Ave/Instantateous Velocity

[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.

 

[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.)

 

[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.

 

[robphy]

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.

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".

 

[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.

 

[Fermat]

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

 

[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?

 

[Fermat]

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

...

Is this correct?

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

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

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

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