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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!
a(t)= \frac{d(s')}{dt} =4s^2
Move over dt integrate, set the limits, and solve.
Can you explain a little further please?
Which part?
What dynamics book are you using? There should be a problem similar to this given somewhere as an example.
after moving over dt to integrate, how did you approach it from there
what do you mean? Show me some work please.
how can i post the work here?
Type out what you got and what you did.
You can use [tex] and [ / tex] tags to make it look nicer.
integral 4s^2 = (4s^3)/3
s=10
4(10^3)/3 = 1333.33
1333.33/10 = 133.33 = particle velocity at x position
Are you familiar with the definition of acceleration?
Which class is this HW for?
It is for a dynamics class, I just want to get a head start on a set of problems that were given to me
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.
Sure, tell me what the definition of acceleration is, and look for your mistake.
I can point out mistakes, but im not going to teach you fundamentals. :wink:
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)
Ah, Im 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.
Do you know what limits of integration are?
yes integral from one value to another
so what are your value you used?
What happened to the limits on the integral of the velocity?
When integrating, you always have two limits. Two for position, two for velocity. What are the two limits for the integral of velocity?
\int^v_{-100}
Remember, I said you must have an upper and lower limit of integration.
can you just show me how you approac it please?
Sure, give me a second. I made another mistake reading the problem. Sorry!
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?
What do you mean "plug in"?
With these problems, you have to read them very carefully, or you will make mistakes. (Like I did when I misread it twice).
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.
how can i determine that :)
determine the particles velocity as a function of position
Integrate that function I gave you.
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