# Ave/Instantateous Velocity

1. Sep 16, 2005

### scorpa

Hello Everyone,

Here is a question that is so stupid I am embarrassed to ask it, I know it should be easy, but for some reason I can't get it :grumpy:

1) An object starting from rest has a constant acceleration of a=8m/s^2. Find the average velocity from t=0 to t=10s using the definition of ave velocity. Show that it equals instantaneous velocity at the midtime t=5s. (Hint use the forumulae for constant a)

Ok, so I know that average velocity is change in position over change in time, and that average acceleration is change in velocity over change in time. I also know by to find the instantaneous velocity I must use change in position over change in time of the point immediately before and immediately after the midpoint (5). Now for some reason I can't seem to get anywhere with this, I hate the beginning of the school year my mind is always so useless

Thanks for any suggestions you can give me to get me on the right track.

2. Sep 16, 2005

### robphy

This is really an algebra problem.
First write what you have to show...
$$\displaystyle\frac{x_{10}-x_{0}}{t_{10}-t_{0}} \stackrel{?}{=} v_{5}$$
Use the hint: "the formulae for constant acceleration". (My hint: use one of these formulas on the left for $$x_{10}$$, and use the other on the right.)

3. Sep 17, 2005

### scorpa

Ok, I'm afraid I'm still a little bit lost. Would it be possible to say that if acceleration is 8m/s^s, take the integral of that to get velocity? No wait, that wouldn't work because that would give me instantaneous velocity right? I think I'm still a bit confused.

4. Sep 18, 2005

### robphy

It's fine. Since the acceleration is constant, the instantaneous velocity has a simple form... which you will use on the right-hand side. By doing another integral, you'll get an expression for the instantaneous position that you'll use on the left-hand side. Suggestion don't use the specific value of 8m/s^2 now... stick to the algebra and keep it as "a".

5. Sep 18, 2005

### scorpa

So do you think it would work it I went ave v=instantaneous v which equals (change in pos/change in time)= (v^2/2)?

I'm also not sure how to find the average velocity when you are given time but no position.

6. Sep 18, 2005

### Fermat

If an object starts from rest and moves with a constant acceln of 8 m/s², how far does it travel in 10s?

7. Sep 18, 2005

### scorpa

Ok, the average velocity is just 8m.s^2 x 10s = 80m/s

So to show that the ave v is equal to inst v;

We now know that the average velocity is 80m/s. If this is true we can find how far it travels in 5s. In 5 seconds we find it travels 400m. So we can now put the numbers into both sides of the equation:

(80m/s^2)(10s)=(400m)/(5s)
80m/s=80m/s

Is this correct?

8. Sep 18, 2005

### Fermat

Nope.
Sorry, but that is the final velocity.

One of the eqns of motion is,

vf = vi + at

where vf is the final velocity
vi is the initial velocity
a is the accln
t is the time of travel

Since we're starting from rest, then vi = 0. The accln = 8 m/s² and t = 10s, so
vf = 0 + 8*10
vf = 80 m/s
=========

Getting back to the problem though, you have to find the distance travelled in 10s. You should use the formula,

$$x_f = x_i + v_i*t + \frac{1}{2}at^2$$

$$\mbox{Here,}\ x_i = 0\ \mbox{and}\ v_i = 0,\ \mbox{so the formula reduces to,}$$

$$x_f = \frac{1}{2}at^2$$

Putting in a = 8 and t = 10, you will get

$$x_f = 400m$$
=========

$$\mbox{But}\ x_f\ \mbox{is the distance travelled at }t = 10$$
$$\mbox{i.e. }x_f = x_{10}$$
$$x_0 = 0$$
$$t_{10} = 10$$
$$t_0 = 0$$
$$v_{av} = \frac{(x_{10} - x_0)}{(t_{10} - t_0)}$$
$$v_{av} = \frac{(400 - 0)}{(10 - 0)}$$
$$v_{av} = 40 m/s$$

$$\mbox{Now you have to use one of the eqns of motion to show that the instantaneous velocity at }t = 5,\ v_5,\mbox{ has the same value as }v_{av}$$