Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble understanding what a problem I have is seeking.

To simplify the problem:

A particle reaches a speed of 1.6 m/s in a 5.0 micrometer launch. The speed is reduced to zero in 1.0 mm by the air. Assume constant acceleration and find the acceleration in terms of g during a) the launch and b) the speed reduction.

The basic strategy to find acceleration I am using is to calculate two velocity equations: one between (0 m/s, 0 m) and (1.6 m/s, 5.0 micrometers); the second between (1.6 m/s, 5.0 micrometers) and (0 m/s, 1.0 mm). Then I will derive the acceleration value for each. Because acceleration is constant I can expect a linear velocity equation.

What is confusing me is that we are to assume constant acceleration. Thus the acceleration equation will merely be some real number. So, what exactly is expected if it is to be in terms of g? Is my strategy to find acceleration incorrect?

share|cite|improve this question
up vote 5 down vote accepted

$g$ denotes the local acceleration due to gravity near Earth's surface. $g = 9.8 \, \mathrm{m}/\mathrm{s}^2$. Whatever acceleration you find, you should express it as a multiple of this value.


$$54 \, \mathrm{m}/\mathrm{s}^2 = 54 \, \mathrm{m}/\mathrm{s}^2 \times \frac{1g}{9.8 \, \mathrm{m}/\mathrm{s}^2} = 5.5g$$

share|cite|improve this answer
Thank you. My thinking was confined to the mathematical sense of "in terms of..." When I acquire the requisite reputation I will upvote this. – d0rmLife Jan 27 '13 at 22:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.