M1 23/1/2013 Thread.

anyone want to explain in jan 2012 why on question 5c the time is taken away from one another :l I just just left the two possible times i found :P

I remember doing the question and I believe it asked for the time it was above that height. So the two times you found represent it passing that point on both the way up and way down. Therefore in between them two times is the total time it was above that height. So subtract :smile:

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Reply 21

MedMed12

21

Original post by Mouth

I remember doing the question and I believe it asked for the time it was above that height. So the two times you found represent it passing that point on both the way up and way down. Therefore in between them two times is the total time it was above that height. So subtract :smile:

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thanks

i dont think im that sharp to pick that up ahah :P I guess I can hope

Reply 22

Joshmeid

5

Original post by MedMed12

anyone want to explain in jan 2012 why on question 5c the time is taken away from one another :l I just just left the two possible times i found :P

The question asks for the length of time the stone is above 6.6m. I'm assuming you have values

\frac{22}{7}

and

\frac{3}{7}

. Those are the times for which the stone is at 6.6m.

When the stone is projected vertically upwards, it travels to its greatest height and then falls down to the point A. So it must cross the point 6.6m above A twice, when it is travelling upwards, and then again when it is travelling back down to A.

The first time it crosses is when

t = \frac{3}{7}

, the second time it crosses,

t = \frac{22}{7}

, so the time it's above that point is

\frac{22}{7} - \frac{3}{7}

.

(edited 11 years ago)

Reply 23

MedMed12

21

Original post by Joshmeid

The question asks for the length of time the stone is above 6.6m. I'm assuming you have values

\frac{22}{7}

and

\frac{3}{7}

. Those are the times for which the stone is at 6.6m.

When the stone is projected vertically upwards, it travels to its greatest height and then falls down to the point A. So it must cross the point 6.6m above A twice, when it is travelling upwards, and then again when it is travelling back down to A.

The first time it crosses is when

t = \frac{3}{7}

, the second time it crosses,

t = \frac{22}{7}

, so the time it's above that point is

\frac{22}{7} - \frac{3}{7}

.

thank you very much! run out of reps :smile:

but that makes sense now!

Reply 24

Dorkins

12

Y'all ready for this? My college decided to do M1 in the winter and C3+C4 in the summer for some reason. :redface:

Reply 25

jessjess

Original post by Dorkins

Y'all ready for this? My college decided to do M1 in the winter and C3+C4 in the summer for some reason. :redface:

Snap! Not joking though, I don't understand most stuff to do with resolving.... I suck at mechanics. Here's to a resit in the summer!

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