Recent content by No1_129848

One big important thing, your formula for a is the (right) formula for an average acceleration. Do you understand what is the (only) acceleration you should worry about in this specific problem? Once you setup your correct equation of motion (and ofc you can derive them if you need eqs for...

Yes, that should be the correct thought process

Oh, I see. I tend to get tricked by problems involving average values because I actually get to practice it much less than other problems, and this time I made the wrong assumption that my math was right.Thanks a lot for the help!

Well, I want to punch myself for not using Δv now that i think about it. My god what a stupid mistake... So, let's say that the problem is just solved if I just use ##a_{avg}= \frac{v_1-v_0}{t_1-t_0}##

I'll do it step by step, I've always used this formula ##a_{avg}=\frac{a_1−a_0}{t_1−t_0}## So, in the first case I get ##a_{avg}=\frac{2−2/3}{2−0}=2/3## I guess this will answer all of your questions, am I calculating aavg wrongly?

I have some questions You say that a in phase 1 is 4/3m/s2, may I ask why? The Δa is 4/3m/s2, but it should be considered during the Δt = 2s if I undestand it correctly, so the total Δv during the 2 seconds is 4/3m/s2, In fact in my calculations I am adding 4/3m/s2 in total for phase 1. Are...

So my thought process was actually correct but I made some misteps in the calculations? If yes, I can't really come up with the algebraic mistakes I made, and at the moment I am not even sure why my original thoughs are actually ok

You probably saw examples about elevators and the "change in weight" they can cause, you can also visualize what happens by just having a pen on top of your hand and then quickly moving your hand down, the pen won't catch up with your hand and it will "float" upwards if your frame of reference...

By the way, I didn't had time to post this earlier but I understood the problem in my logic, of course it's wrong to use the average formula, because the acceleration is not constant, in this case it's hard for me to find a solution that avoids calculus

I could do it this way and I will likely try to do so later, but I'm pretty sure the problem is intended to be solvable without that information, I'd assume you learn about that theorem in the kinetic energy section, which comes later on. Anyway, do you have any idea about solving this problem...

No, if it helps the problem is in the "Force and motion 1" section, and I assume you are supposed to learn what you suggested later on

Homework Statement [/B] Figure 5-55 gives, as a function of time t, the force component Fx that acts on a 3.00 kg ice block that can move only along the x axis. At t = 0, the block is moving in the positive direction of the axis, with a speed of 3.0 m/s.What are its (a) speed and (b) direction...

Consider it this way, as long as the cars have the same velocity, every change in the position of one of them will be matched by the other car, so you should just consider the difference in their speeds. If a car goes 10 m/s and the other goes at 11 m/s and they start in the same position it...

You should focus on the difference in speed between the two cars, that's how car A is going to close the gap

Firstly, there are some issues about your problem, could you please just copy-paste the problem here? The main issues are: 1) what's the "maximum distance", you should be able to tell us if we are talking about the displacement on the x axis, the y-axis or both of them 2) Reading you problem...