I was working through a physics problem today and ended up having to look up the answer because I got stuck. I noticed that the way I worked it was a bit different from the way the textbook worked it, and am curious to know if the method I used is mathematically valid. (Not sure if this belongs in the physics or math section, so my apologies if it is in the wrong one.)
For some context, the problem statement provided numerical values for $P_{av}$ (time-averaged power), $m$ (mass) and $d$ (distance).
Where the textbook and I agree:$$P_{rad} = \frac{I}{c}$$$$F = P_{rad}A = \frac{IA}{c} = \frac{P_{av}}{c} = ma_x$$$$\frac{P_{av}}{mc} = a_x$$
Where the textbook and I diverge:My Solution:
$$a_x = \frac{dv}{dt}$$$$v = \frac{dx}{dt}$$$$a_x = \frac{d\frac{dx}{dt}}{dt} = \frac{d^2x}{dt^2} = \frac{P_{av}}{mc}$$$$dt^2 =d^2x\frac{mc}{P_{av}}$$$$t = \sqrt{d^2x\frac{mc}{P_{av}}}$$
Textbook Solution: (assuming acceleration is constant)
$$x-x_0 = v_{0x}t + \frac{1}{2}a_xt^2$$$$t = \sqrt{\frac{2(x-x_0)}{a_x}}$$$$t = \sqrt{\frac{2(x-x_0)}{\frac{P_{av}}{mc}}}$$$$t = \sqrt{\frac{2(x-x_0)mc}{P_{av}}}$$
Does my answer assume acceleration is constant?
Is $d^2$ actually equal to $2(x-x_0)$?
