Recent content by Zsmitty3

Ok so I just drew it wrong. It's actually 8sin(120) which gives you the same answer. The 2m vector is listed first so do you just assume it's A?

Okay. So the angle between the two is going to be 60 degrees because the northbound one is going 30 west of N. The other's going east. If the North on was going straight N they would make a 90 degree angle. Since it's 30 degrees west of that (to the left) they're going to make a 60 degree angle...

The Truck is parked so yes its initial is 0. Sorry I forgot to add that. Conservation of KE?

One vector with mag 2 pointing East. Other one is mag 4 pointing 30° west of North Would you use sin or cos and would it be - or + I did (2*4)cos60°=+4 because they're vectors and we have the A/H sides. I'm worried about that though because we may use sin? Also Cos is (-) in...

A 1689 kg car collides head on with a 2000 kg truck. The collision is elastic. If the velocity of the truck is 17km/h in the same directions as the car's initial velocity, what is the initial speed of the car in km/h? I'm getting an answer but doesn't seem to make sense. So I'm...

-1/2∫1/(u+3)+1/2∫1/(u-1) ends up being 1/2* ln(√x+3)-1/√x+3)+3

So 2=A(u-1)+B(u+3) Let u= 1 2=A(0)+B(4) 2=3B so B=(1/2) Let u=-3 and A=-1/2 When you put those in can you just bring the 1/2's out in front of the fraction or do you have to distribute then through the bottom of the integral?

Can you make the u2+2u-3 = (u+3)(u-1) Then A/(u+3) + B/(u-1) ?

You mean partial fraction decomposition or just split them up into separate fractions?

-1/u+ln(2u)-u/3 Just plug in √x+3 back in for u. So that gives you -(1/√x+3)+ln(2√x+3)-(√x+3)/3) ?

So you can pull out the 1/2 and make it .5∫(1/u) The integral of that is .5(ln absvalue(u)) Sub u back in and get .5(ln(√x+3)) abs value of √x+3 of course that's still not giving me a right answer though when I check it.

If I sub the other way and put x in for u^2-3 then that gives me 1/x. Integral of that is ln absvalue x. When I plug in numbers to make it a definite integral and check it on my calc. that doesn't work.

So plug it into the original integral and get ∫1/(2√(u2-3+3) ? So the 3's equal 0 Gives you: ∫1(2√u2) Sqrt and 2 cancel leaving you 1/2u but then that gives you 2u-1 integrating that gives you u0 which is where I keep getting stuck.

1. ∫ 1/(2√(x+3)+x) 2. Not sure if I'm beginging this correctly or not but I get stuck. 3. Let u= √x+3 then u2 = x+3 2udu=dx dx=2√[(x+3) Therefore: ∫1/u2-3 Not sure where to go from here?

Well it's multiple choice and that's once so thanks for hanging in there with me :)