2007-09-05
Physics homework due tomorrow.
Still working with the Sally and Joe problem. Sally is walking and it takes it 60 seconds to go distance d. Joe takes 120 seconds to go distance d. How long for Sally if she walked twice as fast as she walked in the first place? What if Sally doubles her speed? The distance is two times the speed of the belt. The distance is also 2*V_{sally}.
We are going to be doing some other problems out of the book. #31. Walk in the class room with a negative velocity and positive acceleration (increasing velocity). Deceleration is not negative acceleration.. Any time it becomes less negative, that means it is increasing. Think of an example where you have negative acceleration and negative velocity- like when you drop something off a cliff where g at all times is -32.2 ft/sec^{2}. Say you throw it up at 96.6 ft/sec. After one second it's going only 64.4 ft/sec upwards. And after two seconds it's going 32.2 ft/sec. And after three seconds it's going 0 and it's reached the top and then in the fourth second it will go downwards 32.2 and then in another second it will be going down at 64.4 ft/sec and another second it will be going down at 96.6 ft/sec.
You throw a ball in the air and it's in the air for ... Dakota throws a ball in the air and it lands 8 seconds later. Bryan didn't understand that he was supposed to catch it, so he throws it and 8 seconds later and it lands a ways away. Which ball went higher? They land at the same time, both in the air for eight seconds. Does larger initial velocity indicate that the ball will stay in the air longer? Yes, because it will take longer to decelerate. If it takes the same time, then they went the same height. Dakota threw the ball straight up, but Bryan threw it at an angle because it's going off to the side. The y-components are the same. If the time takes the same, then it takes the 4 seconds to go up and another 4 seconds to go down. So they went the same height, since the time is the same. The distance that the ball lands away is due to the horizontal vector, which Dakota had of value zero.
At the top of the ball being tossed up, the velocity is zero, but the acceleration is still -32.2ft/sec.
Is it possible for a body to have zero velocity and nonzero acceleration? Yeah, just like a ball at the top of being tossed. So that's instaneously zero of course.
True or false? If the acceleration is zero, the body cannot be moving. False- because constant velocity means zero acceleration. The derivative of a constant is zero.
True or false? If the acceleration is zero, the x-vs-t graph must be a straight line- not necessarily horizontal. Sure, because velocity is constant, so there's constant slope, so there's the x-vs-t graph is a line.
State whether the acceleration is positive, negative or zero for the following functions x(t).
Where is acceleration positive, negative or zero? Acceleration is positive where it's concave up, negative when concave down,
There's a homework problem similar to this, the same concept, similar to Sally & Joe problem - if you double the speed, what happens to the result? This one is different. #45. Object projected upwards with initial velocity = v, attains height h.
v_{0} up to H
Another object had initial velocity of 2v_{0} and attained a height up to what? Will it go higher or not as high? It will go higher. How much higher? 4H? 3H? 2H? H? You can throw out H. So which of the other three choices will it be?
s = (1/2)gt^{2} + V_{0}t + S_{0}, and in this scenario S_{0}=0. So in the first scenario:
1) s = (1/2)gt^{2} + V_{0}t
2) s = {1/2}gt^{2} + 2V_{0}t
So get the derivative for #1. So then you set velocity to zero, which is at the top, and that's how you can get your time. And then you can solve for the 's' value again, so the key here is to notice what happens to V_{0}- if you double V_{0}, you're going to have 2V_{0} all squared, and that's 4V_{0}^2 so the answer is going to be 4H.