments/2455
ok I got \(\displaystyle 5150 ft^2\) for this but if this is \(\displaystyle ft / sec\)
and after \(\displaystyle 60 sec\) we shouldn't have traveled this much?

 

[MarkFL]

Just a quibble on your units here...the bases of the trapezoids are speeds given in ft/s and the heights are time given in s, and so the areas would be in ft. You wouldn't want to say the distance you traveled is the square of a linear measure. :D

I haven't checked your calculations, but that's about a mile a minute or 60 mph, which is 88 ft/s and so that seems reasonable. :D

 

[karush]

thanks for all the help

I thot the \(\displaystyle ft/s\) thing was strange for a car on the road
since it is usually is in \(\displaystyle mi/hr\)

also the deriving the quadratic was pretty helpful. In reality the speed of car on the road would have to be represented as a curve with no corners.

 

[Ackbach]

thanks for all the help

I thot the \(\displaystyle ft/s\) thing was strange for a car on the road
since it is usually is in \(\displaystyle mi/hr\)

also the deriving the quadratic was pretty helpful. In reality the speed of car on the road would have to be represented as a curve with no corners.

In which case you could use cubic splines to model it.