Документ взят из кэша поисковой машины. Адрес оригинального документа : http://www.mso.anu.edu.au/pfrancis/phys1101/Lectures/L03/projectile_motion_longer_derivation.pdf Дата изменения: Fri Feb 25 08:12:57 2011 Дата индексирования: Tue Oct 2 17:38:59 2012 Кодировка: Поисковые слова: с р р с с рер рес с с р р р р р рер р р рер

Full derivation of the projectile motion equations
Acceleration is defined as the rate of change of velocity. So, by definition...

dv a= dt
For the projectile motion case, acceleration is constant. So what is the velocity? We know dv/dt and we want to know v. This means undoing the differentiation. To undo differentiation, you need to integrate. So lets integrate both sides of the equation.

The right hand is just the integral of the differential of v - i.e. it takes you back to v. For the left hand side, to integrate a constant, multiply it by whatever variable you are integrating against (in this case, t) and add a constant. So we get



a dt =



dv dt dt

at + C = v
where C is a constant (whenever you do an indefinite integral, you end up with a constant). As usual with constants like this, you work out their value by setting t = 0. In this case, v=C. So C is just the velocity at time zero. Which can be written as v0. Now, lets try to work out the position s. By definition, velocity is the rate of change of position, i.e.

ds =v dt
Once again, we need to reverse the differentiation, so we need to integrate both sides of the equation.


The left hand term is just s (differentiating it then integrating it just takes you back to where you started). The right-most part is just substituting the above equation for velocity in. So we need to integrate at + v0 with respect to t. A is a constant. t is raised to the power 1, so its integral is t raised to the power 2, with a constant of 1/2 in front (usual law for integrating polynomials). v0 becomes v0t, and a new constant is needed (you always add a constant when you do an indefinite integral).



ds dt = dt



v dt =



(at + v0 )dt

12 s = at + v0 t + K 2
Ive written the constant as K rather than C, to make sure it doesnt get confused with the constant C above. As usual you work out the constant by setting the time to zero, in which case you find that s+K, i.e. K is the position at time zero, which I will write S0. So this becomes

12 s - s0 = at + v0 t 2
I hope this makes more sense!