- #1

- 240

- 23

I wonder why we can take the mean value of only those 2 velocity? Thank you.

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- Thread starter agnimusayoti
- Start date

- #1

- 240

- 23

I wonder why we can take the mean value of only those 2 velocity? Thank you.

- #2

- 10,949

- 12,222

Algebraically, you can write down a formula for distance traveled in terms of the initial speed ##u##, the acceleration ##a##, and the time taken ##t##. And you can write down an expression for the final velocity ##v## in terms of the same quantities. You can then eliminate ##a## from the first expression and divide by ##t## to get the average velocity. With a bit of algebra, ##(v-u)/2## should drop out.

Graphically, plot velocity as a function of time. It should look like the red line:

(Forgive the unlabelled axes - the horizontal one is time and the vertical is velocity.) Distance traveled is the area under the line on this kind of graph (because it's ##\int v(t)dt##). The average velocity is the constant velocity you would have to travel at to cover the same distance in the same time. Where would you have to draw a horizontal line (i.e. constant velocity) so that the area under the graph between times 0 and ##t## is the same? (Hint: the area of a triangle is half base times height). Does your answer depend on anything except ##u## and ##v##?

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