Constant Acceleration -- Equation for Vavge

[Danger Mouse]

Homework Statement

I've recently started Sears and Zemansky's "University Physics"... Everything was fine until pg. 42.

Relevant Equations

...

Initially we are given the statement V_av = (x-x₀)/t, so far so good. But, we encounter the following paragraph...

"We can also get a second expression for V_av that is valid only when the acceleration is constant, so that the v-t graph is a straight line (as in Fig 2-14 - [I've omitted the graph here, it's a v-t graph with constant acceleration]) and the velocity changes at a constant rate. In this case the average velocity during any time interval is imply the arithmetic average of the velocities at the beginning and end of the interval."

For the time interval from 0 to t:

V_av = (v₀ + v)/2

I can't for the life of me figure out where the above equation comes from.

My Apologies if homework is the wrong place for this.

 

[lomidrevo]

How would you calculate an arithmetic average of two numbers, let's say 10 and 20?

 

[haruspex]

Homework Statement:: I've recently started Sears and Zemansky's "University Physics"... Everything was fine until pg. 42.
Homework Equations:: ...

Initially we are given the statement V_av = (x-x₀)/t, so far so good. But, we encounter the following paragraph...

"We can also get a second expression for V_av that is valid only when the acceleration is constant, so that the v-t graph is a straight line (as in Fig 2-14 - [I've omitted the graph here, it's a v-t graph with constant acceleration]) and the velocity changes at a constant rate. In this case the average velocity during any time interval is imply the arithmetic average of the velocities at the beginning and end of the interval."

For the time interval from 0 to t:

V_av = (v₀ + v)/2

I can't for the life of me figure out where the above equation comes from.

My Apologies if homework is the wrong place for this.

The easiest way to see it is graphically. Draw a graph of velocity against time. It will be a straight line from height v₀ to height v₁. The average height is (v₀ + v₁)/2.
(The distance covered is the area underneath.)

 

[RPinPA]

They're just giving it to you as an unproven theorem. Even though average velocity is emphatically NOT defined as the average of the start and end velocities in general, in the case of uniform acceleration they are the same.

You can prove it graphically as another answer suggests. You can also derive it algebraically from the equations of uniform acceleration.

$v = v_0 + at \Rightarrow at = v - v_0$

$d = v_0t + (1/2)at^2$

So $v_{avg} = d/t = v_0 + (1/2)at = v_0 + (1/2)(v - v_0) = (1/2)(v + v_0)$

 

[mfig]

These expressions are equivalent. Perhaps it will help you to see how.

Recall that:

$a=constant \implies v(t)=at+v_0 \implies x(t) = \frac{at^2}{2} + v_0t + x_0 \tag{1}$

Now the first expression for the average velocity is:

$\frac{x_2-x_1}{t_2-t_1}\tag{2}$

If we use the third expression in (1) for the x values in (2), then this gives:

$\frac{\frac{a t_2^2}{2} + v_0t_2 + x_0 -\left(\frac{a t_1^2}{2} + v_0t_1 + x_0\right)}{t_2-t_1}\tag{3}$

This simplifies to:

$\frac{\frac{a }{2}(t_2^2 - t_1^2) + v_0(t_2-t_1)} {t_2-t_1}\tag{4}$

Now consider the second expression you give for the average velocity:

$\frac{v_1 + v_2}{2}\tag{5}$

If we use the second expression in (1) for the velocity values in (5), then we have:

$\frac{at_1 +v_0+ at_2 + v_0}{2} = \frac{a(t_1 + t_2)+ 2v_0}{2} = \frac{a(t_1 + t_2)+ 2v_0}{2} \cdot \frac{t_2-t_1}{t_2-t_1} = \frac{\frac{a }{2}(t_2^2 - t_1^2) + v_0(t_2-t_1)} {t_2-t_1}\tag{6}$

So the expressions (4) and (6) are equivalent, and therefore they both represent the average velocity over some time interval given constant acceleration.

HTH

 

[Danger Mouse]

Thanks! It does...