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(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?
(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
(I) What force is needed to accelerate a sled (mass = 55 kg) at 1.4 m/s$^2$ on horizontal frictionless ice?
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welcome to our last example. Video on two dimensional motion in this video will be looking at how toe handle a vector that is written with an independent variable and we'll do a little bit of calculus with it. So say we have a vector defined by negative 40 In the I had direction. We almost forgot our units here. So let's make sure we put in our units. If we're gonna multiply it by t, we want velocity out of, then we're gonna need negative 4 m per second squared times. T the I had direction plus two meters per second cubed times t squared in the J hat direction. Okay, so we've got our units, right, because we're multiplying by time and time squared. We've got I hat J hat. So here's our vector V. And then the question is, if we are given a situation where we have, um, initial are position that says we're at to I have plus three j hat at the time T equals three seconds. Where will we be and what will there are? Acceleration. Be clearly. We could quickly find the velocity for t equals three seconds by simply plugging it in here. So this is a little bit more complicated. Um, let's go ahead and solve for the acceleration first cause we know all we have to do There is a derivative. So acceleration is equal to the time derivative of velocity, which means we'll end up with an answer that looks something like negative four meters per second squared Tried to put the units underneath the number there. That's not quite right. So we have negative 4 m per second squared in the eye hat, plus 4 m per second cube times T in the J hat. So there's our acceleration. We can plug in t equals three seconds. This will be a constant. This one will be plus 12 m per second squared in the J had direction. Now eso we've solved a set. Now we want to solve for where we will be in the are now ours a little more difficult because we have this are not and we're gonna have to take an integral of V f t in order to find are so when we take our integral A V f t, What we're going to find is we have the integral of this term, which we're going to end up with a T squared. We need the one half in front to make sure we get our units. Right. So we're gonna have 4 m per second squared times t squared, plus and this when we already have a t squared. So it'll go to T Cube. Well, 2 m per second cube times, T cube, And then it's gonna be plus are are not vector okay, which we've been given already. Okay, because we always know that the constant that's left over after taking a indefinite integral is going to be wherever we were when time equals zero. Because we can see that T is in each of these terms. Which means that AT T equals zero. These terms are both zero. So all that would be left is wherever we were at the beginning. That means that we're going to have are is a function of time which will look like this. We'll say that we start out at two minus for over two meters per second squared, T squared, all in the I had direction. Rope. I dropped my hat and J hat previously, and then we'll have plus three Waas two thirds meters per second cube times t cute in the J hat direction. So here is our position vector. We can go ahead and plug anti equals three and solve for where we will be At T equals three seconds. So the trickiest part, as I said about this question, is remembering to use are not appropriately in this particular case because we had a time variable in each of our terms, we knew that AT T equals zero. They were zero, therefore, are not is all that would be leftover. Sometimes it could be a little trickier than that, but usually it comes out something like this. Try plugging in. T not see what's left over.
Newton's Laws of Motion
Applying Newton's Laws