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-   -   Stopping distance (http://www.physicsforums.com/showthread.php?t=48874)

Lannie Oct20-04 10:05 PM

stopping distance
 
I don't actually have a specific homework question, but I'm wondering if anyone could explain to me how you solve a stopping distance problem. I've encountered these questions in very different ways, once with very little information given, and one with a considerable amount of calculating to do. I'm still not understanding how to calculate stopping distance on level ground (ie, where the angle is zero). If anyone could give me some general information, I'd appreciate it.

Lalo1985 Oct20-04 11:12 PM

For a body in movement to stop means there has been a change in its Kinetic Energy. Kinetic enery is given by 1/2mv(f)^2-1/2mb(i)^2, where m = mass of object, (f) is final speed, which is zero, and (i) is the initial speed at which the object is traveling. Therefore, the change in Kinetic energy = -1/2mv(i)^2. However, Newton's 1st Law states that a body in movement tends to stay in motion, and his second law, F = ma states that if all the forces (which are vectors) acting on the object cancel out, then the body does not accelerate and therefore stays in constant motion. So, if an object comes to a stop, that means there is a force causing it to stop. In other words, there is a force acting on the object.

Now, the Work-Enery theorem states that the total work on an object equals its change in kinetic enery (dW = dK). And, remember that work is given by the force times the displacement in the direction of the force. Then, we can set the formula:

F•delta(x) = -1/2m*v^2.

So, to find the stopping distance, simply divide both sides by the force magnitude (F).

delta(x) = (-1/2m*v^2)/F

As an example, pretend there is a toy car of mass 10kg traveling at 5 m/s, and that the wind exerts a force of 3N on the car. Since the force of wind is going against the displacement, its sign is negative.

so, we have

delta(x) = (-1/2*10kg*5m/s^2)/-3N.

You can check to see if this makes sense by cancelling out your units. Notice you get meters, which is what we're looking for.

By solving for x, we get that the stopping distance = 41.67m.

Hope this helps.


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