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Physics A level dynamics question

I am confused about the mark scheme answer for part d(ii) in the following question:

Screenshot 2021-04-08 at 7.26.08 pm.png
Screenshot 2021-04-08 at 7.25.56 pm.png
This is the mark scheme answer:

Screenshot 2021-04-08 at 4.44.11 pm.png

Instead of using v = at as the mark scheme does, I used v = sqrt(2*a*s), where s = 1.0m and a = 9.8/s. But my method gives 4.4m/s and the mark scheme says 4.1m/s. Why is my method incorrect here?
(edited 3 years ago)
Screenshot 2021-04-08 at 4.42.18 pm.png24.3KB
Screenshot 2021-04-08 at 4.42.33 pm.png4.8KB
Screenshot 2021-04-08 at 4.42.33 pm.png4.8KB
Original post by Orlqndo
I am confused about the mark scheme answer for this following question:

Screenshot 2021-04-08 at 4.42.18 pm.png
Screenshot 2021-04-08 at 4.42.33 pm.png

This is the mark scheme answer:

Screenshot 2021-04-08 at 4.44.11 pm.png

Instead of using v = at as the mark scheme does, I used v = sqrt(2*a*s), where s = 1.0m and a = 9.8/s. But my method gives 4.4m/s and the mark scheme says 4.1m/s. Why is my method incorrect here?

Maybe this is me being stupid, but the mark scheme you've screenshotted shows the answer to Part 2, but the question is for Part 3
Reply 2
Avatar for 0le
0le
21
Original post by Kerchoo95
Maybe this is me being stupid, but the mark scheme you've screenshotted shows the answer to Part 2, but the question is for Part 3

Yup, part 1 and 2 are missing which may be needed to answer the OP's question.
Reply 3
Avatar for Orlqndo
Orlqndo
OP
7
Original post by 0le
Yup, part 1 and 2 are missing which may be needed to answer the OP's question.

Sorry my bad - I've edited the original post so it has the entire question. My question is about part d(ii).
Original post by Orlqndo
Sorry my bad - I've edited the original post so it has the entire question. My question is about part d(ii).

Be careful reading the question.

Part d states 'after leaving the ramp at B the skateboarder lands at C 0.42 seconds later.

The question wants you to use the time value stated and hence the appropriate suvat is v = u + at
Original post by Orlqndo
Sorry my bad - I've edited the original post so it has the entire question. My question is about part d(ii).


The information given in the question is not consistent with suvat.

To descend 1m with zero initial velocity would take 0.452s (3sig.fig.) using g=9.8m/s/s - not 0.42s. And that's why you get a different answer if you use v2=u2+2asv^2=u^2+2as

As @uberteknik said, they're expecting you to use v=u+atv=u+at

There's nothing wrong with your working, IMO - it's just a very poorly constructed question.
(edited 3 years ago)

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