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Help with distance catch up problem?

A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Could you please give a few different examples of how to solve this? Best answer awarded to most helpful. Thank you.

6 Answers

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  • 2 months ago
    Favorite Answer

    You know the distance formula is d = rt.

    Let t be the usual time it takes (in minutes)

    Let r be its usual speed

    d = rt

    Then when it is going slower:

    The time it takes is t + 30.

    The rate is r/3.

    d = (t + 30)r/3

    Equate these two:

    rt = (t + 30)r/3

    rt = rt/3 + 30r/3

    rt = rt/3 + 10r

    rt - rt/3 = 10r

    rt(1 - 1/3) = 10r

    rt(2/3) = 10r

    Divide both sides by r (assuming it is non-zero):

    (2/3)t = 10

    t = 10 * 3/2

    t = 30/2

    t = 15

    That's a longer, more algebraic way to solve this.

    But here's an intuitive way to solve it.

    The speed and the time are inversely proportional. In other words, if the speed is 1/3 the usual speed, the time will be 3 times as much as usual.

    Let t be the usual time.

    Let 3t be the new time.

    3t = t + 30

    2t = 30

    t = 30/2

    t = 15 minutes

    Summary:

    Normally it takes 15 minutes, but when it goes 1/3 as fast it will take 3 times as long (45 minutes) which is 30 minutes longer.

    Answer:

    15 minutes

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  • 2 months ago

    Recall: s = d/t → where s is the speed, d is the distance, t is the time

    First case: usual speed

    s = d/t → where d is the distance to reach the destination

    s = d/t → where t is the required time to reach the destination at normal speed

    Second case: usual speed divided by 3, time is 30 minutes more, i.e. (1/2) hours

    s = d/t → adapt this formula to the present case

    s/3 = d/[t + (1/2)] → recall: s = d/t

    (d/t)/3 = d/[t + (1/2)]

    d/(3t) = d/[t + (1/2)] → you can simplify by d both sides

    1/(3t) = 1/[t + (1/2)]

    3t = t + (1/2)

    2t = 1/2

    t = 1/4 ← in hours

    t = 15 minutes

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  • 2 months ago

     usual time to cover the same distance is 15 minutes.

    Step-by-step explanation:

    Required Formula:

    Time = Distance / Speed

    Let's assume

    The usual speed of the train be "s" km/hr

    The distance travelled by the train be "D" km.

    The usual time taken by the train to cover the same distance be "t" hr = D/s

    Given that,

    The reduced speed of the train = km/hr

    Therefore, according to the question, we can write the eq.

     

    - D/(s/3) -D/s=30min/60min

    3D/s-D/s=1/2

    D/s(3-1)=1/2

    D/s(2)=1/2

    D/s=1/4

    Thus,

    The usual time "t" taken by the to cover the same distance is,

    = D/s

    = 1/4*60min

    = 15 minutes

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  • 2 months ago

    Let s = Usual speed in mph

    Let s/3 = Speed now in mph

    Let t = Usual time in hours

    Let t + 1/2 = Time now in hours

    Let d = Distance in miles.

    Then.

    t = d / s. This is the "usual" situation.

    t + 1/2 = d / (s / 3). This is the current situation.

    We can now solve these equations for t.

    t = d / s

    2t + 1 = 2d / (s / 3) = 6d / s = 6t

    4t = 1

    t = 1/4 hour or 15 minutes.

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  • 2 months ago

    Let t min be the usual time to cover the same distance.

    When velocity = 3v units/min, time taken = t min:

    Hence, distance = 3v • t units ... [1]

    When velocity = (1/3)(3v) = v units/min, time taken (t + 30 min):

    Hence, distance = v • (t + 30) units ... [2]

    [1] = [2]:

    3v • t = v • (t + 30)

    3t = t + 30

    2t = 30

    t = 15

    Usual time to cover the same distance = 15 min

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  • Anonymous
    2 months ago

    distance = speed * time.  Let  V be the speed and T be the time

    distance = V * T

    now with the new information remembering that the distance is the same no matter how fast or slow you travel at

    distance = V / 3 * (T + 30)

    setting the two equations equal to each other because distance is the same gives you

    V * T = V/3 * (T + 30)

    VT = VT/3 + 10V

    2/3VT = 10V

    T = 15 minutes or 0.25 hour

     

    Try some numbers and see if it works. Say the speed = 60 miles/hr.  Distance = 15 miles.  If the speed is dropped to 20 mile/hr then it takes 0.75 hours, 0.5 hr or 30 minutes longer.

    Try 10 miles/hr.  Distance = 10 mile/hr * 0.25 hr = 2.5 miles.   Drop the speed to 1/3 and it now takes 0.75 hours.  Again, 30 minutes faster.

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