Christos
  • Christos
Physics, Can you help me with (C) please. https://www.dropbox.com/s/f9v9w4z5fk9u2vz/Screenshot%202015-11-15%2000.44.17.png?dl=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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wmj259
  • wmj259
Do you have a little understanding for integration from calculus or is this Physics class algebra based?
wmj259
  • wmj259
Integrate velocity to find displacement.
wmj259
  • wmj259
Which basically means find the area under the curve of each portion of the graph to find the total area which is the total displacement.

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Christos
  • Christos
how about I use V = V_0 + at
Christos
  • Christos
remember I only want part (C)
Christos
  • Christos
the one asking for accel
Christos
  • Christos
so there is this intepretation https://www.dropbox.com/s/j358aun6hma5fkl/Screenshot%202015-11-15%2003.13.30.png?dl=0
Christos
  • Christos
and then there is mine: working with the formula V = V_0 + at
Christos
  • Christos
However both give different result. So. who's correct and why ?
Christos
  • Christos
@zepdrix
wmj259
  • wmj259
you should find the equation for the speed from 4 seconds to 6 seconds, and then take the derivative of that function. And evaluate it at t=5sec
wmj259
  • wmj259
That is to find the instantaneous acceleration.
Christos
  • Christos
how about V = V_0 + at
wmj259
  • wmj259
That equation is to find final velocity as acceleration causes the object to speed up or slow down as time progresses with an initial velocity.
SolomonZelman
  • SolomonZelman
Which parts do you need hep with at this point?
SolomonZelman
  • SolomonZelman
I would assume you have for the least done a and b, is that so?
wmj259
  • wmj259
@SolomonZelman, Please let me know if my replies were incorrect so I can learn from this also.
SolomonZelman
  • SolomonZelman
(c) Acceleration is the rate at which velocity is gained or lost, or the second derivative of position, or first derivative of velocity ──> (the sope of velocity)... can be defined in various ways. So, acceleration in this case would simply be the instantaneous slope (just line, so you don't need calculus). (d) Average velocity would be the average value of the function. (e) slope of secant between t=1 and t=5.
SolomonZelman
  • SolomonZelman
Laggy today :(
Christos
  • Christos
i though I could use V = V_0 + at and solve for a to find a (PART C)
SolomonZelman
  • SolomonZelman
Well, you are given the function, and the acceleration at x=c is the instantaneous slope at x=c. In a case of a line you can easily find this slope
SolomonZelman
  • SolomonZelman
I mean instantaeous slope of velocity.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle A(x)= V'(x) }\) \(\large\color{black}{ \displaystyle A(5)= V'(5) }\) A - acceeration V - velocity Aternatively, \(\large\color{black}{ \displaystyle A(5)= }\) the slope at t=5
SolomonZelman
  • SolomonZelman
I don't think it would be valid in this case to use the formula you proposed, unless you redefine things. Define t=4 as t=0 and equivalently, t=5 as t=1 Then, V = V_0 + at (where V denotes the final velocity) V_0 = V(4)=14 V = V(5)=7 14 = 7 + a•(5-4)
SolomonZelman
  • SolomonZelman
You can't use it just V = V_0 + at V(5) = V(0) + a•5 7 = 0 + a•5 7/5 = a but that would be INCORRECT.
SolomonZelman
  • SolomonZelman
what I did was basicaly defined the function as: |dw:1447552423252:dw|
Christos
  • Christos
@SolomonZelman are you here?
Christos
  • Christos
how v(4) = 14
SolomonZelman
  • SolomonZelman
well, that is just a point on your velocity-function given

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