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#1
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| I have trouble understanding the answer of the following question: A toy rocket of mass 0.2kg is projected vertically upwards from rest by means of a force which decreases uniformly in 2 seconds from 2kg wt to zero and thereafter ceases. Write equations of motion for the rocket for the case t < 2. Answers: dV/dt = (9 - 5t)g. Why is that so? Thanks! |
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#2
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| F = ma = m dv/dt You are told the propelling force decreases linearly in time, so it is of the form F = a + bt, and b is negative because it is decreasing. At t = 0, the force is "2 kg wt". I assume that means 2 kg * g. Plug in t = 0 and set it equal to 2g: F(0) = a + b*0 = a = 2g. So now you know the constant a. You are told that the force is 0 at t = 2. F(2) = a + 2b = 0. Solving, b = -a/2 = -2g/2 = -g Now you know b, so F = a + bt = 2g - gt One more piece: Don't forget gravity, which acts downward and has magnitude mg. So the total force on the rocket is 2g - gt - mg. This equal to mass times acceleration and that gives you your equation of motion: 2g - gt - mg = m dv/dt Since you know m = 0.2, you can substitute that to get 2g - gt - 0.2 g = 0.2 dv/dt 1.8g - gt = 0.2 dv/dt And that is your equation of motion. If you multiply both sides by 5 to just get dv/dt on the right, you'll have the equation you were told is the answer. |
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#3
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| Are you making an equation for: Motion, Acceleration or Instantaneous Velocity? |
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