- #26
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I converted everything to meters beforehand, and the solution came out to be -1.42 m/s^2 this time...
Can you explain the above equation? You've got the the distance traveled for car A as the average of the two car velocities multiplied by time. How does that come about?Sorry, I try to write neatly on loose leaf so I can take pictures of it to show my work, but it always comes out to chicken scratch. And it takes me forever to format the LaTeX.
So here is my work:
[tex]X_B - 676 = 8.0556t + \frac{1}{2}(0)t^2 \\ X_B = 8.0556t + 676[/tex]
[tex]X_A = 8.0556t + 676 = \frac{1}{2}(44.72 + 8.0556)t [/tex]
I used the 1D Kinematic formula x - x_0 = 1/2(v_0 + v)t. The intial velocity of the car A is 44.72 m/s, and the final velocity of car A 8.0556 m/s if it is to avoid collision with car B going at the same constant velocity of 8.0556 m/s.Can you explain the above equation? You've got the the distance traveled for car A as the average of the two car velocities multiplied by time. How does that come about?
Okay, that's fine. What value did you find for t?I used the 1D Kinematic formula x - x_0 = 1/2(v_0 + v)t. The intial velocity of the car A is 44.72 m/s, and the final velocity of car A 8.0556 m/s if it is to avoid collision with car B going at the same constant velocity of 8.0556 m/s.
See my most recent edit of my last postSo yea, I thought I covered all my grounds yet arrived at a solution that's pretty off...how is my calculation different from the ones mentioned earlier in this thread that got pretty much the exact answer they were supposed to get?
[tex]8.0556 + 676 = 44.72(36.875) - \frac{1}{2}a(36.875)^2[/tex]See my most recent edit of my last post