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# Physics Unit 2 Question, help needed Watch

1. Sorry if this is a bit of a 'noob' question, but how will I go about calculating this? I can't find any instructions in my revision guide, which was made by AQA themselves :/
2. (Original post by Gabzinc)
Sorry if this is a bit of a 'noob' question, but how will I go about calculating this? I can't find any instructions in my revision guide, which was made by AQA themselves :/
A hint - look at the y and x axis, and think about details of the graph you can analyse.

Further hint:
Spoiler:
Show

Thnk about the gradient and what it represents.
3. I knew that gradient represents speed but I just couldn't tell how to calculate maximum speed, as opposed to average speed and speed over a period of time.
Thanks for the help in the end
4. (Original post by Gabzinc)
Sorry if this is a bit of a 'noob' question, but how will I go about calculating this? I can't find any instructions in my revision guide, which was made by AQA themselves :/
We did this question in class and it's one of those sneaky questions. 0-6 seconds it's accelerating and 6 seconds onward it's constant speed hence the straight line. Hope that will make it easier to understand the Q and work it out.
5. (Original post by JTran38)
We did this question in class and it's one of those sneaky questions. 0-6 seconds it's accelerating and 6 seconds onward it's constant speed hence the straight line. Hope that will make it easier to understand the Q and work it out.
Worked it out, thanks!! The only thing is that its hard to see where it becomes a straight line, so there should be a range of answers really. It's these type of questions that are 100x harder in exam conditions when you cant think straight :/

no pun intended lol
6. (Original post by Gabzinc)
Worked it out, thanks!! The only thing is that its hard to see where it becomes a straight line, so there should be a range of answers really. It's these type of questions that are 100x harder in exam conditions when you cant think straight :/

no pun intended lol
I don't think you have to know where does it become straight. You can just choose a random interval where the line is straight and calculate dy/dx
7. (Original post by InneRs)
I don't think you have to know where does it become straight. You can just choose a random interval where the line is straight and calculate dy/dx
doing that gets me a really close answer, but not close enough according to the mark scheme :/
8. (Original post by Gabzinc)
doing that gets me a really close answer, but not close enough according to the mark scheme :/
I think you can get the exact answer, for example you choose the interval of 8 and 10 and you get (170-125)/(10-8)=22.5, which should be the answer. If the mark scheme requires you to choose the interval of 6 and 10, they are too harsh.
9. (Original post by InneRs)
I think you can get the exact answer, for example you choose the interval of 8 and 10 and you get (170-125)/(10-8)=22.5, which should be the answer. If the mark scheme requires you to choose the interval of 6 and 10, they are too harsh.
Yeah, 8 and 10 works. But I've been doing 7.8 and 10 all along, which should still give me 22.5, but it doesn't (170-120)/(10-7.8) = 22.72727272...

Unless I'm doing something wrong :/
10. (Original post by Gabzinc)
Yeah, 8 and 10 works. But I've been doing 7.8 and 10 all along, which should still give me 22.5, but it doesn't (170-120)/(10-7.8) = 22.72727272...

Unless I'm doing something wrong :/
Try and go for a clean triangle to get an answer that is fully accurate so you don't have to estimate values.

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