Finding 'a' when 'x' is given as a function of 't'.

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In summary, the conversation revolves around finding the acceleration of a particle at a given time using the position equation x^2 = t^2 + 1. The group discusses the steps for finding the answer, including the use of derivatives and the chain rule. One person also asks for clarification on a previous algorithm mentioned.
  • #1
Noob245
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Hello everyone! I have this question given to me by a friend and the question is:
Position of a particle x is given by x^2=t^2 + 1, t is time in second, find acceleration for time t.

So this is what I decide to do:
x^2 = t^2+1
so
x = sqrt(t^2+1)
if you double differentiate the function on RHS
You get:
(t^2+1)^-3/2 or 1/(t^2+1)^3/2

I'd be grateful if you give me the steps for finding the answer (including the steps for finding out the derivative, because I used an online calculator).
 
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  • #2
Noob245 said:
Hello everyone! I have this question given to me by a friend and the question is:
Position of a particle x is given by x^2=t^2 + 1, t is time in second, find acceleration for time t.

So this is what I decide to do:
x^2 = t^2+1
so
x = sqrt(t^2+1)
if you double differentiate the function on RHS
You get:
(t^2+1)^-3/2 or 1/(t^2+1)^3/2

I'd be grateful if you give me the steps for finding the answer (including the steps for finding out the derivative, because I used an online calculator).

Are you saying you don't know how to determine the derivative of the right hand side? If not, show us how you get the first derivative?
 
  • #3
Chestermiller said:
Are you saying you don't know how to determine the derivative of the right hand side? If not, show us how you get the first derivative?

No, I do not know how. :(
 
  • #4
It's the standard derivation of a function... hint: think how you can express roots as powers and think what the derivative of a power is...
Good work
 
  • #5
Noob245 said:
No, I do not know how. :(
Have you had calculus yet?
 
  • #6
Yes; I tried double differentiating it but my answer didn't match the one given in the calc. the answer I got was

-1/(t^2+1)^-3/2

This mistake is all the more reason I want the steps. I know that if you double differentiate 'x' I get 'a' but for some reason by answer does not even come close to the original answer. It'd be really kind of you guys if you could just tell me if I was correct or wrong, and if wrong where I went wrong and how I should proceed further.
 
  • #7
Have you heard of the chain rule?
 
  • #8
Noob245 said:
Yes; I tried double differentiating it but my answer didn't match the one given in the calc. the answer I got was

-1/(t^2+1)^-3/2

This mistake is all the more reason I want the steps. I know that if you double differentiate 'x' I get 'a' but for some reason by answer does not even come close to the original answer. It'd be really kind of you guys if you could just tell me if I was correct or wrong, and if wrong where I went wrong and how I should proceed further.
From this answer, it just looks like all you did was make a mistake in algebra. So us the details so we can see if we can spot it.

chet
 
  • #9
GUYS GUYS! Forget about the calculus, please tell me if the algorithm I gave for sum is correct or not? If not, tell me the correct algorithm and the solution so I can try it out and match my answer.
 
  • #10
What algorithm?

How about you try the first derivative first?
##x = \sqrt{t^2 + 1}## - what is dx/dt? You are right that the derivative of ##\sqrt{u} = u^{\tfrac12}## is ##\tfrac12 u^{-\tfrac12} = \frac{1}{2 \sqrt{u}}##. But as the square root is over t² + 1, not u, you will need the chain rule like SteamKing said.
 
  • #11
Noob245 said:
GUYS GUYS! Forget about the calculus, please tell me if the algorithm I gave for sum is correct or not? If not, tell me the correct algorithm and the solution so I can try it out and match my answer.

You didn't give any detail about what you did... haw are we supposed to know where and why and what you did wrong?
 

Related to Finding 'a' when 'x' is given as a function of 't'.

1. What is 'a' in the given function?

The variable 'a' in the given function represents the acceleration of the object in the given system. It is a measure of how quickly the velocity of the object is changing over time.

2. How do you find 'a' in the given function?

To find 'a', you can use the formula: a = d²x/dt², where dx/dt is the first derivative of x with respect to t, and d²x/dt² is the second derivative of x with respect to t. Alternatively, you can use a graph or table of values to estimate the value of 'a' at a specific point in time.

3. Can 'a' be negative in this function?

Yes, 'a' can be negative in this function. A negative acceleration indicates that the object is slowing down or decelerating.

4. What are the units of 'a' in this function?

The units of 'a' in this function depend on the units of x and t. For example, if x is measured in meters and t is measured in seconds, then the units of 'a' would be meters per second squared (m/s²).

5. How does changing 'x' or 't' affect the value of 'a' in this function?

Changing 'x' or 't' can affect the value of 'a' in different ways. For example, if 'x' is changed while 't' remains constant, 'a' may increase or decrease depending on the specific function. Similarly, if 't' is changed while 'x' remains constant, 'a' may also increase or decrease. It is important to analyze the specific function to determine the relationship between 'a' and the other variables.

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