Instantaneous velocity and cos, sin

In summary, the conversation discusses the derivation of the function of velocity (v(t)) for a particle moving in one-dimension with an acceleration of a(t) = 3t + 6sin(0.5t). The conversation includes the attempt at solving the problem and the correct solution, with a focus on the use of integration and the importance of using the correct units (radians vs degrees) when calculating the answer. The expert summarizer also mentions the importance of careful consideration of limits when using definite integrals.
  • #1
fishface1959
3
0
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

Homework Statement


A particle is moving in one-dimension, and its acceleration in m/s[tex]^{2}[/tex]as function of time can be written as:

a(t) = 3.0m/s^3 t + 6.0m/s[tex]^{2}[/tex]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[tex]\int[/tex] 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)

The answer states:
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
 
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  • #2
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
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
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 website too.

thanks
 
  • #5
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.
[tex]v(2s)[m/s]=15.5 + 1.5\cdot 4 - 12 \cos(1)\approx 21.5 - 12 \cdot 0.54\approx 15.02[/tex]
 
  • #6
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
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.
 

FAQ: Instantaneous velocity and cos, sin

1. What is instantaneous velocity?

Instantaneous velocity refers to the speed and direction of an object at a specific moment in time. It is the rate of change of an object's position with respect to time at a specific point.

2. How is instantaneous velocity different from average velocity?

Instantaneous velocity is the velocity at a specific point, while average velocity is the total displacement of an object over a certain time period. Instantaneous velocity gives a more accurate representation of an object's motion at a particular time, while average velocity gives an overall picture of an object's motion over a certain time period.

3. What is the relationship between instantaneous velocity and cos, sin?

Cosine and sine are trigonometric functions that relate the sides of a right triangle to its angles. In the context of instantaneous velocity, cosine and sine can be used to find the horizontal and vertical components of the velocity, respectively. This can then be used to calculate the magnitude and direction of the instantaneous velocity.

4. How is instantaneous velocity calculated?

Instantaneous velocity can be calculated by finding the derivative of the position function with respect to time. This means finding the slope of the position-time graph at a specific point. Alternatively, it can also be calculated by finding the horizontal and vertical components of the velocity using trigonometry and then using the Pythagorean theorem to find the magnitude.

5. Why is instantaneous velocity important in physics?

Instantaneous velocity is important in physics because it helps us understand the motion of objects at a specific moment in time. It allows us to analyze the changes in an object's velocity and acceleration, which are crucial in understanding the laws of motion and predicting the future motion of an object.

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