Physics: Decreasing Speed Parabola Equation

  • #1
kalcorn
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0
Homework Statement:
• What is the equation of this motion?

• According to the equation, what was the car’s…

i. initial position (= constant term)

ii. initial velocity (= t coefficient)

iii. acceleration (= double the t2 coefficient)
Relevant Equations:
X=A*T^2+BT+C
Screen Shot 2020-05-19 at 3.01.42 PM.png
 

Answers and Replies

  • #2
i. initial position (= constant term)
ii. initial velocity (= t coefficient)
iii. acceleration (= double the t2 coefficient)

Right, but do you know why those quantities are those parts of the equation you stated? If ##x = At^2 + Bt + C##, then can you see how you find ##x(0)##, ##\frac{dx}{dt}(0)## and ##\frac{d^2 x}{dt^2}(0)##?

It should be easy enough to read the actual coefficients off of the little auto-fit box
 
  • #3
kalcorn
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I am seriously lost I never took physics or calculus in high school and this is my first college course regrading either.
 
  • #4
I am seriously lost I never took physics or calculus in high school and this is my first college course regrading either.

That's okay. What do you know about how velocity and acceleration are defined in terms of rates of change?
 
  • #5
kalcorn
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I don't, I guess my big question overall is what is "t" like is it an actual number I need to plug into the equation or is it just a variable?
 
  • #6
It's a parameter, if you choose any number for ##t## (so long as the time of the measurement is actually defined), and plug it into the equation, you will get out the position at that time.

Have you covered some calculus?
 
  • #7
kalcorn
9
0
No, The highest level of math I ever had was algebra 2 and geometry.
 
  • #8
In that case, it's going to be hard to convince you of why the things you put in the brackets are true. So I think you just need to take them as true, for now at least.

Ignoring the uncertainties for the moment, can you see what those three quantities are if the relation is $$x = -14.19t^2 +79.75 t+0.1982$$
 
  • #9
kalcorn
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0
Thats what I kinda thought was the equation I didn't put the negative or the t^2. It just looks very different from the "example" equations given in our powerpoint.
 
  • #10
haruspex
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hard to convince you of why the things you put in the brackets are true
Not sure if you are reading it differently, but it looks to me that the descriptions in parentheses were provided as hints in the question, not added by @kalcorn. If so, there is no evidence of any work by the OP.

@kalcorn , am I right about that? Do you understand what is meant by a coefficient? If I write an equation like y=ax+b, can you identify the coefficient of x?
 
  • #11
kalcorn
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Not sure if you are reading it differently, but it looks to me that the descriptions in parentheses were provided as hints in the question, not added by @kalcorn. If so, there is no evidence of any work by the OP.

@kalcorn , am I right about that? Do you understand what is meant by a coefficient? If I write an equation like y=ax+b, can you identify the coefficient of x?
You’re correct those were given by the professor. I do not know what a coefficient is but I would guess a since it’s attached? Math and science are not my strengths. I’m going to school to be a severe interventions special education teacher and this just happens to be a required class, that I have no background knowledge in. Also the class is online with no lecture so I’m teaching myself due to covid.
 
  • #12
kalcorn
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You’re correct those were given by the professor. I do not know what a coefficient is but I would guess a since it’s attached? Math and science are not my strengths. I’m going to school to be a severe interventions special education teacher and this just happens to be a required class, that I have no background knowledge in. Also the class is online with no lecture so I’m teaching myself due to covid.
I did make the graph myself with a given video on loggerpro.
 
  • #13
haruspex
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I do not know what a coefficient is but I would guess a since it’s attached?
Yes, it's the constant multiplier in the term.
The equation ##y=ax^2+bx+c## has three 'terms' on the RHS. If x is the independent variable and y the dependent variable (i.e. its value changes when x changes), and a, b and c are constants, we say that a is the coefficient in the quadratic (i.e. x2) term and b is the coefficient in the linear term.

In your equation x is the dependent variable and t the independent variable.
By convention, we start time at t=0, so the initial position is the value of x when t=0.
We can get the velocity (rate of change of position) at any instant by differentiating the equation with respect to t. The rules of differentiation give ##v=2at+b##. The initial velocity is the value of v when t=0.
To get the acceleration (rate of change of velocity) we differentiate again to get ##accn=2a##.
 
  • #14
kalcorn
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Yes, it's the constant multiplier in the term.
The equation ##y=ax^2+bx+c## has three 'terms' on the RHS. If x is the independent variable and y the dependent variable (i.e. its value changes when x changes), and a, b and c are constants, we say that a is the coefficient in the quadratic (i.e. x2) term and b is the coefficient in the linear term.

In your equation x is the dependent variable and t the independent variable.
By convention, we start time at t=0, so the initial position is the value of x when t=0.
We can get the velocity (rate of change of position) at any instant by differentiating the equation with respect to t. The rules of differentiation give ##v=2at+b##. The initial velocity is the value of v when t=0.
To get the acceleration (rate of change of velocity) we differentiate again to get ##accn=2a##.
Thank you. I do understand that a bit better now!
 

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