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## Homework Statement

How to get Vaverage ( V ) = V1 + V2 / 2?

## Homework Equations

How to get Vaverage ( V ) = V1 + V2 / 2?

## The Attempt at a Solution

As I know Vaverage ( V ) = X2-X1 / t2-t1 = (delta)X / (delta) t

- Thread starter Indranil
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- #1

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How to get Vaverage ( V ) = V1 + V2 / 2?

How to get Vaverage ( V ) = V1 + V2 / 2?

As I know Vaverage ( V ) = X2-X1 / t2-t1 = (delta)X / (delta) t

- #2

berkeman

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Seems reasonable. Is there a problem with what you've written?As I know Vaverage ( V ) = X2-X1 / t2-t1 = (delta)X / (delta) t

- #3

berkeman

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Are you asking how to get that the average velocity is given by $$\bar v = \frac{v_1+v_2}{2}$$ starting from the definition $$\bar v=\frac{x_2-x_1}{t_2-t_1}~?$$## Homework Statement

How to get Vaverage ( V ) = V1 + V2 / 2?

## Homework Equations

How to get Vaverage ( V ) = V1 + V2 / 2?

## The Attempt at a Solution

As I know Vaverage ( V ) = X2-X1 / t2-t1 = (delta)X / (delta) t

- #5

Mark44

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Absolutely not!## Homework Equations

How to get Vaverage ( V ) = V1 + V2 / 2?

Consider a section of a road that is 1 mile long that goes up a hill. A driver in a car goes up the hill at an average speed of 30 miles per hour. How fast must the driver go back down the hill to have an average speed for both sections of 60 miles per hour?

The intuitive, but wrong, answer is that if the driver goes down the hill, the average speed for both sections will be 60 mph. If you work it out using the formula ##d = r \cdot t##, and solving for time t, you'll find that 90 mph isn't fast enough.

As already mentioned, you need to use parentheses!Indranil said:## The Attempt at a Solution

As I know Vaverage ( V ) = X2-X1 / t2-t1 = (delta)X / (delta) t

Your first formula (which is incorrect) means ##V_{ave} = V_1 + \frac {V_2} 2##, which, besides being wrong, isn't what you intended.

Your second formula means ##V_{ave} = X_2 - \frac {X_1}{t_2} - t_1##, which I'm sure isn't what you mean.

- #6

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Yes you are right. I am asking the same thing.Are you asking how to get that the average velocity is given by $$\bar v = \frac{v_1+v_2}{2}$$ starting from the definition $$\bar v=\frac{x_2-x_1}{t_2-t_1}~?$$

- #7

berkeman

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Using vector velocities or scalar speeds?Yes you are right. I am asking the same thing.

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both of themUsing vector velocities or scalar speeds?

- #9

berkeman

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What differences could there be between those two different results? Can you show us some of your reading and understanding on that question? What is the difference between vector velocity and scalar speed?both of them

- #10

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I don't understand the formula you mentioned d = r.t. Could discuss the formula I mean what does this formula mean how to use this formula and where to use this formula?Absolutely not!

Consider a section of a road that is 1 mile long that goes up a hill. A driver in a car goes up the hill at an average speed of 30 miles per hour. How fast must the driver go back down the hill to have an average speed for both sections of 60 miles per hour?

The intuitive, but wrong, answer is that if the driver goes down the hill, the average speed for both sections will be 60 mph. If you work it out using the formula ##d = r \cdot t##, and solving for time t, you'll find that 90 mph isn't fast enough.

As already mentioned, you need to use parentheses!

Your first formula (which is incorrect) means ##V_{ave} = V_1 + \frac {V_2} 2##, which, besides being wrong, isn't what you intended.

Your second formula means ##V_{ave} = X_2 - \frac {X_1}{t_2} - t_1##, which I'm sure isn't what you mean.

- #11

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Velocity = Displacement / time. It expresses both the direction and the magnitude. As it expresses direction, it's a vector quantity.What differences could there be between those two different results? Can you show us some of your reading and understanding on that question? What is the difference between vector velocity and scalar speed?

Speed = Distance / time. It expresses magnitude only. As it expresses magnitude only, it's a scalar quantity.

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Still, I don't understand how to get V = (V1 + V2) / 2 ? as I know V = (X2-X1) / (t2-t1) = delta X / delta t?Yes you are right. I am asking the same thing.

- #15

mfb

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You know that under constant acceleration, ##x=v_0t+\frac{1}{2}at^2##. Write expressions for ##x_2## ad ##x_1##, take the difference and divide by ##t_2-t_1## as the expression for ##\bar v## suggests. You will need to use the velocity equation ##v=v_0+at## to eliminate terms such as ##at_2## and ##at_1##. The identity ##a^2-b^2=(a+b)(a-b)## also comes into play.Still, I don't understand how to get V = (V1 + V2) / 2 ? as I know V = (X2-X1) / (t2-t1) = delta X / delta t?

- #18

Mark44

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Which really was the point of my example. Also, if all you have is a couple of velocities, and don't know the times involved, then merely taking the average of the two velocities leads to an incorrect answer.If they apply to different times (as in your example) you need a weighted average.

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