# Instantaneous velocity and cos, sin

1. Sep 28, 2010

### fishface1959

I'd like to apologize for not using the symbols but I spent over 1 1/2 hours trying and after using one of the symbols it wouldn't format correctly for any others.

The s^2. s^3 etc... are seconds

1. The problem statement, all variables and given/known data
A particle is moving in one-dimension, and its acceleration in m/s$$^{2}$$as function of time can be written as:

a(t) = 3.0m/s^3 t + 6.0m/s$$^{2}$$sin(0.5/s t)

Assume: v(0) = 3.5m/s and x(0) = 5.0m

Derive the function of velocity v(t)
Calculate the instantaneous velocity v(t) when t = 2.0s

The attempt at a solution

This is what I come up with for v(t)

v(t) =integral a(t) + Vi
= 3.5m/s + (3.0m/s^3$$\int$$ t) + (6.0m/s^2 integral sin(u)du)

= 3.5m/s + 1.5m/s^3 t^2 - (6.0m/s^2)cos(.5/s t)(2)

v(t) = 15.5 m/s + 1.5m/s^3 t^2 - 12.0 m/s cos(.5/s t)
v(2s) = 15.02m/s

I do not know where the 15.5 comes from. I realize that at t=0, cos(0) = 1 , I'm not sure how or if this plays into the V(t) equation.

Using the equation to the answer v(2s) answer I come up with V(2s) = 9.5 - 3.5=6 (approx), not what the answer is supposed to be.

Could someone please tell me what I am missing and redirect me to what I need to learn.

Thank you
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 28, 2010

### rock.freak667

a(t) = 3t + 6sin(0.5t)

v(t)= ∫a(t) dt

v(t) = ∫ ( 3t + 6sin(0.5t)) dt

integrating the sin(0.5t) will give you a cosine term in which putting t=0, will give a number and not just zero.

The 15.5 comes from this fact. Do over the integral and you will see.

3. Sep 28, 2010

### betel

You found the point already. It's the cos(0).
You have to be careful what limits you use for your integration. Either you use no limits and fit the initial data by hand. Or if you add the initial velocity your integrals have to go from 0 to t, so that for t=0 all contributions from integrals are zero.

4. Sep 28, 2010

### fishface1959

Thank you for your wonderful help in explaining the integration when using the definate integral and your quick response. Part of my problem is that I haven't had calculas for at least 15 years and I'm taking physics.

I still don't understand where the answer for v(2) comes from.

v(t) = 15.5 m/s + 1.5m/s^3 t^2 - 12.0 m/s cos(.5/s t)
v(2s) = 15.02m/s

I would appreciate it if you couild respond to this. The answer would be great but what I need is to know what I'm missing, also. I'm going to check out the calc part of this web site too.

thanks

5. Sep 28, 2010

### betel

If you calculate the result with your calculator it depends on what argument the calculator expects for the cosine, either in radians or degrees. For radians you will get the correct result.
$$v(2s)[m/s]=15.5 + 1.5\cdot 4 - 12 \cos(1)\approx 21.5 - 12 \cdot 0.54\approx 15.02$$

6. Sep 28, 2010

### jhae2.718

Now that you understand how you get v(t), evaluate v(2), where (0.5 /s t) gives an angle in radians.

Edit: Too late..

7. Sep 28, 2010

### fishface1959

I GOT IT!!!

v(2) = 15.02 m/s

My calculator is in DEGREES and when I enter cos I am getting degrees instead of radians.

Dang I hate this crap :)

Thanks to helping I truly appreciate it. It's these little things that really drive a girl crazy.