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How do you calculate a final velocity of an object when given its initial velocity and the object is accelerating between an initial and final acceleration over some given distance?

Qmechanic
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Tivity
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    Does this answer your question? What are the equations for motion with constant jerk? – Agnius Vasiliauskas Apr 01 '22 at 06:41
  • thanks, I'll take a closer look but at first glance, it's not very clear to me how to incorporate a jerk to calculate a final velocity. – Tivity Apr 01 '22 at 06:55
  • Well, that answer gives you all kinematic required equations when acceleration is function of time, just use it for your calculations. – Agnius Vasiliauskas Apr 01 '22 at 07:00
  • The answer depends on how the acceleration changes over that distance. Do you know what the acceleration is as a function of distance ? Or as a function of time ? – gandalf61 Apr 01 '22 at 07:59
  • starting from some initial velocity the change in acceleration occurs over some given distance. For example, starting a 1000 m/s then instantly accelerating from 2 m/s^2 to 9.8 m/s^2 over 20000 meters. What would the final velocity be? – Tivity Apr 01 '22 at 08:10
  • You need to know more than just the initial velocity and the initial and final values of acceleration. My example below shows that with only this information the problem is underdetermined. You need to know exactly how the acceleration varies during the motion, either as a function of time or of distance. Your comments below suggest that you do know how acceleration varies as a function of distance, but you did not give this information in your question. – gandalf61 Apr 01 '22 at 09:56

3 Answers3

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Your problem here is that your equation has the form:

$$ \frac{dv}{dt} = f(x) $$

i.e. on the left side you have a derivative wrt time, but on the right side you have a function of distance. Solving this requires one of the (many) tricks that physicists only learn from experience. You need to use the chain rule to rewrite:

$$ \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v $$

So now you can rewrite your equation as:

$$ v \frac{dv}{dx} = f(x) $$

and then integrate:

$$ \int v dv = \int f(x) dx $$

John Rennie
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  • thanks, but shouldn't it be $\frac{da}{dt} = f(x)$? As its a change is acceleration over distance? – Tivity Apr 01 '22 at 08:21
  • @Aneikei No, the acceleration is given as a function of distance i.e. a(x). Or at least that's what I understood your question to mean. – John Rennie Apr 01 '22 at 08:29
  • Thank you, I'm trying to understand. If it's not too much trouble would you provide an example For instance, starting a velocity of 1000 m/s then instantly accelerating from 2 m/s^2 to 9.8 m/s^2 over 20000 meters. What would the final velocity be? – Tivity Apr 01 '22 at 08:40
  • @Aneikei I don't understand what you mean by instantly accelerating over a distance. If the acceleration changes over a distance of 20000m doesn't that mean it changes smoothly over this distance? – John Rennie Apr 01 '22 at 08:43
  • correct, what I meant was that the object has an initial velocity of 1000 m/s and then accelerates to 2 m/s^2 and over a distance of 20000 meters has a final acceleration of 9.8 m/s^2. What would the final velocity be at the end of that 20000 meters? – Tivity Apr 01 '22 at 09:02
  • @Aneikei OK, now we know the initial and final accelerations, but we don't know how the acceleration changes between those two points. It could change linearly with time $a(t) = kt$ for some constant $k$, or linearly with distance $a(x) = kx$, or indeed it could be a more complicated function of time or distance. Unless the question specifies how the acceleration changes we cannot make any progress. – John Rennie Apr 01 '22 at 09:06
  • I should have specified the acceleration changes as a function of the gravitational acceleration which itself is a function of distance or radius from a gravitating mass which is $a=GM/r^2$. So at r1 the acceleration is 2 m/s^2 and at r2 the acceleration is 9.8 m/s^2 and the distance between the two distances is 20000 meters for example. Does that help? – Tivity Apr 01 '22 at 09:19
  • Then my answer applies because $a = f(r)$ where $f(r) = GM/r^2$. So you end up with $\int v dv = \int GM/r^2~dr$. – John Rennie Apr 01 '22 at 09:22
  • Thanks, but how do you solve it using the example I gave? Sorry to be a pest but math is not my forte. – Tivity Apr 01 '22 at 09:25
  • @Aneikei I'd be happy to continue this in the Physics chat room. If you want to discuss it there this is the link for the chat room. – John Rennie Apr 01 '22 at 09:27
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Here is an example to show that the problem as stated in the original question is underdetermined.

Suppose the distance $s$ travelled by the object at time $t$ is given by

$s(t) = 91t^3 -49t^4 +21t^5$

Then the object's velocity and acceleration are

$v(t) = 273t^2 - 196t^3 + 105t^4 \\ a(t) = 546t - 588t^2 + 420t^3$

So we have $s(0)=v(0)=a(0)=0$ and $s(1)=63$, $v(1)=182$, $a(1)=378$

But now suppose the object's distance, velocity and acceleration are:

$s(t) = 51t^3 +21t^4 -9t^5 \\ v(t)=153^2+84t^3-45t^4 \\ a(t)=306t+252t^2-180t^3$

So now we have $s(0)=v(0)=a(0)=0$ and $s(1)=63$, $v(1)=144$, $a(1)=378$

So we have two different scenarios where the object travels a distance of $63$, its initial velocity and initial acceleration are both $0$, its final acceleration is $378$, but its final velocity is $182$ in one case and $144$ in the other.

gandalf61
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  • In the detail section of the question I state - "How do you calculate a final velocity of an object when given its initial velocity and the object is accelerating between an initial and final acceleration over some given distance?" – Tivity Apr 01 '22 at 11:36
  • @Aneikei Yes, I read that. But just knowing the distance, the initial velocity and the initial and final accelerations is not enough to determine the final velocity. My examples shows two scenarios where the distance, the initial velocity and the initial and final accelerations are all the same but the final velocity is different. – gandalf61 Apr 01 '22 at 13:07
  • that's not true. With the help of (john rennie) the solution was discovered - vf = √(vi² - 2GM/ri + 2GM/rf) where vf is the final velocity and ri is the initial position and rf is the final position. – Tivity Apr 02 '22 at 17:22
  • @Aneikei John Rennie's solution is for the specific case $a(r) = \frac{GM}{r^2}$. You did not give this information in your original question. If you don't specify how acceleration depends on distance or on time then the problem is underdetermined, as my example shows. – gandalf61 Apr 02 '22 at 17:34
  • sorry about that. I'll try to be more thorough in the future. – Tivity Apr 02 '22 at 18:28
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As others have said, if you have an acceleration function given explicitly as a function of distance you can use mathematical tricks to do it, or use the relation between work and kinetic energy

$1/2mv^2 = \int F(x) dx$

$1/2v^2 = \int a(x) dx$

However given that you actually have acceleration as a function of time, we can do this to solve:

$V=\int a(t) dt$

Plug in $(t=0, v=v_{0})$

Then use this expression for velocity as a function of time to calculate

S(t)= $\int v dt$

Because we want to find out the velocity over a specific distance, we can use this equation to find the time at which the particle reaches a certain distance by setting s(t) to be a certain value, and solving the equation for t

Defining the inverse of s(t) that returns a time for a given distance $S^{-1}(s_{0})$

We can then plug this value for time back into our original equation for velocity

jensen paull
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