×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Converting power into acceleration

## Converting power into acceleration

(OP)
Currently, i have a set of output response on vibration caused by traffic from a distance. The output response are in acceleration but i have a software which can only read values in terms of power. I would like to ask if any one out there knows how to convert acceleration into power and vice versa? Is there a formula to it or a software to do it?

### RE: Converting power into acceleration

Obviously the units are not the same in Physics. Then the program will be dealing with specific assumptions non derivable just from acceleration readings. For example, a program tailored to an experiment in a specific bridge could read the power of a current in an electrical sensor and then produce nice graphics of, say, the instantaneous change of strain.

### RE: Converting power into acceleration

Dear MD7

I am not sure that I understand this but you might...

[copied from http://www.krysstal.com/calculus.html]

The derivative measures rates of change for functions that are continuous and variable. Functions like this are used extensively in science.
If there is a relationship between distance travelled (s) and time (t), then the derivative of distance with respect to time, ds/dt, gives the velocity (v) at any time.

Example: A particle moves so that its distance (in m) from a fixed point is given by
s = 2t2 - 3t + 1, where t is time in seconds. Find its velocity after 4s.

The velocity, v, is given by ds/dt so we differentiate the above equation with respect to t:

v = ds/dt = 4t - 3. When t = 4s, v = 13m/s.

If there is a relationship between the velocity of a particle (v) and time (t), then the derivative of v with respect to t, dv/dt, gives the acceleration (a) at any time.

Example: The above particle's velocity is given by v = 4t - 3. What it is acceleration after 1s.

The acceleration, a, is given by dv/dt so we differentiate the above equation with respect to t:

a = dv/dt = 4. The acceleration is constant at 4m/s2.

If there is a relationship between Energy (E) and time (t), then the derivative of E with respect to t, dE/dt, gives the power (P) at any time.

Example: A device uses up energy in a manner dependent on time: E = t3, where E is energy in Joules and t is the time in seconds. Find the power being used after 2s.

The power, P, is given by dE/dt so we differentiate the above equation with respect to t:

P = dE/dt = 3t2. The power after 2s is therefore 12W.

Good luck!!

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!