Solving Worked Problems: Understanding How to Get the Answer | Help Needed!

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Homework Help Overview

The discussion revolves around a worked problem involving the calculation of time in a physics context, specifically related to motion under gravity. Participants are trying to understand how a specific time value of 2.86 seconds was derived from the given equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rearranging the equation to isolate the variable for time, with some questioning the steps taken to achieve this. There are attempts to clarify the algebraic manipulation needed to solve for time.

Discussion Status

The conversation is ongoing, with various participants providing guidance on how to approach the problem. Some express confusion about the algebraic steps, while others emphasize the importance of showing work to identify errors. There is no explicit consensus on the correct method yet.

Contextual Notes

Participants are reminded of forum rules that prohibit providing direct answers, emphasizing the need for original work and reasoning in solving the problem.

Jacob T Anderson
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Member advised to use the homework template for posts in the homework sections of PF.
This is a worked problem, so they're just basically giving examples on how you would solve this type of problem. My issue is... how did they get 2.86s? Please someone help this has been driving me crazy.
 

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Hi, Jacob. Can you rearrange the equation so that it has just t on one side?
 
Bandersnatch said:
Hi, Jacob. Can you rearrange the equation so that it has just t on one side?
It mentions about rearranging the equation but I don't think it shows you. Do you know what it would look like if it was rearranged? If so please show me.
 
The forum rules prohibit us from giving you the answer. You have to show your work.

If you're working from this textbook, you should already know how to rearrange algebraic equations.
The equation is
$$40=0+\frac{1}{2}9.8*t^2$$
How would you move all the numerical values to the left?
 
Jacob T Anderson said:
It mentions about rearranging the equation but I don't think it shows you. Do you know what it would look like if it was rearranged? If so please show me.

No, we cannot do that; it is contrary to the PF rules. You MUST attempt to solve the problem first, and show your work.
 
Bandersnatch said:
The forum rules prohibit us from giving you the answer. You have to show your work.

If you're working from this textbook, you should already know how to rearrange algebraic equations.
The equation is
$$40=0+\frac{1}{2}9.8*t^2$$
How would you move all the numerical values to the left?
Wouldn't you just 0.5 * 9.81 + 0 = 4.905t2. Then that would give you 40 = 4.905t2
 
Jacob T Anderson said:
Wouldn't you just 0.5 * 9.81 + 0 = 4.905t2. Then that would give you 40 = 4.905t2
But that still doesn't make sense because when you divide afterwards it still doesn't give you the answer.
 
Jacob T Anderson said:
Wouldn't you just 0.5 * 9.81 + 0 = 4.905t2. Then that would give you 40 = 4.905t2
This is not algebraically correct.

The rule is this:
Let's say you have an equation with four variables or constants: a,b,c,d or the form:
$$a*b=c*d$$
if you want to find d, you must eliminate everything but d from one side of the equation. This can be done by dividing both sides by the same value. Since we want to get rid of c on the right hand side, we can divide both sides by c:
$$\frac{a*b}{c}=\frac{c*d}{c}$$
##\frac{c}{c} = 1##, so we end up with:
$$\frac{a*b}{c}=d$$

Can you apply the same transformation to the equation given in the problem?
 
40=4.905t2 is correct and if you keep going it does give you the right answer.

Show your working so we can see where you are going wrong.
 
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  • #10
@Jacob T Anderson CWattters is right. I missed what you did there. (it's just not how you write it out correctly)
 
  • #11
I still don't understand what do I do next? I tried dividing both sides by 4.905 and get t = 8.15, then I square rooting 4.905 which gives me 24.05 then tried dividing both of them again but this time with 24.05 which gave me t = 1.66. I feel retarded
 
  • #12
Jacob T Anderson said:
I tried dividing both sides by 4.905 and get t = 8.15
Show your work.
 
  • #13
I got 8.15 not 8.13 oops
 

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  • #14
Sorry about the picture
 
  • #15
It's all good until the very last step, where you wrote t=8.13. What happened to the ##t^2## you had in the previous step?
 
  • #16
I got it! I forgot about the Quadratic Formula, thanks guys for the help.
 

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  • #17
Jacob T Anderson said:
I got it! I forgot about the Quadratic Formula
Simpler than that. Take the square root of both sides.
 
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  • #18
+1 No need to use the quadratic equation.
 

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