Q13 (1 Viewer)

[mike12345678]

but why is that?
its asking for the blood concentration when it reaches 0.3

so wouldnt you be trying to get an answer closer to .3 and you get closer then the original approximation??

 

[_pikachu]

you have to minus 0.3 in the equation

you guys are assuming that conc. reaches 0mg/L
however, they specify 0.3mg/L

 

[Timske]

you have to minus 0.3 in the equation

This, i had 26 at first but then i saw the 0.3 i was like aw shiet

 

[timeslowsdown]

i got 26hours 40mins for 13dii. anyone ?

I got this too but i guess it's wrong

 

[kingkong123]

Q13 Cii (0 Displacement question)

Yo,

Did you guys sub in x=0 or x=5 to calculate the time when the particle first has 0 displacement?
Having been asked to convert the particle's acceleration in a form that reveals its centre of motion in part (i), i used the definition that DISPLACEMENT is the particle's distance from its CENTRE OF MOTION. So basically for part (ii) i solved when x=5 (so that it's displacement from centre of motion [ie x=5] is 0).

What did you guys do?

 

[iampeterr]

Re: Q13 Cii (0 Displacement question)

the equation was x = 5 + sometrig

i changed the sometrig part into acos(something) so my equation was x = 5 + 10cos(something) then solved for t

 

[Sindivyn]

Re: Q13 Cii (0 Displacement question)

x is equal to displacement. So you sub in x = 0. If you were asked to find when the particle passes the point where it oscillates about, then it would be x=5.

 

[nucgaek]

Re: Q13 Cii (0 Displacement question)

Displacement is distance from origin, and the centre of motion is 5 units away from the origin, so 0 displacement=/=centre of motion

 

[mike12345678]

Re: Q13 Cii (0 Displacement question)

will i get full marks for this section? i got the right answer but i put it in degrees??

so i end up with 86 degrees 34 minutes or something like that because the whole way through i used degrees

 

[HeroicPandas]

Re: Q13 Cii (0 Displacement question)

When x=0, solving for t, u get 1.5...etc

 

[chriss95]

For geometry, last part. I just said how tangents from an external point X are equal (SX=CX), thus SX must be a tangent. And T is collinear with X, therefore ST is tangent. Still get that 1 mark yeh?

 

[mike12345678]

you guys are assuming that conc. reaches 0mg/L
however, they specify 0.3mg/L

ohhhhhhh that explains it. thankyou

 

[Timske]

For geometry, last part. I just said how tangents from an external point X are equal (SX=CX), thus SX must be a tangent. And T is collinear with X, therefore ST is tangent. Still get that 1 mark yeh?

No

 

[_pikachu]

For geometry, last part. I just said how tangents from an external point X are equal (SX=CX), thus SX must be a tangent. And T is collinear with X, therefore ST is tangent. Still get that 1 mark yeh?

I proved that angle MSX is 90 through equal angles of congruent triangles
then said that radii meets tangent at 90 and hence ST is a tangent

 

[_pikachu]

Usually for these questions, you have to use results you obtained from the previous question
In this case, you proved congruent triangles and hence should somehow manipulate it to answer the question

 

[chriss95]

i did. both those sides are equal (matching sides in congruent triangles). therefore st must be tangent, as tangents from an external point are equal

 

[_pikachu]

you should get the mark
im just showing an alternative approach to the question

 

[choiyokbao]

anyone remember how many marks the newtons method of approximation was worth? Q13dii) that part?

 

[get_back23]

fkkkkkkkkkkkkkkk for d ii) lost a fkn mark for not converting into hours

 

[_pikachu]

anyone remember how many marks the newtons method of approximation was worth? Q13dii) that part?

it was a 2 marker