# Average velocity vs. average speed - Videos

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Average velocity vs. average speed

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1. Dalibor Maksimovic

First

2. Emanuele Usai

Interesting topic

3. Haron Lua
4. ayush goyal

Nice revision , solved it in 45 seconds #YAY

5. Cobalt314

So, is it true that speed would just be the absolute value of the velocity? Like, if you were to integrate velocity divided by the time interval to get average velocity, would you integrate average velocity divided by the time interval to get average speed?

6. misotanni

Nice, except the unit of time is seconds, not secants. Abbreviated as s. It doesn't cause confusion with the symbol for distance because one of them is a quantity and the other is a unit. Like this: t[s], s[m].

7. Snejpu

8. Varad Mahashabde

Backpack? That's pretty forgetful

9. CrayNz

Best example is v(t)=sin(t)
If you integrate v(t) from 0 to 2Pi and devide it by 2Pi you'll get average velocity of 0.
If you want the average speed, you need to integrate from 0 to Pi and from Pi to 2Pi. Add the absolute values of these integrals and devide by 2Pi to get the average speed.

10. Sergio H

Now do it using vector calculus 😀💯

Also nice video I thought it was very intuitive <3

11. Dalibor Maksimovic

Do some IMO, pls

12. history buff03

The first part of the problem nicely illustrates the procedure for finding the average value of a function. Analytically, to find the average value of the velocity function, one integrates the velocity function to get the distance function and then plugs in the given limits of integration and then divides by the difference between these limits. This is equivalent to finding the ratio of the change of position to the change in time.

13. Joe Potillor

For speed we use v as well, velocity usually is v with a vector arrow (physicist pet peeve :p). And yes seconds have symbol s. But otherwise perfect.

14. Przemofonix

that's actually pretty smarty.. #YAY

15. 9898 nawaf

we want more physics

16. Kuol Perveilov

Well, how can you make an equation from you walking? I've wonder this since school

17. 9wyn

Thank you!

18. ungratefulmetalpansy

This is why China is taking over the world. Westerners are too busy tattooing and entertaining themselves.

19. J Bottero

Your house better be close as the v function will rapidly exceed light speed.

20. Pedro

You could have used the sign function, so you don't need to bother with grafics:
∫ |v(t)| dt = Speed
v(t)=d/dt(S(t)) and the upper bound you can set to be just "t" and the lower the initial time, or whatever you want.
|v(t)|=v(t)sgn(v(t))
∫ |v(t)| dt = S(t)*sgn(v(t))
But it was a great analysis and your videos are really awesome!

21. WisdomVendor1

this is fairly easy. The hard part would be measuring the distance an object traveled and changed directions and measuring again distance and stopping and changed directions and recording the time involved all along and then writing a position function to represent that path followed. I don't think it can be done which is why I lend little merit to 'solving' a position function which is "given".

22. Garand Chua

At 0:45 why avg speed is |instantaneous velocity|? Shouldn't it be |average velocity|?

23. rorygo456

Do the calc 2 way #YAY

24. Ryan Roberson

V: integral(velocity)dt /t
S: integral(||velocity||)dt /t
Works in any number of dimensions, right? ||x||=e^real(ln(x))

25. Skeleton Rowdie

13:40 YEAH BRO

26. Stefan Klepzig

So the formula for the average speed from t=t₀ to t₁ would be:
(∫|s‘(t)|dt for t=t₀ to t₁)/(t₁- t₀)

s‘(t) = 3t²-12t+9
|s‘(t)| = 3t²-12t+9 if t<1, -3t²+12t-9 if 1≤t≤3, 3t²-12t+9 if t>3
∫|s‘(t)|dt =
t³-6t²+9t if t<1,
-t³+6t²-9t-[-1³+6·1²-9·1]+[1³-6·1²+9·1] if 1≤t≤3,
t³-6t²+9t-[3³-6·3²+9·3]+[-3³+6·3²-9·3-[-1³+6·1²-9·1]+[1³-6·1²+9·1]] if t>3
∫|s‘(t)|dt =
t³-6t²+9t if t<1,
-t³+6t²-9t-(-4)+4 if 1≤t≤3,
t³-6t²+9t-0+[0-(-4)+4] if t>3
∫|s‘(t)|dt =
t³-6t²+9t if t<1,
-t³+6t²-9t+8 if 1≤t≤3,
t³-6t²+9t+8 if t>3

∫|s‘(t)|dt for t=0 to 5 = [t³-6t²+9t+8]_t=5 – [t³-6t²+9t]_t=0 = (5³-6·5²+9·5+8) – (0³-6·0²+9·0)
= 28 – 0 = 28

(∫|s‘(t)|dt for t=0 to 5)/(5-0) = 28/5 = 5.6 m/s

And for example for t=2 to 4 it would be:

∫|s‘(t)|dt for t=2 to 4 = [t³-6t²+9t+8]_t=4 – [-t³+6t²-9t+8]_t=2 = [4³-6·4²+9·4+8] – [-2³+6·2²-9·2+8]
= 12 – 6 = 6

(∫|s‘(t)|dt for t=2 to 4)/(4-2) = 6/2 = 3 m/s

27. Papai Pal

My new Physics Teacher!