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    Please see the picture attached, an explanation of the answer would be much appreciated. Thanks in advance.
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    (Original post by ThomsonM98)
    Please see the picture attached, an explanation of the answer would be much appreciated. Thanks in advance.

    Welcome to TSR physics.

    What is the answer you want explained?
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    (Original post by Stonebridge)
    Welcome to TSR physics.

    What is the answer you want explained?
    The answer is B, but I don't understand how they arrived at that answer.
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    (Original post by ThomsonM98)
    The answer is B, but I don't understand how they arrived at that answer.
    The object accelerates for 4 seconds and decelerates for 2 seconds according to that graph.

    Use v=u+at to find the final velocity after the 1st 4 seconds. (u=0 and you know a and t)
    Then use the formula again.
    This time u = the answer to the 1st part. You know a and t again. Find v
    (Remember that a is negative in the 2nd part as we are dealing with deceleration.)
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    A simple way to remember it is to calculate the area underneath the graph. For acceleration-time graphs, the speed is equal to the area under the graph. For velocity-time graphs the displacement is equal to the distance under the graph. As Stonebridge said, it's important to remember that between 3 and 5 seconds acceleration is negative.

    Using this technique:

    Spoiler:
    Show
    for the first rectangle: acceleration = 4ms-1
    time = 3s
    4x3=12ms-1

    for the second rectangle: acceleration = -2ms-1
    time = 2s
    -2x2 = -4ms-1

    add the two sections together to get the speed after 5 seconds:

    12 + (-4) = 8ms-1

    Therefore the answer is B
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    I was hoping the poster would try to work this out for himself.

    That is the point of this website.
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    (Original post by james153)
    A simple way to remember it is to calculate the area underneath the graph. For acceleration-time graphs, the speed is equal to the area under the graph. For velocity-time graphs the displacement is equal to the distance under the graph. As Stonebridge said, it's important to remember that between 3 and 5 seconds acceleration is negative.

    Using this technique:

    Spoiler:
    Show
    for the first rectangle: acceleration = 4ms-1
    time = 3s
    4x3=12ms-1

    for the second rectangle: acceleration = -2ms-1
    time = 2s
    -2x2 = -4ms-1

    add the two sections together to get the speed after 5 seconds:

    12 + (-4) = 8ms-1

    Therefore the answer is B
    Thank you this is very clear and helpful!
 
 
 
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