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Force vs. time graphs

David explains how to use a force vs. time graph to find the change in momentum and solves an example problem to find the final velocity of a spaceship. Created by David SantoPietro.

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• If the net impulse becomes negative what will happen to the rocket? Will it fall or just do something else?
• Yes! If the net impulse becomes negative, it means the change in momentum is happening in the apposite direction of the initial momentum. Here the initial momentum was away from the planet, hence the negative impulse will change the momentum of the alien rocket towards the planet. In simple terms, KABOOM!
• Would it be possible to calculate the distance the rocket travelled using this graph? If so, how?
• I think so.
In the first 4 seconds, the acceleration is constant(the force is constant) and can be found by using F=m*a which in this case is 3=2.9*a so a = 1 m/s^2

For seconds 3 to 7, we can find the acceleration by finding the mean force, which is 3/2= 1.5 N and then, 1.5=2.9*a so a=0.5 m/s^2

For the last part, the mean force is -2/2 = -1 N and so, -1=2.9*a a=-0.34 m/s^2

Now all we have to do is calculate the distance travelled in each part using the kinematic equations, (especially S = So Vo*t+a*t^2/2) and add them.
• i would want to know how do the varying force-time graph when we try to estimate the aveage impulsive force is somehow similar to mean's value theorem?
(1 vote)
• Yes you would take the mean value of the force if you happen so come across such a uniform slope. You can view it similar to speed and time graph where the area is the distance travelled.
So if there is a slope in such a graph then we know that the average speed is total distance travelled / total time taken. That is total area under the graph(area of traingle) / total time take(length of the base)= mean(average) value of the speed = half the value of height of the triangle.
Therefore the mean value of such a slope is the mid point of the slope or half the value of the height.
So I think
Impluse = average force * time
if force increase or decrease at constant rate then
impulse = 1/2 total change in the magnitude of force * time take.
• Can anyone tell me how to approach a similar problem when mass is not constant? I have a basic knowledge of integration an differentiation.
• At the end of the video, David divides by 2.9 kg, and the left side of the equation goes from Ns to m/s. Can anyone explain why?
• Force is equal to rate of change of momentum.
F = Δp/Δt
or Δp = F*Δt (units of N*s)
since Δp = mΔv (units of kg*m/s)
F*Δt = mΔv
Δv = F*Δt/m or units of N*s/kg is same as m/s
• If I have a F_g (N) vs Time (s) graph, where F_g equals the gravitational force on the person by the Earth(weight) and t = time, what would the slope represent?
Would it represent Net Force?
• Um… not really sure whether this is a valid question. This is because F_g is a constant, that is F_g = 9.8 m/s^2 anywhere on Earth. So the slope of this graph is 0. I don’t know of any quantity that is equal to Force/time.
• What is the slope of at graph?
• When you say "at graph" are you asking about an acceleration vs time graph? The slope of an acceleration vs time graph is the rate of change of acceleration. The automatic filter for insults will not allow me to put the name of rate of change of acceleration which is "j e r k"
• Is there a standard name for units of impulse other than Newton-Seconds? (I might suggest 1 Ns = 1 Kirk or 1 Sulu).