physics – How laborious to brake to completely easily attain the vacation spot?

physics – How laborious to brake to completely easily attain the vacation spot?

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When an object brakes with a relentless deceleration, then its velocity over time seems to be like this:

Velocity over time

Distance traveled is velocity multiplied with time. So the orange space within the graph above truly represents the space traveled. And the world of a proper triangle is half that of a sq. with the identical width and peak. So:

$$ distance = (velocity * time) / 2 $$

And the time it takes to get from the present velocity to 0 is the rate divided by the deceleration, so:

$$ distance = (velocity * velocity / deceleration) / 2 $$

However we all know the space and want the deceleration. So we have to rearrange the system:

$$ distance * deceleration = (velocity * velocity ) / 2 $$
$$ deceleration = (velocity * velocity ) / ( 2 * distance ) $$

There you’ve gotten it.🡅🡅🡅

Nevertheless, this system assumes that you’re utilizing a physics engine that is ready to mannequin acceleration by integration. Some physics engines sadly simply assume that velocities are fixed throughout physics ticks after which adjustments the rate between ticks. So in actuality the deceleration curve in your engine may truly seem like this:

Velocity by time in a tick-based game engine

The mathematics nonetheless principally works out over the long run, however can result in inaccurate ends in the short-term. Particularly when the engine operates on a variable tick-rate. So in case you are utilizing a physics engine that works like that, you may obtain outcomes which can be “adequate” however not completely correct.

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