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    First question: A train is travelling at 55m/s away from a wall. When the train is 660m from the wall the train blows its whistle. A short time later, passengers on the train hear the echo of the whistle, the sound having bounced off the wall. How far is the train from the wall when the echo reaches the train? (the speed of sound is 330m/s)

    What I did:
    Time taken for the sound to travel to travel to the wall AND back=660/330=2*2=4ms-1.
    Distance travelled by the train during 4 seconds=55*4=220m.

    I have NO idea what to do next. I can calculate the distance the train is from the wall after 4 seconds (220+660=880m) but it's not taking me anywhere.

    Second question:

    Of course, we can rule out A & D. What after that? Could somebody enlighten me as to the method of doing such questions? Obviously there's some out-of-the-earth strategy or way or some such. Are we supposed to calculate the difference between their respective speeds? If so why, and what next?
    Thank you.
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    You won't solve it in the way you are currently pursuing (pun) the problem.

    You need to think of the relationship between the train and the sound starting point. The train and the sound start off in opposite directions then after two seconds (660/330 = 2 seconds), the sound will reflect off the wall. But in that two seconds the train is now a further 110 metres (2 x 55) further from the wall when the echo starts on its race to catch up with the train.

    Think about expressing the relationships between the speed of the train moving away from the wall at 55m/s starting at 770 metres (660 + 110) from the wall as an equation and, a different equation for the speed of the echo moving away from the wall at 330 m/s.

    Spoiler:
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    Think of SUVAT (you don't need acceleration) and which equation you could use?

    That will give you two equations in which case you will then solve them simultaneously.
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    A big hint to the first question is given in the second question, which points another way to look at the first problem: draw a graph.

    If you do this it needs to be a distance vs time graph.

    The origin becomes the echo starting from the wall which, because it's a constant speed, will produce a straight line with slope of 330m/s.

    A second line (the train) will start at a point somewhere along the distance axis: 770 metres along the distance axis in fact (660+55*2) and with a slope of 55m/s.

    Notice the slopes of the two graphs are different. At some point the two lines will cross. This is the point at which the sound catches up with the train. The answer can be read off the distance axis.
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    (Original post by uberteknik)

    Spoiler:
    Show
    Think of SUVAT (you don't need acceleration) and which equation you could use?

    That will give you two equations in which case you will then solve them simultaneously.
    but all the SUVAT equations require an acceleration. :confused:
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    (Original post by Oh, me! Oh, me.)
    First question: A train is travelling at 55m/s away from a wall. When the train is 660m from the wall the train blows its whistle. A short time later, passengers on the train hear the echo of the whistle, the sound having bounced off the wall. How far is the train from the wall when the echo reaches the train? (the speed of sound is 330m/s)

    What I did:
    Time taken for the sound to travel to travel to the wall AND back=660/330=2*2=4ms-1.
    Distance travelled by the train during 4 seconds=55*4=220m.

    I have NO idea what to do next. I can calculate the distance the train is from the wall after 4 seconds (220+660=880m) but it's not taking me anywhere.

    Second question:

    Of course, we can rule out A & D. What after that? Could somebody enlighten me as to the method of doing such questions? Obviously there's some out-of-the-earth strategy or way or some such. Are we supposed to calculate the difference between their respective speeds? If so why, and what next?
    Thank you.
    First question - it takes sound two seconds to reach the wall. You can calculate the position of the train at this point. From that moment on the sound catches up with the train at a net speed of (330-55)m/s so you can calculate how long it will take to reach the train at this relative speed. You can use this time to calculate how far the train has moved and therefore what position it's at when the sound catches up, which is when the echo is heard on the train.

    second question - the lead varies by the difference between the two speeds. You can calculate from the individual speeds the times when the runner and the cyclist reach the end of the flat, mud and uphill sections:

    cyclist - 2 mins (distance/speed) on flat; 30 mins on mud; 15 mins uphill; 1.5 mins downhill
    runner - 10 mins; 22.5 mins on mud; 10 mins uphill; 12.5 mins downhill

    You can calculate the rate at which the lead varies by the difference between the speeds of the two competitors at each stage so you should be able to identify the graph that best fits the variation.
 
 
 
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