VCE Specialist 3/4 Question Thread!

[zvezda]

Hey,
Just a q with exercise 12C q6a in essentials.
r(t) = 2ti + 16t^2(3-t)j where t is greater than or equal to 0. Find when the velocity and acceleration vectors are perpendicular. I got t=0,1 and 2. Except in the answers, they havent included t=1??

[lzxnl]

r'(t)= 2i + (96t-48t^2) j
r''(t) = (96-96t)j
When t = 1, the acceleration vector is zero. According to wiki zero vectors are parallel and perpendicular to all vectors, so perhaps it should be counted as a solution. You'd probably have to ask your teacher for what VCAA want.

[zvezda]

Yeah thats what i thought.
Cheers

[saba.ay]

Could someone please help with the questions attached?

Please and thanks.

[lzxnl]

For the first one, find an equation for the slope with y in terms of x. I think you have to assume the existence of a turning point somewhere. Then, you have dy/dt. Use dy/dx and dy/dt to find dx/dt

For the second one, draw up a triangle. You have the height, and the rate of change of the horizontal distance. You can then work out an expression for the rate of change of the angle from the triangle. tan angle = horizontal distance/constant height. Use that.

[saba.ay]

For the first one, find an equation for the slope with y in terms of x. I think you have to assume the existence of a turning point somewhere. Then, you have dy/dt. Use dy/dx and dy/dt to find dx/dt

Had not considered the presence of a turning point. I think that's where I'm going wrong.

For the second one, draw up a triangle. You have the height, and the rate of change of the horizontal distance. You can then work out an expression for the rate of change of the angle from the triangle. tan angle = horizontal distance/constant height. Use that.

I was doing tan angle = constant height/horizontal distance and so was finding another angle entirely. -.-

Thank you so much. This definitely helps

[Aelru]

Answer was: -0.64m/s^2

Annnnnd: (part b) would be enough from then on. Thanks!)

Answer was: 590.7m.

[Hancock]

For 11:

If we assume that the initial velocity is 0m/s - 



After 12 seconds:



In order to get back to the origin:








12:b -

If it reached the ground in 12 seconds, with an initial velocity of 10m/s:





At t = 12, y = 0 i.e. you've reached the ground. Therefore: y_0 is the height at which the stone is released initially.

Time to reach the top (after releasing it initially  - reached the apex when velocity = 0m/s):







Therefore maximum height reached is 5.1 + 585.6 =590.7m

[zvezda]

A particle travels in a straight line with velocity v at time t. If the velocity of the particle is given by v=2/sqrt(1-x^2), for 0<x<1, then find the acceleration of the particle in terms of x.
Really stumped on this one.
Help is much appreciated

[lzxnl]

a = dv/dt = dv/dx * dx/dt = v*dv/dx

So differentiate your velocity function with respect to x, then multiply by the original velocity function.

[Alwin]

A particle travels in a straight line with velocity v at time t. If the velocity of the particle is given by v=2/sqrt(1-x^2), for 0<x<1, then find the acceleration of the particle in terms of x.
Really stumped on this one.
Help is much appreciated

What we do, is use a different rule. I'll quickly prove it for you here:


You sub it in and solve for acceleration

EDIT: beaten

[Aelru]

For 11:

If we assume that the initial velocity is 0m/s - 



After 12 seconds:



In order to get back to the origin:








12:b -

If it reached the ground in 12 seconds, with an initial velocity of 10m/s:





At t = 12, y = 0 i.e. you've reached the ground. Therefore: y_0 is the height at which the stone is released initially.

Time to reach the top (after releasing it initially  - reached the apex when velocity = 0m/s):







Therefore maximum height reached is 5.1 + 585.6 =590.7m

Ah, could you explain how you obtained the acceleration?
That is, Q11A

[SocialRhubarb]

The particle first travels 48m at 4m/s.

Afterwards, the particle goes 48m backwards after being subject to a constant acceleration for 20 seconds.

Taking the original direction to be positive:

[saba.ay]

hey, could someone please explain how to do q. 19d) . Please and thanks

[SocialRhubarb]

The height of the water in the vase goes from 30cm to 0cm.



Not sure if correct, could probably use a check.