# What does it mean to find acceleration in terms of g?

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?

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