Originally Posted by avito009 So now how can F=ma be true because the formula contains acceleration and in an inertial reference frame there is no acceleration or the acceleration is zero. 
It's totally fine for inertial frames to contain objects that accelerate. It's accelerating
frames (actually... just noninertial frames) that cause issues with the physics and therefore require additional treatment (e.g. Coriolis terms in rotating frames).
An inertial frame of reference is one in which acceleration can be measured. So what it means is this: Lets take an example: A ball is at rest and you take the ball and throw it horizontally to another person, so initially the body is at rest so acceleration is 0. Now when the ball is thrown it accelerates 3 meters per second. So the force is equal to: say ball has a mass of a kg so F=ma= 1*3= 3 kg m/s^2. So Force is measured when there is acceleration.How?

Well, force is defined as the rate of change of momentum which, for particles with mass, causes acceleration. Therefore, when acceleration is observed, a force can be associated with it. Note that forces can be associated with acceleration regardless of what kind of frame you are in (e.g. the centrifugal force and Coriolis force associated with dynamics in rotating reference frames).
What this second law states is that initially the body is at rest but later it accelerates but after it accelerates its speed remains constant. From the above example the ball it at rest so acceleration =0. But later it accelerates 3 meters per second and stays at this speed. So this is an inertial reference frame since acceleration is measured and stays constant later.
But if after the ball accelerated and later it decelerated and again it accelerated then newtons second law does not hold and force is not measurable since acceleration is not measurable.
This is in laymans terms, is this correct?

No, because what you have here is a measurement of an
average acceleration over a given duration and measuring a value of zero for your average acceleration doesn't necessarily mean that the instantaneous accelerations are zero. Even so, measurements of zero don't invalidate laws of physics; they're just zero measurements.
Let's put some numbers to your last point. Let's say that at time t = 0, a 1 kg ball accelerates for 0.001 seconds at 1 m s$\displaystyle ^{2}$ and then 0.008 seconds later (t=0.009) the ball decelerates for 0.001 seconds at 1 m s$\displaystyle ^{2}$. At t = 0 and t = 0.01 the ball's speed is zero but between times t = 0.001 and t = 0.008, the ball's speed is 0.001 ms$\displaystyle ^{1}$. The velocities in the initial and final phases look like linear ramps that connect the 0.001 value to 0.
Now, let's assume we have this particle in an inertial frame and you have an accelerometer that measures the acceleration of an object every 0.01 seconds. The measurement of the average acceleration is made by sampling the velocity at two points in time and then plugging into the following formula:
$\displaystyle a = \frac{\Delta v}{\Delta t} = \frac{v_2  v_1}{0.01}$
Since the velocities are zero for both times, the measured acceleration is:
$\displaystyle a = \frac{0  0}{0.01} = 0 m s^{2}$
However, remember that a deceleration is the same as a positive acceleration, so the two accelerations can be written down as
$\displaystyle a_1 = 1 m s^{2}$
$\displaystyle a_2 = 1 m s^{2}$
So is it surprising that our average acceleration over the interval is 0?
However, let's say we upgraded our accelerometer so it can now make measurements every 0.001 seconds (so it can make 10 measurements in the same that the original accelerometer made 1). For the first 0.001 seconds we get
$\displaystyle a = \frac{0.001  0}{0.001} = 1 m s^{2}$
and for the last 0.001 seconds, we get
$\displaystyle a = \frac{0  0.001}{0.001} = 1 m s^{2}$
so it seems our new accelerometer is now capable of measuring the changes.
Small changes that occur over very short time intervals are called perturbations. Perturbations happen all the time in reality and are considered in physics theory like everything else. In fact, perturbation theory allows for the exploration of what can happen to certain systems if a perturbation does occur.
Now... let's say that our frame of reference is not a accelerating frame (i.e. noninertial) and it accelerates at a constant 0.001 m s$\displaystyle ^{2}$. In the period t = 0 to t= 0.001, the object will look stationary in the frame since it's acceleration is the same as the frame of reference. However, in the period t = 0.001 to t = 0.009, the object will appear to accelerate relative to it's inertial frame at 1 m s$\displaystyle ^{2}$, so the final velocity at t = 0.009 is
$\displaystyle v_2 = v_1 + at = 0  1 \times 0.008 = 0.008 m s^{1}$
The final part will then have the ball accelerate at 2 m s$\displaystyle ^{2}$ for 0.001 seconds, giving a final velocity of 0.01 m s$\displaystyle ^{1}$
Okay... now let's look at forces. In the first situation we looked at (inertial frame) we have the forces for the accelerating and decelerating phases as
$\displaystyle F = ma = 1 \times 1 = 1 N$
and
$\displaystyle F = ma = 1 \times 1 = 1 N$
and zero for the constant velocity bit. However, for the second situation (noninertial frame) we have
$\displaystyle F = ma = 0 N$
for the first phase and
$\displaystyle F = ma = 1 \times 1 = 1 N$
for the second phase. The last phase has an even greater force:
$\displaystyle F = ma = 1 \times 2 = 2 N$
So it is clear that the formula gives incompatible results between inertial and noninertial frames. The equations that describe mechanics in noninertial frames does exist though(consider, for example,
https://en.wikipedia.org/wiki/Mechan...article_motion)