# Deriving acceleration from Position vector help

## 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?

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?

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.

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.