2007-09-12 - Test review
Today's nonsense
- Spiller will not keep a pig pet
- Goats and pigs
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#A12 - An object that is dropped at the same velocity as it would be thrown horizontally, would hit the ground at the same time, if dropped or thrown from the same height.
#A16 - For airbag 2, approximately how much distance does the dummy cover between the moment the car crashes and the moment the dummy first makes contact with the airbag? Remember with the airbag graphs, this is a graph of the velocity of the person, not of the cars or anything like that. You're looking at the area under the graph from the first part until it starts to change (which is sort of like at 0.01). This was an interesting question, but you will not have this question on the test.
#B1CD. A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s^2. At 10 sec after launch, the first stage of the rocket (now without fuel) is released. The second stage now has an acceleration of 6 m/s^2. What will be the maximum height attained by the first stage after separation? What will be the distance between the first and second stages 2 sec after separation? You find the height of the first stage (in this case, 200 m) because the average velocity is 20 m/s, and acceleration was 4 m/s for 10 seconds, so the average (0+40 / 2) = 20 m/s for the average velocity, and then you multiply that by the 10 seconds in order to get how far it traveled in the first stage. When you release the first stage, the rocket does not just fall down, there's still velocity, so it's going to go up and it no longer has acceleration, so the first stage will fall while the second stage keeps going up, and the second stage now has an acceleration of 6 m/s^2 according to the setup of the problem. What will be the maximum height attained by the first stage after separation -> you can start at the 200 m point. The first stage (after that 10 seconds), you have T=0 at that point and so s=-4.905T^2 + 40T + 200----- and the 40 is because the velocity is 40 m/s. So how will you find maximum height? Take the derivative, to get velocity, find where velocity equals zero, and then find time, and plug that time back into the position function, and that will tell you the maximum height. You could use the quadratic equation, and then take the average of the two zeroes in order to get the maximum height, or you could take the derivative (as previously explained). So the highest position is something like 281.549 meters. So what will be the distance between the first and second stages two seconds after separation? Stage2 at this moment has an s=(1/2)(6 m/s^2)T^2 + 40T + 200. If they tell you the acceleration, then g is already out of there.. So now you find the distance between T=2 for both of those 's' or position functions.
#B7. Initial velocity is 20 m/s upwards. Be careful, if they say that an object is thrown downward with a speed of 20, the velocity would be -20, so be careful. But this one is indeed +20. The maximum height? Well, s=(1/2)gT^2 + 20T + 0. And so: V=(-9.81)T + 20. Maximum height occurs when V=0 so you solve for T and then plug it into the position function in order to determine the height at that time. We're talking about free fall and neglecting air resistance.
#B6. A car starts from rest heading due east. It first accelerates at 3.0 m/s^2 for 5 sec and then continues without further acceleration for 20.0 sec. It then brakes for 8.0 sec in coming to rest. What is the car's velocity after the first 5.0 seconds? What is the car's acceleration over the last 8.0 sec interval? What is the total displacement? How fast is it going after 5 seconds? It's going 15 m, because 3*5 = 15. To get how far it travels? And then continues without acceleration for 20 seconds. So here, it's velocity is going to be 15 m/s but it is without acceleration for the average velocity is the velocity because the velocity stays the same the whole time. In the first part, the average velocity was something like 7.5 m/s. The third part of this: it brakes for three seconds and comes to a rest, for 8 seconds, and so we don't know the acceleration. So the question. Part A = 15 m/s. What's the car's acceleration over the last 8 seconds? Alright, how fast was it going? It was going 15 and it ended at 0, so acceleration = change in velocity all over change in time, so we're looking at -15/8, and then of course we have our units: m/s^2. Because, and things that you can do to remember, you know that acceleration is the derivative of velocity, well, we hope you know, and v' has to be dv/dt, which is your change in velocity over your change in time, so all of those concepts help each other, and then part C, what's the total displacement. Now, all of these are moving forward, so how far does it go during this first little part? So the total displacement, this first little section had 5 * 7.5 m/s, and the second part was a constant 15 for 20 seconds so that's 300 m, and then the third part was from 15 to 0 so its average velocity was 7.5. So get the total displacement. The average, now we back up, it's going different velocity, so displacement will be velocity * time, so if you're going 15 m/s for 20 sec, you multiply, and these are different velocities, so you find the average velocity over that time period, and take the average velocity and multiply by that last 8 seconds, and when the time is changing at a constant rate and you want to know how far it traveled you can take the average by adding them together and dividing by two.
Problems Spiller "likes":
Traveling so much in the last second. If you're going on a trip, and the first half of your trip you travel at this speed, what would you have to travel at the second part to get a certain average. You drop something, you throw something, they land at the same time, how tall was the ...