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Answer ( 1 )

  1. All the graphs show positive velocities (they are all above the x-axis).

    Note that 2 graphs have time on the x-axis and 2 graphs have distance on the x axis. So you have to be careful.

    For velocity-time graphs (graphs (1) and (2)):

    The gradient at any time is the acceleration at that time. A constant negative acceleration means a constant negative gradient. This would give a straight line going downhill on a v-t graph.

    So graphs (1) and (2) are wrong.

    For velocity-distance graphs (graphs (3) and (4)):

    There are different ways to think about it.

    I’d simply say it can’t be graph(4) because of ‘common sense’. Velocity will become less positive. And at low speed it will take a long time to cover a given distance, so the velocity will decrease a lot over a short distance. Graph (4) gives the correct shape, but graph (3) makes sense.

    Another way is this:
    v² = u² + 2as.
    Keep things simple by taking u²=1 and a = -1/2 (whatever units you want).
    v² = 1 + 2(-1/2)s
    . . = 1 – s
    v = √(1-s) gives the general shape.

    Sketch the graph (like y = √(1-x)) and you get the shape of graph (3).

    (Note that in this example, the maximum value of s = 1 because you can’t take the square root of a negative quantity. What this means is the max. distance (max. value of s) = 1 in this example. The object reverses direction when it has reached s=1, like throwing a ball up and it reaching max. height.)

    Answer: graph (3)

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