Vector Question Analysis: Did My Lecturer Make a Mistake?

[KiNGGeexD]

I was given an example in a lecture of a vector question which is typical of this module.

Question

The position of mass m is described as a vector r, from an origin such that r= 7t i + (4t-3t^2) j metres.

Find the magnitude and directions of r and dr/dt when dy/dt=0Now I have worked through this and know what to do but when I looked at the model answer I noticed what to me looks like a mistake?

I've attached the solution done by my lecturer but he deduces when 4t-3t^2 = 0

That t=2/3 which it can't be it has to be 4/3?I'm not sure if I am mistaken that's why I'm asking and if I am wrong I would like to know why:) thanks guys:)

 

[Chestermiller]

I was given an example in a lecture of a vector question which is typical of this module.

Question

The position of mass m is described as a vector r, from an origin such that r= 7t i + (4t-3t^2) j metres.

Find the magnitude and directions of r and dr/dt when dy/dt=0


Now I have worked through this and know what to do but when I looked at the model answer I noticed what to me looks like a mistake?

I've attached the solution done by my lecturer but he deduces when 4t-3t^2 = 0

That t=2/3 which it can't be it has to be 4/3?


I'm not sure if I am mistaken that's why I'm asking and if I am wrong I would like to know why:) thanks guys:)View attachment 65683

I don't know whether these notes are in your own handwriting or in your lecturer's handwriting, but whoever wrote them got the equation for dy/dt wrong. The equation for dy/dt is

dy/dt =4-6t

This component of velocity is zero when t = 2/3.

 

[KiNGGeexD]

It was the lecturers handwriting

 

[KiNGGeexD]

So then the second part where he solves for the magnitude of r must be wrong as the t is added back in?

 

[collinsmark]

So then the second part where he solves for the magnitude of r must be wrong as the t is added back in?

No, that part is okay. Everything looks okay to me in the attachment except for one thing.

The attachment has in it:

$$\vec r = 7 t \ \hat \imath + (4t - 3t^2) \ \hat \jmath$$
$$x = 7 t$$
$$y = 4t - 3 t^2$$
$$\frac{dy}{dt} = 4 t - 3 t^2 = 0$$
That last line has the problem. I think your instructor meant to write:

$$\frac{dy}{dt} = \frac{d}{d t} \{ 4 t - 3 t^2 \} = 0$$
but neglected to write in the d/dt operator.

Everything looks okay after that though.

If you take the derivative of y with respect to t and set it equal to zero, you'll find that t = 2/3, which is shown on the attachment correctly.

 

[AR1399]

Yeah, as collinsmark said, it might just be a prob of not writing d/dx as a step,
The ans is still supposed to be 2/3 seconds.

 

[Chestermiller]

The last previous post on this thread was over a year and a half ago. I'm closing this thread.

Chet