 Well, let's continue to track the motion of our object. Well, suppose that after it leaves the tube, it goes into another tube, 10 meters in length, and while there, its velocity decreases by kt meters per second squared, where t is the number of seconds after it enters the second tube. Here we want to find k so the object stops before it splats into the end of the tube. Well, finding when the object stops, that says something about its velocity, and so our velocity is going to be the anti-derivative of this acceleration, and we do know how rapidly the object was moving at the start, so c is going to be 13.4. So the object stops when its velocity is zero. So we'll set up our equation and solve for t. And since t has to be positive, we only need the positive square root. And so depending on the value of k, the object will stop sooner or later, and the important question is, will the object stop in time? What will determine whether the object stops in time is whether or not it's traveled more or less than 10 meters by the time it stops. So we need to know the distance the object travels. Well, distance is measured in meters. We have a velocity in meters per second, so distance will be the anti-derivative of velocity. And again, we can calibrate and find this constant since t is the amount of time since the object entered the tube. We know that s of zero is zero, and so our constant is also zero. So remember that we found that the object stopped moving when t is equal to the square root of 26.8 over k. So let's figure out where the object is at the time it stops. So if we want the object to stop before it splats against the far end of the tube, then we might reason as follows. If the object doesn't stop until t equals square root of 26.8 over k, and it only has 10 meters in which to stop, we need to find a value of k so that the distance of the object at this time is less than 10. So we'll drop this value into our formula for the distance, and we'll do a little bit of algebra and arithmetic to solve for k. And after all the dust settles, we find that square root of k has to be larger than 4.6247, and so k has to be greater than 21.3876. And if k is greater than this amount, then where the object stops is short of the 10 meter distance, and so the object will stop in time.