3 ms·
Indeed, the optimal trajectory is to accelerate as hard as possible then decelerate as hard as possible so you land at 0 with dx=0. It's more exciting to calcu
by nullc 3y ago
Indeed, the optimal trajectory is to accelerate as hard as possible then decelerate as hard as possible so you land at 0 with dx=0.
It's more exciting to calculate the dynamics if you need to be jerk limited.