2013 CSSA Mathematics Thoughts? (1 Viewer)

[Kostie100]

For the one in the multiple choice: 17/38
Other one: 190/351 ... I hope I got that in the exam, if not, so screwed.

Got both those as well I think

 

[Carrotsticks]

Distance traveled...

Nope, area under the velocity curve is DISPLACEMENT.

Oh Fuck it actually could be because the area would always be a positive number therefor distance not displacement. If it had the integral symbol it would be definitely displacement hmmm

It is still displacement.

However, in the case where the velocity curve is always positive, the displacement is the SAME thing as distance.

 

[SharkeyBoy]

what did u guys get for the last question of mc? a?

Yep, i got a, the inequalities yeah?

 

[fatima96]

It was fine at the start, but by the end it became very difficult.
Hoping for at least 80-82% (band 6 aligned)

Did anyone find it was harder than past papers? I think 8 marks allocated to the last question (c) was too much

I definitely found it harder than other papers and LONGER. I normally finish in 1.5 hrs however this exam took the full 3.

 

[fatima96]

Guys I actually think that the multiple choice answer was distance. The question said circle the Best answer and the option was the displacement AFTER 2 seconds or the distance in the first 2 seconds. Obviously the best answer is the distance, all you need is an Absolute value of it to get the distance. Idk. It said after 2 seconds so I was hesitant.

 

[bedpotato]

No, I think it's still displacement (Idk why I put distance in the exam). The value of the shaded region would've been how much the object has been displaced during the first two seconds. So, if it was let's say 50 (well, -50), the object has been displaced 50 units during the first two seconds, so AFTER two seconds, the object's displacement is 50 units from the origin.

 

[mac1996]

No, I think it's still displacement (Idk why I put distance in the exam). The value of the shaded region would've been how much the object has been displaced during the first two seconds. So, if it was let's say 50 (well, -50), the object has been displaced 50 units during the first two seconds, so AFTER two seconds, the object's displacement is 50 units from the origin.

its distance travelled, 100% sure

 

[wagig]

its distance travelled, 100% sure

x = displacement
dx/dt = velocity

integrating velocity (value of shaded region) --> displacement

 

[ebbygoo]

Further then that, it said the shaded VALUE. A value is a NUMBER. The value was negative. You can't have a negative distance. And the 'after 2 seconds' thing, makes sense for displacement, as it is a measure from the origin. Therefore, after 2 seconds, how far is it from the origin.

 

[mac1996]

no, u can have a negative distance
negative distance is the distance when a particle is returning to the origin.

 

[nerdasdasd]

You can have a negative distance but not a negative displacement .... A negative distance means you are on the left of the origin .

 

[bedpotato]

You can have a negative distance but not a negative displacement .... A negative distance means you are on the left of the origin .

Distance can't be negative. But you can have a negative displacement. If the displacement was -5m, it means the object is 5m to the left of the origin...

 

[iBibah]

You can have a negative distance but not a negative displacement .... A negative distance means you are on the left of the origin .

You have them the wrong way around. Displacement can be negative.

 

[Siddy123]

distance is a scalar quantity( just a value)
displacement is a vector quantity ( can be negative- just means direction)

 

[RealiseNothing]

Not its y (greater than or equal to sign) 1/2

This, but should probably put in

 

[Carrotsticks]

no, u can have a negative distance
negative distance is the distance when a particle is returning to the origin.

Not quite.

Refer to bedpotato's post.

 

[mRym]

Further then that, it said the shaded VALUE. A value is a NUMBER. The value was negative. You can't have a negative distance. And the 'after 2 seconds' thing, makes sense for displacement, as it is a measure from the origin. Therefore, after 2 seconds, how far is it from the origin.

what if it didnt start at the origin?

 

[wagig]

what if it didnt start at the origin?

It doesn't matter whether it started at the origin or not, the displacement represented by that shaded value still occurred over that time period since the velocity was constantly negative.

 

[mRym]

It doesn't matter whether it started at the origin or not, the displacement represented by that shaded value still occurred over that time period since the velocity was constantly negative.

but if it didnt start at the origin, the displacement from the origin after 2 seconds is different to the distance travelled

 

[wagig]

but if it didnt start at the origin, the displacement from the origin after 2 seconds is different to the distance travelled

I see what you mean, but in this particular question the answer relating to displacement was "the displacement of the particle after 2 seconds" (or something similar), which made no reference to the displacement from the origin