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Math Help - Simple speed/distance/time questions

  1. #1
    Senior Member Mukilab's Avatar
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    Simple speed/distance/time questions

    Hello I'm stuck on a particular 'genre' of sorts of speed/distance/time questions where you have to incorporate two objects travelling.

    Apart from trial and error I don't know how to do these.


    For example:
    A car travelling at 90km/h is 500m behind another car travelling at 70km/h in the same direction. How long will it take the first car to catch the second?

    or

    A train of length 180m approaches a tunnel of length 620m. How long will it take the train to pass completely through the tunnel at a speed of 54km/h?


    I have several of these questions so I would please like a method...



    I've tried different tactics but I can't seem to figure out how to do these type of questions.

    Please help.
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  2. #2
    Senior Member Mukilab's Avatar
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    Furthermore I have a simpler question of a different ty[pe that I can't do.

    This is the question:
    A car must complete a journey at an average speed of 40mph. At what speed must it travel on the return journey if the average speed for the complete journey is 60mph?

    I started off:

    Average speed = \frac{2D}{T1+T2}

    Then because T1 = \frac{D}{40}

    Average speed = \frac{2D}{1}\cdot \frac{40+T2}{2D}

    To which I think I can cancel the 2D's. I'm left as T2 = 20mph but that's not the answer.
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    Hello I'm stuck on a particular 'genre' of sorts of speed/distance/time questions where you have to incorporate two objects travelling.

    Apart from trial and error I don't know how to do these.


    For example:
    A car travelling at 90km/h is 500m behind another car travelling at 70km/h in the same direction. How long will it take the first car to catch the second?

    or

    A train of length 180m approaches a tunnel of length 620m. How long will it take the train to pass completely through the tunnel at a speed of 54km/h?


    I have several of these questions so I would please like a method...



    I've tried different tactics but I can't seem to figure out how to do these type of questions.

    Please help.
    Hi Mukilab,

    For the car catching up to the one in front,
    it's travelling 20km/hr "faster" than the one in front.

    Hence, you can imagine the one in front stopped
    and the car in behind travelling at 20km/hr effectively.

    Hence it will bridge the 500m gap in t=\frac{d}{v}=\frac{0.5}{20} hours.

    For the train,

    the nose of it has to travel a distance it's own length plus the tunnel's length
    in order for the tail of it to exit the tunnel.

    You then have the speed and distance,
    so you can calculate the time.

    v=\frac{d}{t}

    d=vt

    t=\frac{d}{v}
    Last edited by Archie Meade; March 22nd 2010 at 01:16 PM.
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  4. #4
    Senior Member Mukilab's Avatar
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    Thanks for those but I'm confused with this one know x_xx

    An earthworm of length 15cm is crawling along at 2cm/s. An ant overtakes the worm in 5 seconds. How fast is the ant walking?


    I'm just confused. I thought I wrote something worthwhile but it turns out I didn't because I assumed the... nevermind, no point in me rambling. Please could you show me?
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    Quote Originally Posted by Mukilab View Post
    Thanks for those but I'm confused with this one know x_xx

    An earthworm of length 15cm is crawling along at 2cm/s. An ant overtakes the worm in 5 seconds. How fast is the ant walking?


    I'm just confused. I thought I wrote something worthwhile but it turns out I didn't because I assumed the... nevermind, no point in me rambling. Please could you show me?
    If the ant overtakes the worm in 5 seconds,
    then given that the worm will have moved 10cm in those 5 seconds, due to it's own speed,
    and is 15cm long, the ant will have moved 25cm in those 5 seconds,

    so the ant's speed is 5cm/s
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  6. #6
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Archie Meade View Post
    If the ant overtakes the worm in 5 seconds,
    then given that the worm will have moved 10cm in those 5 seconds, due to it's own speed,
    and is 15cm long, the ant will have moved 25cm in those 5 seconds,

    so the ant's speed is 5cm/s

    sorry, I thought thise question was the same but it's totally differnt, I need to call upon your help again.

    A train of length 100m is moving at a speed of 50km/h. A horse is running alongside the train at a speed of 56km/h. How long will it take the horse to overtake the train?


    Also please could you help me with the question I posed on post #2?
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  7. #7
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    Quote Originally Posted by Mukilab View Post
    Furthermore I have a simpler question of a different ty[pe that I can't do.

    This is the question:
    A car must complete a journey at an average speed of 40mph. At what speed must it travel on the return journey if the average speed for the complete journey is 60mph?

    I started off:

    Average speed = \frac{2D}{T1+T2}

    Then because T1 = \frac{D}{40}

    Average speed = \frac{2D}{1}\cdot \frac{40+T2}{2D}

    To which I think I can cancel the 2D's. I'm left as T2 = 20mph but that's not the answer.
    You haven't handled the division properly here, Mukilab

    Av=\frac{2D}{\frac{40}{D}+T_2}=\frac{2D}{\frac{40+  DT_2}{D}}=\frac{2D^2}{40+DT_2} etc

    However, because we are dealing with average speeds, the situation is linear, so we can choose a distance of 40 miles for convenience.

    \frac{2D}{1+x}=60

    since the car will cover the 40 miles in 1 hour, so since D=40

    \frac{80}{1+x}=60

    80=(1+x)60=60+60x

    20=60x

    x=\frac{1}{3} hours = 20 minutes.

    On the return journey the car covers 40 miles in 20 minutes,
    so what is it's average speed on the return journey?
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  8. #8
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Archie Meade View Post
    You haven't handled the division properly here, Mukilab

    Av=\frac{2D}{\frac{40}{D}+T_2}=\frac{2D}{\frac{40+  DT_2}{D}}=\frac{2D^2}{40+DT_2} etc

    However, because we are dealing with average speeds, the situation is linear, so we can choose a distance of 40 miles for convenience.

    \frac{2D}{1+x}=60

    since the car will cover the 40 miles in 1 hour, so since D=40

    \frac{80}{1+x}=60

    80=(1+x)60=60+60x

    20=60x

    x=\frac{1}{3} hours = 20 minutes.

    On the return journey the car covers 40 miles in 20 minutes,
    so what is it's average speed on the return journey?

    I got lost when you put 40+DT2. No idea where you got D*T2...


    I also have no idea where you get 1.

    or + x
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    Quote Originally Posted by Mukilab View Post
    sorry, I thought thise question was the same but it's totally differnt, I need to call upon your help again.

    A train of length 100m is moving at a speed of 50km/h. A horse is running alongside the train at a speed of 56km/h. How long will it take the horse to overtake the train?


    Also please could you help me with the question I posed on post #2?
    This one (horse and train) is the same type as the one with the two cars.
    The "effective" speed of the horse is 6km\hr relative to the train,
    (ie as if the train was stopped).

    Have a shot at it.
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  10. #10
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    Quote Originally Posted by Mukilab View Post
    I got lost when you put 40+DT2. No idea where you got D*T2...


    I also have no idea where you get 1.

    or + x
    As it's convenient to choose D=40 miles, then T_1=1\ hour

    The return time is x.

    You could choose D as other distances too, but as this is a linear model,
    you will keep getting the same answer for the return average speed.

    You chose a more complicated way.

    \frac{2D}{\frac{40}{D}+T_2}=\frac{2D}{\frac{40}{D}  +\frac{DT_2}{D}} getting a common denominator under the line.

    =\frac{2D}{\frac{40+DT_2}{D}}

    When you divide by a fraction, just turn the fraction upside-down and multiply instead.

    Divide by 2 is multiply by \frac{1}{2}

    Divide by \frac{1}{2} is multiply by 2 etc.

    Hence

    \frac{2D}{\frac{40+DT_2}{D}}=2D\frac{D}{40+DT_2}
    Last edited by Archie Meade; March 22nd 2010 at 01:37 PM.
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  11. #11
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Archie Meade View Post
    As it's convenient to choose D=40 miles, then T_1=1\ hour

    The return time is x.

    You could choose D as other distances too, but as this is a linear model,
    you will keep getting the same answer for the return average speed.

    You chose a more complicated way.

    \frac{2D}{\frac{40+DT_2}{D}}

    When you divide by a fraction, just turn the fraction upside-down and multiply instead.

    Divide by 2 is multiply by \frac{1}{2}

    Divide by \frac{1}{2} is multiply by 2 etc.

    Hence

    \frac{2D}{\frac{40+DT_2}{D}}=2D\frac{D}{40+DT_2}


    I understand everything apart from how you got T1 as 1hour.

    Thank you for your complete perserverance

    to get the time for the first one you would do:

    T1=\frac{D}{40}

    You have two unknowns and no simultaneous equation so how could you solve?
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    Quote Originally Posted by Mukilab View Post
    I understand everything apart from how you got T1 as 1hour.

    Thank you for your complete perserverance
    Hi Mukilab,

    To obtain a "pain-free" answer, I choose D=40 miles,
    so that the outward journey would take 1 hour exactly.

    You could choose D as any other distance
    and you really should verify that regardless of D (within reason)
    you will keep getting the same answer for reverse journey average speed.

    So my choice of D=40 miles was to simplify everything.
    If you choose D=80 miles, you get the same answer, or D=100 miles etc.

    for practice, you should experiment with a few different distances,
    calculate the average return speed and verify that the entire journey average speed is 60mph.
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  13. #13
    Senior Member Mukilab's Avatar
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    120minutes
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    Hi Mukilab,

    If the outward journey is 40 miles in 1 hour,
    the return journey is 40 miles in 20 minutes,

    that's one hour and 20 minutes, which is 60+20 minutes
    (rather that 100+20 minutes).
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    Quote Originally Posted by Mukilab View Post
    I understand everything apart from how you got T1 as 1hour.

    Thank you for your complete perserverance

    to get the time for the first one you would do:

    T1=\frac{D}{40}

    You have two unknowns and no simultaneous equation so how could you solve?
    Yes, it gets a bit complex,
    so it's better to experiment by choosing a value for D.
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