Averages and Other Nice Things | Quick Calculus 6 of 6 | Doc Physics

[@adwerq1]

i can't express just how much this helped. thank you for explaining what you're doing. crashcourse really left me confusedly juggling equations. hope you keep being able to makr these.

[@DocSchuster]

+Josiah Sprute-Boyajian Thanks! Please tell your friends.

[@adwerq1]

Doc Schuster sure thing.

[@DocSchuster]

See - that's just it. SINCE I can calculate the area under the curve (using the definite integral), it's trivial for me to find the height of the rectangle that has the same base and area.

[@ahappyimago]

Doc Schuster you da man, baby!!

[@kevalexkill]

how do you know the area of THAT rectangle is the same as the area under THAT curve? is the rectangle supposed to be half the height of the curve? does this only work with a velocity that is constantly accelerating?

[@Christian-en7vb]

I do not think the height of the rectangle which has the same area (displacement) as the area underneath the curve, is half of the height of the curve, since we concluded in the video the height was 13 m/s. I do not think your velocity is accelerating, I believe it's better to say the velocity is changing or the object is accelerating. Moreover, the graph shows that the acceleration is changing since the slope of the graph is increasing at every point along the positive x-axis. I hope it helps for someone that also has the same question.

[@strawberry7loves]

Thank you sooooo much! the best playlist on calc physics out there!:)