Calculate the change in velocity over a given distance.

Question

I am trying to calculate the maximum change in speed over a given distance with a known constant acceleration, current speed and of course, distance. I do not know the time.

However, I run into the issue that the speed is undefined if the starting speed is zero.

Given a=∆v/t, ∆v=a*t, and ∆t=d/v, I came up with ∆v=a*(d/v). However if the current speed is zero, I get a undefined value for the delta speed.

I feel as if I am missing something obvious, or that the algorithms I am using are simplified versions of more complete ones.

Thanks in advance for any help.

You can't use $\Delta t=\frac dv$ when the velocity is not constant. — Ross Millikan, Dec 16 '19 at 16:52
What values are you allowed to choose? Do you choose the initial velocity and acceleration or just one of them? Please edit this and the fact that the acceleration is uniform into the question. — Ross Millikan, Dec 16 '19 at 16:56
If acceleration does not change , shouldn't the velocity be constant ? — The Demonix _ Hermit, Dec 16 '19 at 16:58
You can use calculus to model this problem. Consider the position as a function of time, so that "speed" (velocity, to be more precise) is the first derivative and acceleration is the second derivative. — hardmath, Dec 16 '19 at 17:08

Ross Millikan · Accepted Answer · 2019-12-16T17:34:21.850

1

If we measure from the starting location and the acceleration is constant, the position as a function of time is $$s(t)=v_0t+\frac 12at^2$$ where $v_0$ is the starting velocity and $a$ is the acceleration. We can find the time to get to position $d$ by solving the quadratic $$d=v_0t+\frac 12at^2\\ t=\frac 1{a}\left(-v_0+\sqrt{v_0^2+2ad}\right)$$ Then the change in speed is just $at$

edited Dec 16 '19 at 17:34

answered Dec 16 '19 at 17:04

Ross Millikan

374,822

The change in speed is at . – Jiri Kriz Dec 16 '19 at 17:25
I think I agree, the resulting $$\Delta v=\sqrt{2ad+v_0^2}-v_0$$ seems more accurate from plugging it into the simulation. – Chase Dec 16 '19 at 17:39
@Chase: the units do not match. The square root has units of time while the other two terms have units of velocity. – Ross Millikan Dec 16 '19 at 20:32

amd · Answer 2 · 2019-12-16T23:19:34.067

Use conservation of energy to eliminate those unknown times entirely. The change in kinetic energy is equal to $\frac12mv_{\text{final}}^2-\frac12mv_{\text{init}}^2$. The work done in accelerating the object is $mad$ (using the convention that this will be negative for deceleration). Equate these and solve for $v_{\text{final}}$: $$v_{\text{final}}=\sqrt{v_{\text{init}}^2+2ad}.$$ Note that if there’s a change in direction as a result of this acceleration, you’ll need to split the distance $d$ into two segments.

Calculate the change in velocity over a given distance.

2 Answers2