Learn the difference between the average velocity and the average speed of a moving particle. Remember the average velocity is always the change in distance divided by the change in time but the average speed is the total distance divided by the change in time.
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#blackpenredpen #math #calculus
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Does anyone know that he has put Doraemon them song on background 😂
Speed is the derivate of position, and the average of a function is the intagrate of this function divided by the difference of time, so I thought it was the intagrate from 0 to 5 of thé derivative of the position divided by 5 (so s(5)-s(0))/5 , but I was wrong
Well …..I didn't consider the fact that it is all happening in one direction…..I assumed it to be following some wierd locus while travelling😏
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you enlightened me, sir. Thank you!
i was searching exact these questions explaniation.
thanks.
How make sense t power 3 for speed
very good job educating the youth. thankful for people like you.
making the world a better place…
If s(t) describes the displacement of the particle then why wouldn’t the arc length of s(t) from t=0 to t=5 not describe the total length traveled ? thefore the average speed will be 29.07/5=5.814m/s
This is astonishing 😍😍😍😍😍😍😍😍😍😍😍it's my first time I feel that big difference between them
Profesor de física, wowww…. me dio risa cuando te acercabas y luego retrocedías jajaja …. 🤣…. eres un heroico 💪💪💪
I was having a huge trouble with my specialist maths and this video helps a lot😭💖
How did you derive a function of speed?
😨 thumbnail formula does not match with the formula xd
at 3:00 Is it what we call as Lagrange's mean value theorem?
What happen if S(3) not equal to 0 ? (from the video)
You're best at both maths and physics.
I'm confused. From your definition of velocity, if someone travels in a circle, their average velocity is 0 km/h. Is this correct?
I'm totally ok with this methodology. Velocity is a vector and Speed is a scalar.
Average speed is the integral of the velocity function between t=0s and t=5s over the time interval, that is 4 m/s..
The average speed is a vector and I'm not sure how could be calculated..
In Portuguese, both ''velocity'' and ''speed'' are translated as ''velocidade''
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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
V: integral(velocity)dt /t
S: integral(||velocity||)dt /t
Works in any number of dimensions, right? ||x||=e^real(ln(x))
Do the calc 2 way #YAY
At 0:45 why avg speed is |instantaneous velocity|? Shouldn't it be |average velocity|?