Particle acceleration and velocity Problem

[s4orce]

A particle is moving along a straight line such that its acceleration is defined as a=(4s^2)m/s^2, where s is in meters. If v=-100m/s when s=10m and t=0, determine the particles velocity as a function of position.

Now I'm taking the integral of a but when I plug in the other parts, it doesn't work out. I think I'm doing something wrong.

Someone please comment on how-to approach this problem step by step thanks!

 

[Cyrus]

$$a(t)= \frac{d(s')}{dt} =4s^2$$

Move over dt integrate, set the limits, and solve.

 

[s4orce]

Can you explain a little further please?

 

[Cyrus]

Which part?

What dynamics book are you using? There should be a problem similar to this given somewhere as an example.

 

[s4orce]

after moving over dt to integrate, how did you approach it from there

 

[Cyrus]

what do you mean? Show me some work please.

 

[s4orce]

how can i post the work here?

 

[Cyrus]

Type out what you got and what you did.

You can use $$and [ / tex] tags to make it look nicer.$$

 

[s4orce]

integral 4s^2 = (4s^3)/3

s=10

4(10^3)/3 = 1333.33

1333.33/10 = 133.33 = particle velocity at x position

 

[Cyrus]

Are you familiar with the definition of acceleration?

Which class is this HW for?

 

[s4orce]

It is for a dynamics class, I just want to get a head start on a set of problems that were given to me

 

[Cyrus]

I think you need to review the basics of dynamics. What it means when they say acceleration, velocity, and position.

I think you need to solidify the fundamentals before doing this HW. You should know better that acceleration is the second derivative of position by now.

 

[s4orce]

so you can't help me?

 

[Cyrus]

Sure, tell me what the definition of acceleration is, and look for your mistake.

I can point out mistakes, but I am not going to teach you fundamentals.

 

[s4orce]

can you solve it and I can find my mistake, or post how.

also acceleration is the rate at which a object increases/changes its velocity(speed)

 

[Cyrus]

Ah, I am sorry. Its asking for velocity as a function of position. I was thinking you had to integrate twice.

What you need to do: set up your limits of integration.

 

[s4orce]

?

 

[Cyrus]

Do you know what limits of integration are?

 

[s4orce]

yes integral from one value to another

 

[Cyrus]

so what are your value you used?

 

[s4orce]

0 to 10....

 

[Cyrus]

What happened to the limits on the integral of the velocity?

 

[s4orce]

-100x? that one?

 

[Cyrus]

When integrating, you always have two limits. Two for position, two for velocity. What are the two limits for the integral of velocity?

 

[s4orce]

im confused man...

 

[Cyrus]

$$\int^v_{-100}$$

Remember, I said you must have an upper and lower limit of integration.

 

[s4orce]

can you just show me how you approac it please?

 

[Cyrus]

Sure, give me a second. I made another mistake reading the problem. Sorry!

 

[s4orce]

ok thanks!

 

[Cyrus]

Ok, it was tricky. Had to crack open my dynamics book from a few semesters ago!

Here is the trick:

$$V=\frac{ds}{dt}$$

Also,

$$a=\frac{dv}{dt}$$

If you solve each of these for $$\frac{1}{dt}$$ you will get:

a*ds=v*dv

Now set your limits of integration and replace your relation for acceleration -a

$$\int^s_{10} 4s^2 ds = \int^V_{-100}vdv$$

Is this clear?