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Can someone correct my understanding on this? So, if an object has constant acceleration, and travels a distance $x$, it means its momentum will change by $mv_2 - mv_1$.

Now, if the same object travels a distance $4x$, and undergoes the same constant acceleration does it also mean that the momentum will also change by $4mv$? Since, it is now traveling at 4 times the previous time, therefore it must also gain a velocity of 4 times the amount?

Since momentum does not have a distance dependency, yet velocity and time does, this is just something I wanted to double check.

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  • Action of force over distance relates to work/energy: $$dE=\mathbf{F} \cdot d\mathbf{r}$$

  • Action of force over a period of time relates to momentum change: $$d\mathbf{p}=\mathbf{F} \, dt$$

  • Kinetic energy and momentum can be related by:

$$E=\frac{p^2}{2m}$$

$$\mathbf{v}=\frac{\partial E}{\partial \mathbf{p}}$$

In your case, the energy is four times of the original. If it started from rest, then the velocity or momentum is doubled.

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It is usually easier to consider all quantities (position, velocity, acceleration) as functions of time, not of distance traveled. This often makes equations simpler and also is easier for understanding.

Position, velocity and acceleration are related by time derivatives: $v=ds/dt$, $a=dv/dt$. For constant acceleration, $a=a_0$ you can integrate these equations to get: $$v=a_0 t + v_0$$ $$s=\frac{a_0}{2}t^2+v_0t +x_0$$

As you can see the relationship is not the simple linear relationship you suggest in your question.

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Actually no, momentum won't change 4 times. You are right at that point the time of motion will be 4 times longer, because $\Delta V = a\cdot \Delta t$. But there is no such linear dependance between the distance and the momentum. It's like: $\Delta x \propto a\cdot \frac{\Delta t^2}{2}$, which can be written as $\Delta x \propto \frac{(a\Delta t) \cdot \Delta t}{2}=\frac{\Delta V \cdot \Delta t}{2}$. So if you want to estimate change of momentum within x, you should count not only change of path lenght, but also time what it takes.

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Consider how distance changes with time in a constant acceleration situation: $$\Delta x = v_0t+\frac{1}{2} a t^2.$$ From this we immediately see that to increase from $\Delta x$ to $4\Delta x$ with the same $a$ depends on what the initial velocity, $v_0$, is (and you didn't specify it in the original question), and on the time. If we conveniently assume that the initial velocity is zero, then the time required only doubles to get a distance of $4\Delta x$.

Now momentum change in a constant acceleration situation is easily calculated by $\Delta p = Ft$. This quickly shows us that the momentum change in travelling $4\Delta x$, starting a rest, is simply $2\times$ the momentum change in travelling $\Delta x$.

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protected by Qmechanic Feb 2 '17 at 22:07

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