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Homework Help: Motion in 1D problem, it's not supposed to be this hard.

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data

    To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 91.0 km/hr approaches a car stopped at the red light. When the truck is 115.7 meters from the car the light turns green and the car immediately begins to accelerate at 2.90 m/s^2 to a final speed of 106.0 km/hr. How close does the truck come to the car assuming the truck does not slow down?

    2. Relevant equations

    I might be wrong but I've used this formula:
    Vo = "V knot"
    Xo = "X knot" (initial position)

    X = (1/2)at^2 + Vot + Xo



    3. The attempt at a solution

    I convert the trucks velocity from 91 km/hr into 25.278 m/s, and I've converted the car's final velocity to 29.44 m/s. I tried to do the above equation for both vehicles and set them equal to each other so I'll know what time they meet. I was thinking this way and then when I find TIME, I'll plug it in back into one of the "X = (1/2) etc" equation to find the distance.

    My professor doesn't show us any examples, all he did was just show us how we get the equation. Thanks guys for any help, I really appreciate it.
     
  2. jcsd
  3. Aug 29, 2010 #2

    ehild

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    The problem asks the minimal distance the truck approaches the car. If you find a time when they meet it means that the truck and the car will collide. Just think: The truck goes with uniform speed. The car accelerates from zero speed. Its speed increases, reaches that of the truck and gets even higher up to 106 km/h. Up to what speed will the distance decrease between them?

    ehild
     
  4. Aug 29, 2010 #3
    The truck is gaining on the car right up until the cars speed > the trucks speed. At what time does the cars speed equal the trucks speed? Where are the truck and car at this time?
     
  5. Aug 29, 2010 #4
    Thanks guys, so at some point the car AND the truck both have the same velocity right?
     
  6. Aug 29, 2010 #5

    ehild

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    Yes, and you have to find that moment.

    ehild
     
  7. Aug 29, 2010 #6
    OK, so the car will hit the speed of 25.278 (which is the same like the truck) at some time T right? I find that T and then plug it back into the position equation for each of the vehicles right? After that subtract the two and I get how much farther (closest) they are apart?
     
  8. Aug 29, 2010 #7
    Awesome! I got it! I knew it had something to do with time when they both has the same velocity! You just confirmed it and helped me out! Thank you so much guys!!!!! :)
     
  9. Aug 29, 2010 #8

    ehild

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    Congratulation! Good job!

    ehild
     
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