Leah.J
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Here’s my work
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Hazelly
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(Original post by Leah.J)
Q36 a

Here’s my work
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Leah.J
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(Original post by Hazzabear)
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RogerOxon
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(Original post by Leah.J)
Q36 a
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Size:  146.6 KB
Here’s my work
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I've not looked at your working, but here's the approach:

Use s=ut+\frac{at^2}{2} for the person and the train - it's simpler for the train, as a=0. Equate the two to form a quadratic in time, and solve. You should have a solution of t=0, as they were level then.

If your solution for t is past the time that they would have attained 6 m/s, you need to do a more complicated calculation. In that case calculate the positions and speeds when the person attains their terminal speed, and redo from that starting point.

Fort the second part, use the train's equation for s and plugin your t value from part a. Finally, check it (no marks, but it's always good to verify that you're getting the previous marks) by using your equation for the person - the two should match.
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RogerOxon
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(Original post by Leah.J)
Q36 a
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Size:  146.6 KB
Here’s my work
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Size:  71.0 KB
I'm looking at this now. My first inclination would be to answer 0s and 0m ..

First, we need to see if the person catches the train before they attain their terminal speed (6m/s). My calculation shows that they haven't. They attain their terminal speed at t=\frac{6}{1.4}=\frac{30}{7}. At that time:

s_P=\frac{u+v}{2}t=\frac{90}{7}

s_T=\frac{u+v}{2}t=\frac{150}{7}

This agrees with your calculations.

The difference in distance travelled is therefore \frac{60}{7}m. Now calculate the time required for the person to catch the train - they're approaching at a relative speed of 1 ms^{-1}. Once you have that, you can add it to the acceleration time to get (a) - you look to have got that right (\frac{90}{7}s). For (b), take the time and multiply it by the train's speed - it's simpler than piecing together the distances for the person.
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