1. ## Particle Motion

A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a(t) = ln(1 + 2t). If the velocity of the particle is 2 at time t = 1, what is the velocity of the particle at time t - 2?
- I'm having a tough time getting the antiderivative for a(t)

2. Originally Posted by xxlvh
A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a(t) = ln(1 + 2t). If the velocity of the particle is 2 at time t = 1, what is the velocity of the particle at time t - 2?
- I'm having a tough time getting the antiderivative for a(t)
Do you mean $a(t) = \ln (1 + 2^t)$ ?

3. Whoops! Yes that is what I meant to put.

4. Originally Posted by mr fantastic
Do you mean $a(t) = \ln (1 + 2^t)$ ?
Originally Posted by xxlvh
Whoops! Yes that is what I meant to put.
Is an exact answer required? Because you can express the velocity as

$v(t) = \int_1^t \ln (1 + 2^u) \, du + 2$.

Then

$v(2) = \int_1^2 \ln (1 + 2^u) \, du + 2$

and the integral can be approximately evaluated using technology.

I don't think an exact answer in terms of a finite number of elementary functions is possible (although I may stand corrected on this).

5. Originally Posted by mr fantastic
Is an exact answer required? Because you can express the velocity as

$v(t) = \int_1^t \ln (1 + 2^u) \, du + 2$.

Then

$v(2) = \int_1^2 \ln (1 + 2^u) \, du + 2$

and the integral can be approximately evaluated using technology.

I don't think an exact answer in terms of a finite number of elementary functions is possible (although I may stand corrected on this).
I agree with Mr. F. The symbolic answer to this problem involves the dilogarithim function...which has absolutely no meaning or use here, but thanks to Mathcad we can approximate the answer to $\int_1^2 \ln\left(1+2^x\right)~dx\approx 1.346$. Hope that helps.