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Avoiding vehicular rear-end collision

  • Thread starter PhizKid
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  • #26
475
1
I converted everything to meters beforehand, and the solution came out to be -1.42 m/s^2 this time...
 
  • #27
gneill
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You'll have to show all of your steps and the intermediate numbers. We can't tell what went wrong if there's nothing to see :smile:
 
  • #28
475
1
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 \\ 676 = 18.3322t \\ \textrm{Therefore } t = 36.875[/tex]

[tex]8.0556 + 676 = 44.72t - \frac{1}{2}a(36.875)^2 \\ -964.9944 = 679.8828a \\ a = -1.42 m/s^2[/tex]

(44.72 is the initial velocity of car A in m/s, and 8.0556 m/s is car B's constant velocity)
 
  • #29
1,065
10
If you like to know my method.
The accelerated car displacement contains 2 parts, the triangle and rectangle with speed 161km/hr and 29km/hr
The constant part contains 676m and rectangle which is equal to the accelerated car for minimum deceleration. Actually the answer is the minimum deceleration.

So i only need to calculate the non common area.


0.676=1/2(132)t
t=0.01024hr
a=-132/0.01024= -12890.625km/hr.hr=-0.9944m/s2

add: For my post#13 there's an error(corrected here) in units multiplication as noted by CAF123.
 
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  • #30
475
1
Isn't that just the average acceleration?
 
  • #31
1,065
10
Yes it an average deceleration or constant acceleration as in equation s=s0+v0t+0.5t^2.
 
  • #32
gneill
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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]
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?
 
  • #33
1,065
10
Sketching a graph is really helpful
As you see from the graph, the aim is to reduce the speed to less than the front car to avoid collision. It can be done just behind the front car or anywhere behind the front car

http://imageshack.us/a/img688/5505/60741537.jpg [Broken]
 
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  • #34
475
1
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?
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.
 
  • #35
gneill
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20,781
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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.
Okay, that's fine. What value did you find for t?

Edit: Never mind, I see where you calculated it above.

The value is fine. I think you just have a mathematical error in your calculations for a using that value of t back in post 28. Redo the calculation.
 
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  • #36
475
1
So 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?
 
  • #37
gneill
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So 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?
See my most recent edit of my last post :smile:
 
  • #38
475
1
See my most recent edit of my last post :smile:
[tex]8.0556 + 676 = 44.72(36.875) - \frac{1}{2}a(36.875)^2[/tex]

36.875^2 = 1359.765625
1359.765625 / 2 = 679.8828125
44.72(36.875) = 1649.05
8.0556 + 676 = 684.0556

So 684.0556 = 1649.05 - 679.8828125a

684.0556 - 1649.05 = -964.9944

So -964.9944 = -679.8828125a

-964.9944 / -679.8828125 = 1.41935401551

Damnit, so I got positive 1.41935401551 m/s as acceleration, which would mean it's entirely incorrect then.

Where did I perform the wrong calculation?
 
  • #39
gneill
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On the left hand side of the first equation, you're missing multiplying the speed of car B by time. 8.0556*t + 676 = ....
 
  • #40
475
1
Oh wait I was using the wrong equation. I'm supposed to be using x - x_0 = (v_0)t + (1/2)at^2, not x - x_0 = vt - (1/2)at^2, since 44.72 m/s is the initial velocity of car A. So it does come out to -1.41935401551 m/s^2.

So the displacement of car A should equal the displacement of car B, which means I did forget to multiply the velocity of car B by the time...thanks

Edit: No, I had to use the subtraction formula with the final velocity and change the final velocity to 8.0556 m/s in the equation. This yields the correct -0.994 m/s^2 solution

Thanks
 
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