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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 arctan2(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.

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.

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 acc. has taken place.

a=lateral acceleration (ft/s^2)

v=velocity (ft/s)

dt=timestep duration

My first thought was to use arctan2(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.

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 arctan2(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.