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Thread: uniform acceleration

  1. #1
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    uniform acceleration

    A car, starting from rest and travelling from p to q on a straight level road,
    where | pq | = 10 000 m, reaches its maximum speed 25 m/s by constant
    acceleration in the first 500 m and continues at this maximum speed for the
    rest of the journey.
    A second car, starting from rest and travelling from q to p, reaches the same
    maximum speed by constant acceleration in the first 250 m and continues at
    this maximum speed for the rest of the journey.
    (i) If the two cars start at the same time, after how many seconds do the
    two cars meet ?
    Find, also, the distance travelled by each car in that time.

    i am trying to use uvast equations to solve this but i cant. can someone help me
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  2. #2
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    Re: uniform acceleration

    acceleration has nothing to do with this problem.

    car 1 reaches $25 m/s$ in the first $500m$.

    car 2 also reaches $25 m/s$ but does so in the first $250m$.

    So you have two cars, travelling towards each other at $25 m/s$, separated by a distance of $10000m-500m-250m=9250m$.

    The relative speed of the two cars is just their sum since they are travelling opposite directions so it is $50 m/s$

    see if you can finish from here.

    Spoiler:

    it takes $\dfrac {9250m}{50m/s}=185s$ for them to meet.

    the first car travels $500m + 25 m/s \cdot 185s = 5125m$.

    the second car travels $250m + 25 m/s \cdot 185s = 4875m$.
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  3. #3
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    Re: uniform acceleration

    romsek: I'm afraid I must disagree with your analysis. You have assumed that the two cars reach 25 m/s at the same time. But clearly if one car reaches 25 m/s in 250 meters while the 2nd car requires 500 meters, the first car reaches that velocity in a shorter period of time than the second. It also seems apparent that the slower accelerating car will travel a shorter distance than the faster accelerating one.

    Mark: what you need to do is calculate the acceleration of each car, and then the time it takes for each car to reach 25 m/s. Then for the faster car determine how far it travels at 25 m/s while the slower car is still accelerating. From that you can determine the distance that both cars must cover when both are going at 25 m/s toward each other. That distance divided by 50 m/s gives the time to collision starting from the time when the slower car has reached 25 m/s. I trust you can finish it from there.
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    Re: uniform acceleration

    thanks
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  5. #5
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    Re: uniform acceleration

    Yep, you are correct.

    Letting $v_0 = x_0 = 0$

    we can note that $d = \dfrac 1 2 a t^2$ and $v = a t$ and thus

    $a= \dfrac {v^2}{2 d}$

    $t= \dfrac {2d} v$

    So car 1 takes $40s$ to reach $25m/s$ and $x=500m$ and car 2 takes only $20s$ to travel $250m$ to reach $x=9750m$

    So between $20s$ and $40s$ car 2 is travelling at $25m/s$ and thus travels $20m\cdot 25m/s=500m$ and reaches $x=9250m$ at $t=40s$

    Now from $t=40s$ on both cars are travelling at $25m/s$ in opposite directions and must travel $9250m-500m=8750m$

    it takes $t=\dfrac{8750}{50} = 175s$ to do this.

    Car 1 travels $500m+175s\cdot 25m/s = 4875m$

    Car 2 travels $250m+500m+175s \cdot 25m/s = 5125m$
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