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I am doing a 2D simulation in discrete steps of a vehicle.

The vehicle will desire a certain lateral acc. at any given timestep.

So I would like to know the delta angle of the velocity vector after the lateral acceleration has taken place.

a=lateral acceleration (ft/s^2)

v=velocity (ft/s)

dt=timestep duration

My first thought was to use $\arctan 2(a * dt,v)$, but then I thought it might be rather a isosceles triangle problem, so I thought $ \arcsin (a * dt * 0.5/v)$.

I am unsure what is correct, if any of them.

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  • $\begingroup$ Why is this marked homework and voted down. I even mentioned its for my simulation, I don't understand. Not a nice way to greet me to physics page. I am not a scientist but an engineer, but my physics get rusty sometimes. $\endgroup$
    – Invariant
    Commented Oct 31, 2018 at 21:50

2 Answers 2

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It's easier to treat velocity as a vector of x velocity and y velocity. Then you can just add your acceleration to the x velocity: $v_{x}$ += $a*dt$.

The absolute velocity is the square root of the dot product of the velocity vector.

Then if you still need the angle, normalize the velocity vector (divide by the absolute velocity) and the angle is the arccosine of the normalized x velocity, and the arcsine of the normalized y velocity.

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  • $\begingroup$ Although I am going to use the answer below I am gonna mark this as answer, since it made me understand what lateral acc. really is. I thought it was something more complicated. $\endgroup$
    – Invariant
    Commented Nov 2, 2018 at 14:41
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I found the answer by looking more careful at some documents.

angle_radians = a * dt/v. For small dt.

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