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Except that F = ma, not ma^{2}, F is a vector, not a scalar,c is a speed, not an acceleration.In other words, no, not at all.

I did not ask any of the above Alan , I asked E=mc² is exactly the same as F=ma² when (c) and (a) are accelerations of the same magnitude?

Quote from: Thebox liI did not ask any of the above Alan , I asked E=mc² is exactly the same as F=ma² when (c) and (a) are accelerations of the same magnitude?No. Those aren't the same. In fact F = ma^2 is dimensionally incorrect and not an equality.

I now want to know what is the acceleration of light if you are saying c is only a speed?

Take a simple situation where a = b + c. Say a = 10, b = 4 and c = 6. We can square both sides of the equation as in a^2 = (b + c)^2. To expand (b + c)^2 we first have (b + c)(b + c). This multiplied out gives b^2 + 2bc + c^2. Working this out we get a^2 = 10^2 which is 10 times 10 so that is 100. Now we have to find a value of 100 from b^2 + 2bc + c^2. b^2 is 4^2 which equals 16. c^2 is 6^2 which equals 36. So from those we get 16 + 36 which equals 52. Now we get to 2bc which is 2 times 4 times 6. So two times 4 is 8 and 6 times 8 = 48. Then adding 52 to 48 hey presto we get 100. Squaring values on both sides of an equation can be very useful when solving some types of equation.

As a good first step in advancing your understanding of probability a well known real life example is useful.https://en.wikipedia.org/wiki/Lottery_mathematics

F=m1*aF=M2*aF=ma²does this not describe a collision?If not sorry my misunderstanding.

E=mc² is exactly the same as F=ma² when (c) and (a) are accelerations of the same magnitude?

By the way "c" is speed whereas "a" is acceleration. Why did you think that the "c" in E = mc^{2} was acceleration?