jr question (1 Viewer)

[nsw..wollongong]

does anyone know why the answer is b i don't get it

 

[carrotsss]

One way to think about it: say for 2, to get it to 4 you have to square it, and then likewise for 4, you have to root it (ie power of 1/2) to get to 2. Basically, the power to get it back (or the power when they’re swapped) is the reciprocal of the original power, and hence log(a)(c)=1/log(c)(a) and the same follows for the rest of them

 

[Galactic Drama]

View attachment 40558
does anyone know why the answer is b i don't get it

 

[nsw..wollongong]

i get ittt tysm!!

also im kinda lost with part b of this question, why are they finding the absolute value of the integral?

 

[SB257426]

i get ittt tysm!!

also im kinda lost with part b of this question, why are they finding the absolute value of the integral? View attachment 40560

this is because the integral can yield a negative value which means it can refer to displacement when this is the case. But since we are looking for a distance we must take the absolute value instead.....

 

[carrotsss]

Distance is a scalar quantity, so even when ur moving backwards distance goes up

 

[Big_Kev]

could also think of it like velocity graph is under x axis for a part of that specific interval, so since ur finding "area" under the curve, u obv wanna absolute value the part under the x axis as opposed to finding the "value", where u wouldn't abs 🫣

 

[nsw..wollongong]

could also think of it like velocity graph is under x axis for a part of that specific interval, so since ur finding "area" under the curve, u obv wanna absolute value the part under the x axis as opposed to finding the "value", where u wouldn't abs 🫣

how do u know that the v graph is under the axis tho? like i didn't even think of it that way

 

[Big_Kev]

how do u know that the v graph is under the axis tho? like i didn't even think of it that way

haha coz when u integrate between the bounds & get a negative value it means it's under the axis

 

[nsw..wollongong]

haha coz when u integrate between the bounds & get a negative value it means it's under the axis

am i stupid or am i not seeing a negative value im so sorry where is it

 

[Big_Kev]

am i stupid or am i not seeing a negative value im so sorry where is it

dw it's not easy to see with just the equation - but if u take a look at the graph, you'll notice that there's a small part between 0 & 0.25 which is under the axis, and then the rest (between 0.25 and 0.5 like they ask in the Q) is above the axis so u just absolute that first bit between 0 & 0.25 and normally integrate between 0.25 & 0.5

 

[nsw..wollongong]

dw it's not easy to see with just the equation - but if u take a look at the graph, you'll notice that there's a small part between 0 & 0.25 which is under the axis, and then the rest (between 0.25 and 0.5 like they ask in the Q) is above the axis so u just absolute that first bit
View attachment 40562

OHHH i get it! in the exam tho how would i get that? since i only have the equation. for these types of questions should i always draw the graph to see if any part lies below the axis?

 

[carrotsss]

OHHH i get it! in the exam tho how would i get that? since i only have the equation. for these types of questions should i always draw the graph to see if any part lies below the axis?

pretty much, you don’t have to completely draw a graph but you can just simply sub in t=0 and it becomes clear that it’s less than 0

 

[nsw..wollongong]

i feel so stupid, it did that an got -1 but just moved on not thinking anything abt it lmaooo