VCE Specialist 3/4 Question Thread!

[09Ti08]

Question 8:
Draw your velocity-time graph, it should look like a trapezium.
Now 72km/h=20m/s
The phrase "shortest time" gives you a hint that you need to use the "maximum speed".
Now choose the positive direction as the direction in which the bus moves.
Starting from 0, the time it takes to reach the maximum velocity is: (20-0)/2=10s (1) during which the bus covers a distance of 1/2*10*20=100m
Now it also needs to decelerate, the time taken for deceleration is the same: 10s (2) during which it covers a distance of 100m again
In between, it moves with a constant velocity of 20s, and it needs to go for only 800-100-100=600m. The time taken for this period is 600/20=30s (3)
Now the total time is (1)+(2)+(3)=50s

[Kanye East]

Find the work done by the force F=(-2,-1,3) in moving a particle from the point P to the point Q having position vectors (-1,2,3) and (1,-3,4) respectively

[09Ti08]

if distances are measured in meters and magnitude of force is measured in newtons.

[Aelru]

I'm having problems finding part b) the distance in the 90km/hr phase.
I have obtained 10 seconds as answer for the acceleration phase, which is correct,

But I've obtained... mixed answers for the 2nd.
Textbook's answer is 150m. I got 250m.

[RKTR]

(Image removed from quote.)

I'm having problems finding part b) the distance in the 90km/hr phase.
I have obtained 10 seconds as answer for the acceleration phase, which is correct,

But I've obtained... mixed answers for the 2nd.
Textbook's answer is 150m. I got 250m.

erm i get 150m ..  from 10s-16s it travels at that speed .. 25m/s x 6 s = 150m

[BubbleWrapMan]

Find the work done by the force F=(-2,-1,3) in moving a particle from the point P to the point Q having position vectors (-1,2,3) and (1,-3,4) respectively

For the record, work isn't an examinable concept

[BubbleWrapMan]

Essential Ex 10E Q6

I get v^2 = -2/(2+x) + 1 but the answer says v^2 = x/(2+x)

My cousin told me not to upload anything to ATARnotes that's copyrighted so I can't post question sorry

Thanks in advance

They're both the same

[Deceitful Wings]

Hi guys, could someone answer the question (i have attached to this post).
I am not sure if this is in the spesh course, but if any uni students (or vce maths genius') know how to do it, please help! I have been stuck on this question for hours :\

[brightsky]

a) i shall assume that the distance between the two non-parallel lines is not zero (i.e. that the two lines are skew).

the direction vector of the first line is d1 = (a1, b1, c1) and the direction vector of the second line is d1 = (a2, b2, c2). A (x1, y1, z1) is some random point on the first line, and B (x2, y2, z2) is some random point on the second line. now, to find the shortest distance between two skew lines, all we need to do is find the absolute value of the scalar resolute (yay spesh) of vector BA in the direction of d1 x d2. why? picture to yourself two skew lines. picture also the plane on which each skew line lies. now, if you bring the two 'parallel' planes together, then the two direction vectors will lie on the 'same' plane. the cross product of the two direction vectors (d1 x d2) will, therefore, always be in the direction perpendicular to this new plane we created by merging the two planes together (and hence in the direction perpendicular to both original planes). (notice how d1 x d2 = 0 if the two lines are parallel, since d1 would simply be a scalar multiple of the d2.) this is convenient because all we need to do now is take a point on each line, join them up to form a vector, and find the magnitude of the projection of this newly formed vector on the 'normal' vector (d1 x d2) to find the shortest distance between the two lines.

as for the working:

D = (BA.(d1xd2))/abs(d1xd2) = (x1-x2, y1-y2,z1-z2).(d1xd2)/abs(d1xd2), as req. (this formula won't work if the lines are parallel because then you would be dividing by zero.)

b) the area of the parallelogram with edge vectors (a1, b1, c1) and (x2-x1, y2-y1, z2-z1) is given by norm((a1, b1, c1) x (x2-x1, y2-y1, z2-z1)). now we know that norm((a1, b1, c1)) is sqrt(a1^2 + b1^2 + c1^2).

now, realise that the height of the parallelogram corresponds to the shortest distance between the two parallel lines.

h*sqrt(a1^2 + b1^2 + c1^2) = norm((a1, b1, c1) x (x2-x1, y2-y1, z2-z1))
h = norm((a1, b1, c1) x (x2-x1, y2-y1, z2-z1))/sqrt(a1^2 + b1^2 + c1^2)
therefore, D = norm((a1, b1, c1) x (x2-x1, y2-y1, z2-z1))/sqrt(a1^2 + b1^2 + c1^2)

hopefully this is right...haha.

[Homer]

how would you integrate something like this by hand

[brightsky]

dy/dx=x/y
y*dy/dx = x
int y dy/dx dx = int x dx
int y dy = int x dx
y^2/2 + c1 = x^2/2 + c2
etc

[lzxnl]

Or as a lazy physicist

dy/dx = x/y
y dy = x dx
drawing an integral sign on both sides
1/2*y^2 = 1/2 x^2 + constant
y^2 - x^2 = constant

[barydos]

Argh, I'm always having trouble with these inflow/outflow differential equation setup questions.
Here's the question I'm finding difficult to do haha:

Spoiler

Thanks

[Homer]

Im not too sure myself but wanna give it a go so,

[Homer]

Also another question:

A particle travelling in a straight line has velocity v m/s at time t s. Its acceleration is given by . Its velocity is 50m/s initially and is reduced to 3 m/s. What is the expression for the time taken in seconds for this to occur.

Answer

i understand everything except the limits of the intergration. why did they use them (50 and 3) to get the time?