Recent content by zero_infinity

I just started a numerical analysis class and I'm curious: what are the advantages and disadvantages of the two methods? Do we use numerical methods in situations where getting analytical solutions is possible? If so, why? I just want a better understanding of when each method is used in...

Hi which part are you talking about?

Homework Statement at t=0, a 4kg mass is moving at 2 m/s\hat{i} + 3m/s\hat{j} at t=0, F1 and F2 act upon the object F1= [1 N + (1/2 N/s)t ]\hat{i} F2= -(1/3 N/s2)t2 \hat{j} a) what is the object's velocity when t=3? b) at what time does the object stop moving in \hat{j} direction? c)...

What is wrong with what I am already doing? 250 is supposed to be the V_pw vector? that's it?

right... it has to fly in the blue direction to compensate for the wind. It's asking you to find that direction and speed. Why do the vectors have to be switched? It has to fly north relative to the ground so that has to be the vector relative to the ground. i extended the red vector to show...

Homework Statement a plane flies at an airspeed of 250 km/h. A wind is blowing at 80 km/h toward the direction 60(degrees) east of north. a) in what direction should it fly in order to fly north relative to the ground? b) what is the speed of the plane relative to the ground?Homework...

i'm an idiot -_-

what do you mean?! in that equation it's the time it takes for them to meet at the same point !?

... time the function for the train could have just been .2t^2(1/2*a*t^2), but i rewrote it for t=6 (found its position, velocity) if you're plugging 6 into t, then it's 6 extra seconds on-top of the 6 after it started accelerating. how did you know to plug in 6?

How did you know they were going to meet at t=6(or t=12?)? It doesn't make any sense to me. That's really not intuitive for me, wtf. If the passenger starts running at t=6, how they possibly meet at t=6(t=12?)?

Yes, I have done that too. It's wrong v(t) = .2(t^2 + 12*t + 36)/t v'(t) = .2(t^2-36)/t^2 =0 this gives two extrema: 6 and -6 i already know the answer is 4.8, so... it doesn't work like that. I'm also pretty sure the velocities have to be the same...

Still need help with this. :/ Nobody knows how to do it? Or did I not follow the rules? I'm pretty sure I did..

Um, what? The passenger is going to be moving with constant speed. I know the answer - it's 4.8 m/s (i don't know how to arrive at it). That equation doesn't help me for the passenger, and I've calculated the velocity for the train (at t=6).

Homework Statement A train pulls away from a station with constant acceleration of .4 m/s^2. A passenger arrives at a point next to the track 6.0s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train...