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Physics Unit 2 Question, help needed

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 :/
physics unit 2 max spe.png
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 :/
physics unit 2 max spe.png


A hint - look at the y and x axis, and think about details of the graph you can analyse.

Further hint:

Spoiler

Reply 2
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 :smile:
Reply 3
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 :/
physics unit 2 max spe.png


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. :biggrin:
Reply 4
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. :biggrin:


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
Reply 5
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
Reply 6
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 :/
Reply 7
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.
Reply 8
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 :/
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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