How to get average velocity: Vaverage = V1 + V2 / 2?

[Indranil]

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

 

[berkeman]

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

Seems reasonable. Is there a problem with what you've written?

 

[berkeman]

I'd just mention that you should be using parenthesis to be sure to group the numerator terms and the denominator terms. The way you've written it without parenthesis is ambiguous...

 

[kuruman]

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

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 $$\bar v=\frac{x_2-x_1}{t_2-t_1}~?$$

 

[Mark44]

Homework Equations

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

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.

The Attempt at a Solution

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

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.

 

[Indranil]

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 $$\bar v=\frac{x_2-x_1}{t_2-t_1}~?$$

Yes you are right. I am asking the same thing.

 

[berkeman]

Yes you are right. I am asking the same thing.

Using vector velocities or scalar speeds?

 

[Indranil]

Using vector velocities or scalar speeds?

both of them

 

[berkeman]

both of them

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?

 

[Indranil]

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.

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?

 

[Indranil]

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?

Velocity = Displacement / time. It expresses both the direction and the magnitude. As it expresses direction, it's a vector quantity.
Speed = Distance / time. It expresses magnitude only. As it expresses magnitude only, it's a scalar quantity.

 

[mfb]

@Mark44: It depends on what you average over. If the two velocities are attained for the same time then you can use the simple arithmetic average. If they apply to different times (as in your example) you need a weighted average.

 

[Charles Link]

In general, the formula $\bar{v}=\frac{v_i+v_f}{2}$ works for constant acceleration $a$, so that distance $s=v_i t +\frac{1}{2}at^2$ with $v_f=v_i+at$. This gives the result that $\bar{v}=\frac{s}{t}=v_i+\frac{1}{2}at=v_i+\frac{1}{2}(v_f-v_i)=\frac{1}{2}(v_i+v_f)$. $\\$ There may be isolated cases where it also works, but in general, it applies to constant acceleration.

 

[Indranil]

Yes you are right. I am asking the same thing.

Still, I don't understand how to get V = (V1 + V2) / 2 ? as I know V = (X2-X1) / (t2-t1) = delta X / delta t?

 

[mfb]

You can't consider formulas in isolation, you have to consider what they describe. What is the motion you want to look at? Constant acceleration between t1 and t2? Something else?

 

[Indranil]

You can't consider formulas in isolation, you have to consider what they describe. What is the motion you want to look at? Constant acceleration between t1 and t2? Something else?

Constant acceleration between t1 and t2

 

[kuruman]

Still, I don't understand how to get V = (V1 + V2) / 2 ? as I know V = (X2-X1) / (t2-t1) = delta X / delta t?

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.

 

[Mark44]

If they apply to different times (as in your example) you need a weighted average.

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.