Velocity is speed with direction. Speed is the magnitude of velocity. When you square them, they are both the same. Average speed takes the magnitude of velocity at each point in time, and finds the average value of the function. Average velocity only considers the start and end positions, and calculates the net change in position divided by total time.
When I took physics in college, we had a problem where we had to calculate the average velocity given a velocity function. Nominally, the answer would be (Vf - Vi)/2, but that only works if the acceleration is constant. However, the velocity function wasn't linear, so no constant acceleration. In order to get the average velocity, you had to use the mean value theorem. I'm pretty sure I was the only one in the class that got it right.
For a parabola in general, given by: y = a*x^2 + b*x + c The vertex is located at x = -b/(2*a). This tells us that for this application of a parabola, which tells us the y-position as a function of time, that the time of the maximum y-position at the vertex of the path, happens at t = -b/(2*a). Plug in the given a = -16 and b = 80, and get t = 5/2 seconds as the time at which the projectile reaches its maximum height.
For a bigger picture, for polynomials in general, given as: y = a*x^n + b*x^(n-1) + c*x^(n - 2) + ... other terms ... + k Something interesting will happen at the x-value of x = -b/(n*a). For parabolas, this is the vertex. For cubics, this is the inflection point. The collection of roots will average to this value, and this is the case for all polynomials. It's also strategy to horizontally shift a polynomial, so that the b*x^(n - 1) term disappears completely, and then undo the shift when finished.
How to find the instantaneous velocity at t=1: th-cam.com/video/ApVEVPz-MgY/w-d-xo.htmlsi=rpODZJIWaN0pkwGY
That red marker is going to get it tonight
I m sorry but what the fu-
Beautiful Explanation and Video...❤❤
NOICE Dear Teacher❤❤
What's the difference between the definition of velocity and the definition of speed?
iirc velocity uses the displacement (shortest path between the start and end) while speed uses raw distance (amount traveled)
Velocity is speed with direction. Speed is the magnitude of velocity. When you square them, they are both the same.
Average speed takes the magnitude of velocity at each point in time, and finds the average value of the function.
Average velocity only considers the start and end positions, and calculates the net change in position divided by total time.
When I took physics in college, we had a problem where we had to calculate the average velocity given a velocity function. Nominally, the answer would be (Vf - Vi)/2, but that only works if the acceleration is constant. However, the velocity function wasn't linear, so no constant acceleration. In order to get the average velocity, you had to use the mean value theorem.
I'm pretty sure I was the only one in the class that got it right.
I agree your insight.
If h(t) is vertical height then u should have given the horizontal speed as that would change the avg speed.
Max: t ? How you did this?
For a parabola in general, given by:
y = a*x^2 + b*x + c
The vertex is located at x = -b/(2*a).
This tells us that for this application of a parabola, which tells us the y-position as a function of time, that the time of the maximum y-position at the vertex of the path, happens at t = -b/(2*a). Plug in the given a = -16 and b = 80, and get t = 5/2 seconds as the time at which the projectile reaches its maximum height.
For a bigger picture, for polynomials in general, given as:
y = a*x^n + b*x^(n-1) + c*x^(n - 2) + ... other terms ... + k
Something interesting will happen at the x-value of x = -b/(n*a). For parabolas, this is the vertex. For cubics, this is the inflection point. The collection of roots will average to this value, and this is the case for all polynomials. It's also strategy to horizontally shift a polynomial, so that the b*x^(n - 1) term disappears completely, and then undo the shift when finished.
@@carultch thanks👍🌷
something basic
Why so difficult..... Just use av(f)=(1/(b-a))\int_{a}^{b}f(x)dx
Thats what he did he jist simplified
Original poster, the host is deliberately not using calculus.