# Deriving acceleration from Position vector help

In summary, the conversation discusses finding the magnitude of acceleration for an object at a given time, using the position vector and derivatives. The correct process is to differentiate the vector function component-wise and use the chain rule for non-constant terms. The conversation also addresses common errors in the calculations.

## Homework Statement

An object has a position given by r-> = [2.0 m + (1.00 m/s)t] i + [3.0m−(5.00 m/s2)t2] j, where all quantities are in SI units. What is the magnitude of the acceleration of the object at time t = 2.00 s?

## Homework Equations

All I am thinking here is that I can find acceleration by deriving velocity from the position vector, then taking the deriving the velocity vector.

## The Attempt at a Solution

deriving the original r vector for velocity gave me V=[2.0m + (1.00 m/s)(1)]i + [3.0m - (5.00 m/s^2)(2t)]j

deriving the velocity vector for a gave me A=[2.0m + (1.00 m/s)]i + [3.0m - (5.00 m/s^2)(t)]j

once I have this vector for acceleration, I get the acceleration at t=2.00s to be A(2.00s)=(3m)i + (-7m/s^2)j.

I find the vector A=7.616 m/s^2, which my online software says is incorrect. Can I get some input here?

What is the derivative of 2.0?

voko said:
What is the derivative of 2.0?
Isn't the derivative of 2, 1? I apologize, I thought you left variables in the equation until you were done deriving? As in I would put 2 in for t after I had the equation for acceleration?

Your function is basically ## \vec{r}(t) = (a + bt)\vec{i} + (c + dt^2)\vec{j}##, where ## a, \ b, \ c, \ d ## are some constants. You differentiate the vector function component-wise, so you compute ## \frac {d} {dt} (a + bt) ## and ## \frac {d} {dt} (c + dt^2) ##.

Now, ## \frac {d} {dt} (a + bt) = \frac {d} {dt} a + \frac {d} {dt} bt##, and what is ## \frac {d} {dt} a ## if ## a ## is constant?

Wouldn't the derivative of a be 0?

So then I can find that V= [(0m)+(t)]i + [(0m)-(t^2)]j

So V= (t)i + (-t^2)

Deriving V, I get that A= (1)i + (-2t)j ? I see my answer here is wrong as well... I really how no clue what I am doing wrong, and I can't even find deriving V and A from an R vector.

Ok, I now see here that the derivative of dt^2= 2dt. I'm going to see if this helps me out.

Wouldn't the derivative of a be 0?

Correct.

So then I can find that V= [(0m)+(t)]i + [(0m)-(t^2)]j

You have fixed the problem with constants. Now fix the problem with non-constants. What are ## \frac {d} {dt} (pt) ## and ##\frac {d} {dt} (qt^2) ##, where ## p, \ q ## are constant?

voko said:
Correct.

You have fixed the problem with constants. Now fix the problem with non-constants. What are ## \frac {d} {dt} (pt) ## and ##\frac {d} {dt} (qt^2) ##, where ## p, \ q ## are constant?

d/dt (pt) woud be (t)? correct? and d/dt (q(t^2)) would be 2qt? so my vector for velocity is V= (t)i + [(-10m/s^2)t]j?

Giving the vector A= [d/dt (t)]i + [d/dt (-10(t)m/s^2)]

what I can see:
1)The derivative of t would be 1, as with any lone variable?
2)the derivative of 10(t) is t, since we just remove the constant>

so A= (1m)i - (tm/s^2)

d/dt (pt) woud be (t)? correct?

2)the derivative of 10(t) is t, since we just remove the constant

This is one error that plagues your calculations repeatedly. Review this subject.

voko said:
This is one error that plagues your calculations repeatedly. Review this subject.

Wow, it was so simple that I never even thought about it. I don't know how id made it through math 137 and still make that mistake. Thank you very much.

## 1. How is acceleration related to the position vector?

The position vector is a mathematical representation of an object's location in space, while acceleration is a measure of how an object's velocity changes over time. The two are related through the second derivative of the position vector, which gives the rate of change of acceleration.

## 2. What is the process of deriving acceleration from the position vector?

The process involves taking the second derivative of the position vector with respect to time. This will give the acceleration vector, which represents the rate of change of an object's velocity over time.

## 3. Is the position vector necessary for calculating acceleration?

Yes, the position vector is necessary because it provides information about an object's location in space, which is needed to calculate the change in velocity over time.

## 4. Can acceleration be negative or positive?

Yes, acceleration can be negative or positive. Negative acceleration, also known as deceleration, means that an object is slowing down, while positive acceleration means that an object is speeding up.

## 5. How does the direction of the position vector affect the acceleration?

The direction of the position vector does not affect the acceleration itself, but it does determine the direction of the acceleration vector. For example, if the position vector is pointing in the positive x-direction, the acceleration vector will also be pointing in the positive x-direction.