- #1
erobz
Gold Member
- 3,818
- 1,642
Impossible?!?
By my estimation the equation that describes this motion is given by:
$$Pt = \frac{1}{2}m{ \dot x}^2$$
or
$$\dot{x} = \sqrt{\frac{2P}{m}} \sqrt{t}$$
but this implies:
$$\ddot{x} = \sqrt{\frac{2P}{m}} \frac{1}{\sqrt{t}}$$
So, no matter how small we make the power, we appear to get an infinite acceleration at ##t=0##( or in the case of ##P=0, t= 0##) an indeterminate form ##\frac{0}{\infty}##.
Conclusion: Acceleration from rest under constant power is non-physical. However, with any non-zero initial velocity things seem to be ok with the following:
$$\ddot{x} = \frac{P}{m} \frac{1}{\sqrt{ \frac{2P}{m}t + v_o^2}}$$
Anything interesting here, something I'm missing?
By my estimation the equation that describes this motion is given by:
$$Pt = \frac{1}{2}m{ \dot x}^2$$
or
$$\dot{x} = \sqrt{\frac{2P}{m}} \sqrt{t}$$
but this implies:
$$\ddot{x} = \sqrt{\frac{2P}{m}} \frac{1}{\sqrt{t}}$$
So, no matter how small we make the power, we appear to get an infinite acceleration at ##t=0##( or in the case of ##P=0, t= 0##) an indeterminate form ##\frac{0}{\infty}##.
Conclusion: Acceleration from rest under constant power is non-physical. However, with any non-zero initial velocity things seem to be ok with the following:
$$\ddot{x} = \frac{P}{m} \frac{1}{\sqrt{ \frac{2P}{m}t + v_o^2}}$$
Anything interesting here, something I'm missing?