Position of a Particle as Function of Time (Velocity, Acceleration, etc.)

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  • #1
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Homework Statement


The position of a particle as a function of time (in s) is given by C1 + C2t + C3t2.
Let C1 = 19.0 m, C2 = 39.1 m/s, and C3 = -0.62 m/s2.

(a) What is the velocity of the particle at time t = 13.0 s?
(b) What is the particle's acceleration at time t = 13.0 s?


Homework Equations



C1 + C2t + C3t2.

Let C1 = 19.0 m, C2 = 39.1 m/s, and C3 = -0.62 m/s2.

The Attempt at a Solution



I wasn't sure how to do this problem from the beginning, so I tried playing around with some numbers to see if anything could come up. I tried substituting 13.0 into the formula to create a quadratic equation and got: -104.78x2 + 508.3x + 19. I then used the quadratic formula to get some points: (-0.037, 0.0495) (4.888, 0.06186). If I'm doing anything right so far, I would say from the graph that the velocity is decreasing and the acceleration is negative?

Any help would be greatly appreciated. Thank you.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rock.freak667
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So you have (after substituting the Cs) s=19+39.1t-0.62t2.

You need to get 'v' and 'a' when 't=13'. All you need to do is remember your Relevant Equations

v=ds/dt and a=dv/dt
 
  • #3
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I actually haven't taken calculus and don't know derivatives yet but I'm taking physics and calculus simultaneously right now. I'm going to try to look derivatives up and see what I can come up with and post my solutions.

Thanks for the tip. I'll let you know what I get
 
  • #4
rock.freak667
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Well for your problem, you just need to know that

y=tn then dy/dt=ntn-1 for all n.
 
  • #5
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What does "n" stand for in the equation
 
  • #6
rock.freak667
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What does "n" stand for in the equation

'n' is a constant so it can be any number.
 
  • #7
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Wait never mind. Just saw it.
 
  • #8
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Using the equation I'm getting -1.24? Is that anything?
 
  • #9
rock.freak667
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Using the equation I'm getting -1.24? Is that anything?

That is for the velocity or acceleration?
 
  • #10
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I'm actually not sure. Can you use that equation to find both the velocity and acceleration?
And is this the "Simple Power Rule?"
 
  • #11
rock.freak667
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I'm actually not sure. Can you use that equation to find both the velocity and acceleration?
And is this the "Simple Power Rule?"

I am not sure if that is the name of the rule.

But yes, you will use the initial equation to find the velocity and acceleration.

Remember v=ds/dt (so you will now have a formula for 'v') and then a =dv/dt (you will get a formula for 'a').
 
  • #12
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Oh I didn't even think about substitution. Man I suck at this more than I thought. One sec.
 
  • #13
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Velocity = 22.98 m/s?

Acceleration = -1.24 m/s2?
 
  • #14
rock.freak667
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Yes, those should are correct.
 
  • #15
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Thank you for your help. Very much appreciate it.
 

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