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m1 help :(

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Original post by kiiten
So just use values of t like 1, 2 etc.

Yes, it says velocity when the stone hits the ground. If you use suvat why would s=-2 dont you have to account for the distance from when its thrown up?

If you want you can just consider it from when its first thrown, and therefore s=-2
so uyou're assuming that the displacement when the stone is thrown is 0m below that 0m where it was thrown is a displacement of -2m and above 0m is a distance of 3m which the stone travelled before reaching it's maximum height?
Reply 22
Original post by Zacken
s is displacement not distance. If you got up x amounts and go down x+2 amounts then your displacement is -2.


Ahh of course, i forget that s is displacement and not distance

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Reply 23
Thanks for all your help everyone :smile:

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Reply 24
Original post by kiiten
Ahh of course, i forget that s is displacement and not distance

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Original post by kiiten
Thanks for all your help everyone :smile:

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De nada.
Reply 25
image.jpgI plugged in values to plot the graph and got

5i+10.5j
10i+5j
15i-10.5j
20i-36j

Etc. but its different to the graph in the book?
(edited 8 years ago)
Reply 26
Original post by kiiten
image.jpgI plugged in values to plot the graph and got

5i+10.5j
10i+5j
15i-10.5j
20i-36j

Etc. but its different to the graph in the book?


In this case - you're plotting x v/s y not (x+y) v/s t.
i got tagged here? ????
Reply 28
Original post by Zacken
In this case - you're plotting x v/s y not (x+y) v/s t.


Thats how i plotted it with i on x axis and j on y axis but my graph is different
Reply 29
Original post by thefatone
i got tagged here? ????


Sorry i tagged then untagged you - i forgot that you havent done i and j vectors yet
oh yea :smile: that's fine.
Reply 31
Original post by kiiten
Thats how i plotted it with i on x axis and j on y axis but my graph is different


Can you take a picture and show us your working out for each point?
Reply 32
Original post by Zacken
Can you take a picture and show us your working out for each point?


Oh I used a calc. I just subbed t= 1, 2, 3, 4 and 5 into
R = 5ti + (6+9.5t - 5t^2) j
Reply 33
Original post by kiiten
Oh I used a calc. I just subbed t= 1, 2, 3, 4 and 5 into
R = 5ti + (6+9.5t - 5t^2) j


That should be correct, can you see that if you plug in t =0, you get the point 0i + 6j which is (0,6) on the graph? As shown in the picture, etc...?
Reply 34
Original post by Zacken
That should be correct, can you see that if you plug in t =0, you get the point 0i + 6j which is (0,6) on the graph? As shown in the picture, etc...?


Yeah, but i dont understand why they used random values for i like 9.5, 10 and 12 whereas i got 5, 10, 15 etc.
Reply 35
Original post by kiiten
Yeah, but i dont understand why they used random values for i like 9.5, 10 and 12 whereas i got 5, 10, 15 etc.


They used 5, 10 as well. It's fine.
Reply 36
Original post by Zacken
They used 5, 10 as well. It's fine.


Right, thanks :smile:

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