The discussion centers on Richard Feynman's lectures regarding Newton's Laws of Dynamics, specifically focusing on equations (9.13), (9.14), and (9.15). Equation (9.13) illustrates that when velocity is constant, displacement can be calculated as the product of time elapsed and velocity. However, since velocity is time-dependent, this becomes an approximation, improving with smaller time intervals. Equations (9.14) and (9.15) relate velocity to acceleration, with the latter defined as the negative of position, as indicated in equation (9.12).
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Thank you very much! I agree that Feynman explains it well, but I still have gaps in my knowledge.DrClaude said:It is hard to explain things better than Feynman
For eq. (9.13), if the velocity was constant, then
$$
x(t+\epsilon) = x(t) + \epsilon v_x
$$
would be exact, as the displacement in ##x## is time elapsed (##\epsilon##) multiplied by velocity (##v_x##). Since velocity depends on time, it is only an approximation; the smaller ##\epsilon##, the better.
Eq. (9.14), is the same, but for velocity in terms of acceleration. Eq. (9.15) follows from the fact that in this case acceleration is ##-x##, see eq. (9.12).