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    Hi, I was doing part (d) of this question and was wondering why you are allowed to sub in t = 4 and x = 32 to work out the integration constant for the equation of t is greater then 4.
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    (Original post by Glavien)
    Hi, I was doing part (d) of this question and was wondering why you are allowed to sub in t = 4 and x = 32 to work out the integration constant for the equation of t is greater then 4.
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    Draw a sketch of both graphs on the same coordinate axes. It should make sense from there on. I choose to use triangles to find the distance for the second half of the graph because it's simpler than integrating it twice.
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    (Original post by Glavien)
    Hi, I was doing part (d) of this question and was wondering why you are allowed to sub in t = 4 and x = 32 to work out the integration constant for the equation of t is greater then 4.
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    Why are we doing integration constants? It's a definite integral, is it not? Displacement is the area under the velocity graph.
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    (Original post by Zacken)
    Why are we doing integration constants? It's a definite integral, is it not? Displacement is the area under the velocity graph.
    This was my solution after looking at the mark scheme. The bit I don't get is why they substituted t = 4 and x = 32 to work out the constant. Surely you can't as t=4 is not in the domain of the function.
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    (Original post by Glavien)
    This was my solution after looking at the mark scheme. The bit I don't get is why they substituted t = 4 and x = 32 to work out the constant. Surely you can't as t=4 is not in the domain of the function.
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    I understand your concerns but at A-Level this is pretty much expected without justification. What you're really doing is saying \lim_{t \to 4} x = 32 so \lim_{t \to 4} (16t - t^2 + c) = 32 \iff c=-16.

    For what it's worth, I'd have used areas right away.

    i.e: \int_4^{8} 16-2t \, \mathrm{d}t - \int_8^{10} 16-2t \, \mathrm{d}t instead of faffing about with constants.
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    (Original post by Zacken)
    I understand your concerns but at A-Level this is pretty much expected without justification. What you're really doing is saying \lim_{t \to 4} x = 32 so \lim_{t \to 4} (16t - t^2 + c) = 32 \iff c=-16.

    For what it's worth, I'd have used areas right away.

    i.e: \int_4^{8} 16-2t \, \mathrm{d}t - \int_8^{10} 16-2t \, \mathrm{d}t instead of faffing about with constants.
    Thanks for the help, I prefer your method of definite integrals.
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    (Original post by aymanzayedmannan)
    Draw a sketch of both graphs on the same coordinate axes. It should make sense from there on. I choose to use triangles to find the distance for the second half of the graph because it's simpler than integrating it twice.
    Thanks, I'll try that.
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    (Original post by Glavien)
    Thanks for the help, I prefer your method of definite integrals.
    No problem!
 
 
 
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