note sure if I'm missing something obvious, but i cant seem to find a good way to tackle this question.

A space shuttel of mass 400kg is moving in a straight line where there is no resistance. the engine is working with a constant power of 650kW, the shuttles speed increases from 120 m/s to 160m/s in a time t seconds. Find the value of t.

any help would be grateful


closed as off-topic by ZeroTheHero, John Duffield, David Hammen, AccidentalFourierTransform, Jon Custer Apr 8 '18 at 17:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ZeroTheHero, John Duffield, David Hammen, AccidentalFourierTransform, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ There's not enough information to solve this problem unless a simplifying assumption is made. Can I ask what level of course is asking the question? A complete, correct solution needs more information, and even then the solution is not simple. $\endgroup$ – garyp Apr 8 '18 at 14:05
  • $\begingroup$ I'm doing AS Further maths mechanics $\endgroup$ – H.Linkhorn Apr 8 '18 at 15:03

Take the equation: Power = force times velocity, you can show this is true because the S^-1 on both sides cancel, leaving joules = force times distance which is work done, which definitely is energy.

From there, you can convert that power to a force. Yes this force will change with time because it is velocity dependant, but its a step in the right direction.

Simpler method that I didn't think about earlier: calculate it straight from kinetic energy: Ke = 1/2 times mass times velocity squared, find the energy for 120 and 160m/s, then divide the difference by the power of the energy to get your answer

  • $\begingroup$ I was able to get to this point but it was from here I wasn't sure where to continue from $\endgroup$ – H.Linkhorn Apr 8 '18 at 15:08
  • $\begingroup$ @H.Linkhorn ok i'll try to give some further steps. But something worth noting for the future: its always a good idea to state (and if you can, show) how far you've got with a problem $\endgroup$ – Alex Robinson Apr 8 '18 at 15:11
  • $\begingroup$ Concerning your edit: you have made the simplifying assumption that I mentioned in my comment to the OP. With the information given, this might be the intended course of action, but the question should be worded more carefully. $\endgroup$ – garyp Apr 8 '18 at 15:21
  • $\begingroup$ Cursed, from you edited version of the answer i have been able to reach the answer. thanks $\endgroup$ – H.Linkhorn Apr 8 '18 at 15:26
  • $\begingroup$ could downvoters please explain $\endgroup$ – Alex Robinson Apr 9 '18 at 11:27

First find the difference in kinetic energy (Joules) between the shuttle moving at 120m/s and moving at 160m/s. Then divide that by the amount of energy (Joules) the engine can deliver per second.

  • $\begingroup$ It is contrary to the spirit and rule of this forum to post complete solutions. Please delete it or modify it. Also, this answer makes an assumption not mentioned in the original question, so it might not be the intended solution. $\endgroup$ – garyp Apr 8 '18 at 15:18
  • $\begingroup$ @garyp I can't think of the simplifying assumption you're thinking of unless it's for applying Relativity Theory. Was that the assumption? No application of Relativity Theory? $\endgroup$ – DG123 Apr 8 '18 at 16:03
  • $\begingroup$ The assumption is the mass of the rocket is constant. That it uses no fuel to accelerate. We can assume that the fuel consumption is negligible. There really is no alternative; in order to make progress otherwise we would need to know the efficiency of the engine. But I do not understand the down vote you received. Your approach is probably the one intended. $\endgroup$ – garyp Apr 8 '18 at 17:02
  • $\begingroup$ @garyp Didn't think of that but you're right, and also probably right about the fact the teacher probably didn't mean for this problem to be any more complicated because he didn't give info for fuel usage rate, etc. $\endgroup$ – DG123 Apr 8 '18 at 18:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.