velocity: estimating instantaneous velocity from....

[Sophie]

I am very confused with instantaneous velocity. I have to estimate a sprinters instantaneous velocity at 6 seconds. I have worked out the average velocity at the intervals below, which was requested:

. . .[5, 6] = 8.6m/s
. . .[6, 7] = 10.6m/s
. . .[5.9, 6] = 11m/s
. . .[6, 6.1] = 12m/s

These numbers do not appear to be heading towards anything, which is how I normally work out instantaneous velocity. I am guessing that I would take the average of 11 and 12 to get 11.5m/s. However, this does not feel right.

Could someone explain what I am doing wrong?

Thank you!

sophie

 

[mark07]

Instantaneous velocity at 6 is the derivative of the velocity function at 6.

Without knowing the velocity function but data given at certain points, you can estimate it using the difference quotient

\(\displaystyle \L \frac{v(x+h)-v(x)}{h}\)

If you still can't finish, please post the original problem.

 

[stapel]

Could someone explain what I am doing wrong?

Lacking the rest of the information they gave you, and not being able to see what you did to arrive at your values, I'm afraid it would be difficult to analyze how you responded to what they gave you. Sorry.

Eliz.

 

[Sophie]

Here is the original question:

Suppose that a sprinter reaches the following distances s in the given times, t, below.

T(sec) 5.0 5.5 5.8 5.9 6 6.1 6.2 6.5 7
S(m) 39.7 43 46.2 47.2 48.3 49.5 50.4 53.6 58.9

I used V = Distance traveled / time elapsed
And worked out the average velocity at the 4 following points.

.[5, 6] = 8.6m/s
. . .[6, 7] = 10.6m/s
. . .[5.9, 6] = 11m/s
. . .[6, 6.1] = 12m/s

The next part of the question asks to estimate a sprinters instantaneous velocity at 6 seconds using the above information.

I have used a tangent to find the instantaneous velocity at 6, which I found to be 10.93m/s, however this is not the method I am suppose to use.

Thank you

sophie

 

[wjm11]

I am very confused with instantaneous velocity. I have to estimate a sprinters instantaneous velocity at 6 seconds. I have worked out the average velocity at the intervals below, which was requested:

. . .[5, 6] = 8.6m/s
. . .[6, 7] = 10.6m/s
. . .[5.9, 6] = 11m/s
. . .[6, 6.1] = 12m/s

These numbers do not appear to be heading towards anything, which is how I normally work out instantaneous velocity. I am guessing that I would take the average of 11 and 12 to get 11.5m/s.

Hello, Sophie,

I believe your method and answer are correct.

 

[soroban]

Hello, Sophie!

Suppose that a sprinter reaches the following distances \(\displaystyle s\) (meters} in the given times, \(\displaystyle t\) (seonds).

\(\displaystyle \begin{array}{cc}t & | & 5.0 & 5.5 & 5.8 & 5.9 & 6 & 6.1 & 6.2 & 6.5 & 7 \\ \hline \\
s & | & 39.7 & 43 & 46.2 & 47.2 & 48.3 & 49.5 & 50.4 & 53.6 & 58.9\end{array}\)

Average velocities for \(\displaystyle t\,\to\,6\)

\(\displaystyle \begin{array}{cccccccc}V(5.0,\,6.0): & \frac{48.3-39.7}{6.0-5.0} & = & \frac{8.6}{1} & = & 8.6 &\;\downarrow \\ \\ \\
V(5.5,\,6.0): & \frac{48.3-43}{6.0-5.5} & = & \frac{5.3}{0.5} & = & 10.6 &\; \downarrow\\ \\ \\
V(5.8,\,6.0): & \frac{48.3-46.2}{6.0-5.8} & = & \frac{2.1}{0.5} & = & 10.5 & \;\downarrow \\ \\ \\
V(5.9,\,6.0): & \frac{48.3-47.2}{6.0-5.9} & = & \frac{1.1}{0.1} & = & 11 & \;\downarrow \\ \\ \\
V(6.0,\,6.1): & \frac{49.5-48.3}{6.1-6.0} & = & \frac{1.2}{0.1} & = & 12 & \;\uparrow\\ \\ \\
V(6.0,\,6.2): & \frac{50.4-48.3}{6.2-6.0} & = & \frac{2.1}{0.2} & = & 10.5 & \;\uparrow \\ \\ \\
V(6.0,\,6.5): & \frac{53.6-48,3}{6.5-6.0} & = & \frac{5/3}{0.5} & = & 10.6 & \;\uparrow \\ \\ \\
V(6.0,\,7.0): & \frac{58.9-48.3}{7.0-6.0} & = & \frac{10.6}{1} & = & 10.6 & \;\uparrow\end{array}\)

Estimate a sprinters instantaneous velocity \(\displaystyle t\,=\,6\) using the above information.

It appears to be approaching a value between \(\displaystyle 11\) and \(\displaystyle 12\)
. . . . . maybe \(\displaystyle 11.5\) ?

 

[Sophie]

Thanks