Potentially painful

----------------- Hold on here: Acceleration constant- force constant- rate of consumption of fuel is constant. However acceleration is *rate of change of velocity* so with constant acceleration the velocity is increasing. v=vo +at increasing linearly. Lets assume initial velocity vo is 0 going for a given time T the velocity is aT and the KE is (0.5)*M*(aT)^2 Energy input over that length of time =integral of force times distance. The force is Ma (constant) and the distance is the integral of velocity over the time =0.5*a*(T^2) so energy input is 0.5*M*a*(a)*(T^2) =(0.5)M (aT)^2

Newton lives!

The question is, why is the rate of change of kinetic energy of the rocket linear in time, whereas the energy supply is only burning chemical fuel at a constant rate? On a moment to moment basis, how do you reconcile this? I think we've decided that the chemical energy is changed into kinetic energy of the rocket *and* the fuel. The rate of change of kinetic energy of rocket is linear with time, but the rate of change of kinetic energy of the fuel is decreasing with time, owing to the fact that the rocket is speeding up, so the difference in rocket speed vs thrust speed is decreasing with time. The sum of the two equals a constant.

Learning from other people's mistakes is nearly as effective as learning from one's own.

Your teacher had the opportunity to come back with "Oh no! I don't get it, Help me out here!" but I suspect that teaching 3rd formers turned him into a facist.

If kids hate school then let them leave at fourteen and be brickies, plasterers or plumbers because:

And then the teacher can stay liberal and the geeks get better service. Everyone's a winner.

Cheers Robin

In this case the interaction is between the lorry and the cyclist. The change in KE is related to the change of the relative velocity. The KE as measured with respect to the ground or to the galactic center is not germane to the problem. If it were, then walking into a wall would have deadly consequences. You have to look at the system that is involved. Ground is not involved in this particular collision. Part of the problem is that the KE expression that is most commonly used deals with a single moving object with respect to a stationary reference. This KE expression was originally developed for objects and is an integration of force*distance.

If you grab the lorry not a problem, if you miss *and* fall, the next thing you hit is the stationary ground at a velocity between 100 and 101 mph - that is a different and far more painful thing. This is where you expend nearly all of of the energy that you put in in getting up to 101 mph.

I completely agree with this. At my local HS, the focus is on science and math. Everybody has to pass algebra and science. Thus, you end up with algebra and science classes full of bored senior stoners who could care less, along with freshmen and sophmores who want to learn. You also end up with people who can't pass the exit exams, and so can't get a diploma even if they would be good at non-technical things like plumbing or plastering.

Trade schools are a great solution. Do the French still do this? Unfortunately, at least here in California, the 'trade schools' are really set up for the kids who are troublemakers, and have been tossed out of the normal high schools. Thus, even if they excel, they feel like losers, because they couldn't do algebra.

I gave the solution for constant acceleration and constant mass. It is messier for a rocket.

Initial mass =Mo=Mr +Km where Mr =mass of empty rocket and Km is the initial fuel mass mass of fuel at time t is m(K-t) where m is the mass per second of the exhaust. Assuming the exhaust is at constant speed c with respect to the rocket, then the force is cm ={Mo-mt)dv/dt solving gives velocity =c*ln{Mo/(Mo-mt)} {See Artley -Fields and configurations p262} which is not varying linearly with time so the KE is not changing quadratically with time. Integrate this to find the distance travelled in time t and multiply by the force cm to get the total KE at time t

After t=K/m the fuel is exhausted and velocity is constant at 2c*ln(Mo/Mr) and the mass is Mr. so final KE appears to be 2*(c^2)*Mr*(ln(Mo/Mr))^2

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

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Actually, v=at only applies if F = mass * acceleration, and both mass and acceleration are constant. So relativity is out, and so are rockets (mass changes, as well as the fuel loss eventually means the rocket thrust stops). The atmosphere angle makes the problem harder accept again, the force is velocity dependent, so acceleration is not constant unless the rocket adjusts. So assume a simpler problem: A hammer is dropped on the moon. After t seconds it hits the ground with velocity v, and KE=1/2 mv^2. Why does the velocity increase linearly with time, while the KE goes as t^2?

Answer: It doesn't go as t^2. KE is a function of v and not t, so you have to use the chain rule to find dK/dt. So,

dK/dt = dK/dv dv/dt = 1/2m (2v) dv/dt = m (at) a = ma^2t

Or, more fundamentally, the kinetic energy is increasing because work is being done a the hammer (work-energy theorem). Work is defined as a force exerted over a distance, so

dK/dt = dW/dt = d(Fx)/dt = F (dx/dt) = mav = ma(at) = ma^2t

Craig