Thanks. Ill try to format it better next time... I guess the only part I don't understand is why if were calling 1.18 m/s a velocity initial value... are we plugging it into Vf on the equation? I assume this has something to do with there being two different worlds (post collision and directly...
Ok, after reading that maybe I understand a little better not perfectly though. Is this how you wanted me to proceed?
$$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8)1.18 $$
$$ (796.1*V1)+(10610.05)=1536.242 $$
$$ 796.1Vl = -9073.808 $$
$$ 796.1Vl/796.1 = -9073.808/796.1 $$
$$ Vi = -11.40 m/s $$...
I kind of feel like I am getting lost in the problem now... we are so far down (my fault since I just can't seem to comprehend it)
I'm going to restate this...
$$ (m1v1)+(m2v2)=(m1+m2)Vf $$
$$(796.1∗v1)+(505.8∗21.01)=(796.1+505.8)Vf $$
I really now just need help figuring out if the stuff I...
First off, yeah I definitely made an error (Sorry :() the actual answer would be
$$ Vi = 1.18m/s $$
Now let me just write were in the post-collision world here so I remember. Next I think I have to get back on track and find the VF after the collision if I have the initial velocity
So I think I...
I think
$$ A = ΔV/T $$
$$ -0.54 = ΔV/22 $$
$$ ΔV = -11.88 m/s $$
You said this is my change in velocity... does that mean I can't put this in my VF all the way at the original equation? If so what would I do with this/where would I go now since I thought we needed VF.
Ok the the thing you said about their being two different VF's helps me visualize a little better, i was thinking of it as one big problem when really there's the stuff after the collision when its skidding and then the stuff when it collides.
Ok so let me try to stop messing up on the...
Ok... so I did half of that above i think? Because i need to get average velocity to plug into an acceleration equation?
$$ V=D/T $$
$$ V=13/22 $$
$$ .59m/s $$
Then a step further for acceleration
$$ A=V/T $$
$$ A=.59/22 $$
$$ A = .027 m/s^2 $$
Im not sure where that would leave me to go next...
Ok guess not... what if I try this one
$$ D=½(VF+VI)T $$
$$ 13=½(VF+21.01)22 $$
$$ VF = -19.83 m/s $$
Then I do this thing again
$$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8)-19.83 $$
$$ (796.1∗v1)+(10626.858)=-25816.677 $$
$$ (796.1∗v1)=-36443.533 $$
$$ (796.1∗v1)/796.1=-36443.533/796.1 $$
$$ V1...
Yes it is morbid! My physics teacher is a former funeral director until he became a teacher!
Back to the problem...based on what u two said I think that might mean I have to do
$$ V=D/T $$
$$ V=13/22 $$
$$ .59 m/s $$
Then Maybe I can plug that into this which makes sense
$$...
Homework Statement
Pete got into into a car accident and collided with another vehicle, which resulted in the death of the second vehicles passenger, David. Whether Pete was speeding or not determines if he goes to jail. The speed limit where he got into the accident was 45mph. How fast was...