doubt in past paper question

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How do I do a) iii)? thanks in advance.

[kw0573]

5.6 happens to be 7 * 0.8. We know a drop is made every 0.8s. The the rightmost dot (8th dot) would be the 7th dot after t = 0 which is t = 5.6, when a car is 15 cm or 15*4 = 60m away from the position at  t = 0. 

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But the mark scheme says this :/ IDK why it says 37.6 m.. 

And why the 8th dot instead of the 7th dot?

 

[kw0573]

This question is from May 2008, HL Physics paper 2, Timezone 2. I tried to look for the subject report for that session but I couldn't find it. The subject report usually has explanations to questions that are likely to trip up people. 

[kw0573]

I think the question made a mistake and the answer should be 38.4m.

The dots for part a (ii) should be at 0cm, 0.2 cm and 0.8 cm, which allows constant acceleration as implied by the question. That means the spacing between the drops go like 1, 3, 5 ... 11 squares, when then acceleration stops and it becomes even spacings of 12 squares. 
Then at 1st dot, it's at t = 0, s= 0cm ==> 0m. At 2nd dot, it's at t = 0.8, s = 0.2cm ==> 0.8m. Hence we should go to the 8th dot, which correspond to t = 7*(0.8) = 5.6s, which corresponds to s = 9.6cm ==> 38.4m. So I think IB made a mistake. Usually because IB publishes markschemes to teachers, they can appeal the answer key and will likely change the accepted result to 38.4m or throw away a (iii) in general. 

Side note, because of the series 1 + 3 + 5 + 7 + ... (of n such terms) = n², that's why in relation to the equation s = 0.5 at² the distance increases at 1, 3, 5 progression. 

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On 10/23/2016 at 10:05 AM, kw0573 said:

I think the question made a mistake and the answer should be 38.4m.

The dots for part a (ii) should be at 0cm, 0.2 cm and 0.8 cm, which allows constant acceleration as implied by the question. That means the spacing between the drops go like 1, 3, 5 ... 11 squares, when then acceleration stops and it becomes even spacings of 12 squares. 
Then at 1st dot, it's at t = 0, s= 0cm ==> 0m. At 2nd dot, it's at t = 0.8, s = 0.2cm ==> 0.8m. Hence we should go to the 8th dot, which correspond to t = 7*(0.8) = 5.6s, which corresponds to s = 9.6cm ==> 38.4m. So I think IB made a mistake. Usually because IB publishes markschemes to teachers, they can appeal the answer key and will likely change the accepted result to 38.4m or throw away a (iii) in general. 

Side note, because of the series 1 + 3 + 5 + 7 + ... (of n such terms) = n², that's why in relation to the equation s = 0.5 at² the distance increases at 1, 3, 5 progression. 

thank you! I thought I was really stupid for a second - lol.