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1. 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.
2. (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.
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.
3. (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.
Why are we doing integration constants? It's a definite integral, is it not? Displacement is the area under the velocity graph.
4. (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.
5. (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.
I understand your concerns but at A-Level this is pretty much expected without justification. What you're really doing is saying so .

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

6. (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 so .

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

Thanks for the help, I prefer your method of definite integrals.
7. (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.
8. (Original post by Glavien)
Thanks for the help, I prefer your method of definite integrals.
No problem!

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