Average Velocity: finding average velocity of t on a line

[mattflint50]

How can I find the average velocity of t on a line. Is there a certain formula? I completely forgot.

thank you

 

[pka]

Re: Average Velocity

How can I find the average velocity of t on a line. Is there a certain formula? I completely forgot.

You must post the actual problem!
Otherwise, we have no idea what you are asking.

 

[mattflint50]

the question asks me to find the average velocity of the particle from t=1 to t=3.

The original equation was S(t)=2(t)^3-6(t)^2=14.

Meaning the equation of the velocity =6(t)^2-12(t)

Im am not really sure what to do, but I know I should. I do not know whether I should use the original equation or the velocities equation to find it. Plases help.

 

[pka]

In general, if f is an integrable function on the interval [a,b] the the average value of f on [a,b] is give by
\(\displaystyle \L
ave(f)_{[a,b]} = \frac{1}{{b - a}}\int\limits_a^b {f(t)dt}\).

Now in your case \(\displaystyle \L
\int\limits_1^3 {v(t)dt = s(3) - s(1)}\).

So what is the average velocity?

 

[soroban]

Hello, mattflint50!

Find the average velocity of the particle from t = 1 to t =3.

The original equation was: \(\displaystyle s(t)\:=\:2t^3\,-\,6t^2\,+\,14\)

Usually with an average velocity, the derivative is not involved (well, not directly).

\(\displaystyle \;\;\;\text{Average velocity} \:=\:\frac{\text{Total distance}}{\text{Total time}}\)


But we still must be careful . . .

\(\displaystyle s(1)\;=\;2\cdot1^3\,-\,6\cdot1^2\,+\,14\;=\;10\)
\(\displaystyle s(3)\;=\;2\cdot3^3\,-\,6\cdot3^2\,+\,14\;=\;14\)

It look like the particle moved: \(\displaystyle \,14\,-\,10\:=\:4\) units to the right in 2 seconds . . . WRONG!


Find the critical value(s) of the function.
\(\displaystyle \;\;v(t)\:=\:s'(t)\:=\:6t^2 - 12t\;=\;0\;\;\Rightarrow\;\;6t(t\,-\,2)\:=\:0\;\;\Rightarrow\;\;t\,=\,0,\,2\)

You see? . . . The particle stops at t = 2 (and reverses direction).


We have: \(\displaystyle \,s(1)\;=\;10\)

Then: \(\displaystyle \,s(2)\;=\;2\cdot2^3\,-\,6\cdot2^2\,+\,14\;=\;6\)
\(\displaystyle \;\;\)The particle moved 4 units to the left.

And: \(\displaystyle \,s(3)\;=\;14\)
\(\displaystyle \;\;\)The particle moved 8 units to the right.


Hence, the particle moved a total of 12 units in 2 seconds.

Therefore, its average speed is: \(\displaystyle \,\frac{12\text{ units}}{2\text{ seconds}} \;=\;6\text{ units/sec}\)

 

[mattflint50]

S(3)-S(1) would give me 24

 

[pka]

S(3)-S(1) would give me 24

No S(3)-S(1)=14 – 10=4. Now divide by 2 and get 2.
The average velocity on [a,b] is \(\displaystyle \L \frac{{s(b) - s(a)}}{{b - a}}\)
Look at the graph of v(t) between 1 and 3.

Hello, mattflint50!
\(\displaystyle \;\;\;\text{Average velocity} \:=\:\frac{\text{Total distance}}{\text{Total time}}\)
Therefore, its average speed is: \(\displaystyle \,\frac{12\text{ units}}{2\text{ seconds}} \;=\;6\text{ units/sec}\)

Average velocity is not same as average speed: the question is about average velocity.
The average speed on [a,b] is
\(\displaystyle \L
\frac{{\int\limits_a^b {\left| {v(t)} \right|dt} }}{{b - a}}\) and in this case is 6.