# Thread: no idea what type of question this is

1. ## no idea what type of question this is

ok, im doing a course in Architectural Technology and we were required to do a Maths Module, however i failed it, and now have to do coursework resit.

i have most of the questions in the new piece of coursework, but there is 1 question that I cannot work out....neither can my Father, or my Brother's Girlfriend.

so if anyone can help please do...

the question is

"The distance, x, in meters, of a particle from an origin 0 at time t seconds is given by,

x= 1/3t (cubed) - 4t (squared) + 15t - 2

at what time(s) is the particle at rest?

(a) t= -3 seconds and -5 seconds
(b) t= -2 seconds
(c) t= 1 seconds and t= 2 seconds
(d) t= 2 seconds
(e) t= 3 seconds and 5 seconds"

plus if there is any chance you can show me how it is worked out would also help me to understand it for the next time i come across this sort of question.

sorry if this is too basic a question to be placed here in the university section...but i didnt know where else to put it

2. Originally Posted by DKoK
ok, im doing a course in Architectural Technology and we were required to do a Maths Module, however i failed it, and now have to do coursework resit.

i have most of the questions in the new piece of coursework, but there is 1 question that I cannot work out....neither can my Father, or my Brother's Girlfriend.

so if anyone can help please do...

the question is

"The distance, x, in meters, of a particle from an origin 0 at time t seconds is given by,

x= 1/3t (cubed) - 4t (squared) + 15t - 2

at what time(s) is the particle at rest?

(a) t= -3 seconds and -5 seconds
(b) t= -2 seconds
(c) t= 1 seconds and t= 2 seconds
(d) t= 2 seconds
(e) t= 3 seconds and 5 seconds"

plus if there is any chance you can show me how it is worked out would also help me to understand it for the next time i come across this sort of question.

sorry if this is too basic a question to be placed here in the university section...but i didnt know where else to put it
This is a calculus problem. You'll need to know what a derivative is in order to answer the question, after which it will become easy.

Motion graphs and derivatives - Wikipedia, the free encyclopedia

3. ok, i looked up the wiki page....but i still cant make head nor tail of it.....i even looked in the math textbook i was given for the year and this sort of question isint even in there, so far as i can make out anyway.

any other sort of help???

4. Originally Posted by DKoK
ok, im doing a course in Architectural Technology and we were required to do a Maths Module, however i failed it, and now have to do coursework resit.

i have most of the questions in the new piece of coursework, but there is 1 question that I cannot work out....neither can my Father, or my Brother's Girlfriend.

so if anyone can help please do...

the question is

"The distance, x, in meters, of a particle from an origin 0 at time t seconds is given by,

x= 1/3t (cubed) - 4t (squared) + 15t - 2

at what time(s) is the particle at rest?

(a) t= -3 seconds and -5 seconds
(b) t= -2 seconds
(c) t= 1 seconds and t= 2 seconds
(d) t= 2 seconds
(e) t= 3 seconds and 5 seconds"

plus if there is any chance you can show me how it is worked out would also help me to understand it for the next time i come across this sort of question.

sorry if this is too basic a question to be placed here in the university section...but i didnt know where else to put it
Plot this with t on the horizontal axis and x on the vertical axis, where the tangent is horizontal the particle is stationary.

CB

5. Hello, DKoK!

There is a question that I cannot work out ...
neither can my father or my brother's girlfriend.
. . None of you have taken Calculus I ?

The distance $x$, in meters, of a particle from an origin $O$ at time $t$ seconds

. . is given by: . $x \:=\:\tfrac{1}{3}t^3 - 4t^2 + 15t - 2$

At what time(s) is the particle at rest?

. . $(a)\;t= \text{-}3,\,\text{-}5 \quad (b)\;t= \text{-}2 \quad (c)\;t= 1,\,2 \quad (d)\;t= 2 \quad (e)\; t= 3,\,5$

The particle is at rest when its velocity is zero.

The velocity is given by the derivative: $\dfrac{dx}{dt}$

We have: . $\dfrac{dx}{dt} \;=\;t^2 - 8t + 15 \;=\;0$

Hence: . $(t - 3)(t - 5) \:=\:0 \quad\Rightarrow\quad t \:=\:3,\,5$

6. its not that we havent taken Calculus before, but its not like we are fanatical about Maths and remember every single equation we are taught, when it is taught, and be able to recall these when we need to.
i especially have no love for Mathematics...if i could i would tear my math textbook up, but im pretty sure i'll need it for next year

so I take a little offence to that None of you have taken Calculus I ?
and we managed to work it out in the end thanks to the help of 'undefined' pointing out that it was a Calculus question, and the way we worked it out was a lot clearer than what you put down

7. Originally Posted by DKoK
its not that we havent taken Calculus before, but its not like we are fanatical about Maths and remember every single equation we are taught, when it is taught, and be able to recall these when we need to.
i especially have no love for Mathematics...if i could i would tear my math textbook up, but im pretty sure i'll need it for next year

so I take a little offence to that None of you have taken Calculus I ?
and we managed to work it out in the end thanks to the help of 'undefined' pointing out that it was a Calculus question, and the way we worked it out was a lot clearer than what you put down

It was a perfectly legitimate question, since the assumption that none of you had was the cause of a number of people here wasting their time trying to help without explicitly mentioning calculus.

Also to answer this question does not require that you have memorised every equation that was ever mentioned in a calculus course, just that you have understood the principles behind the main points covered in the course. But then I suppose we are out of our time thinking that education is about education rather than just passing the exams on a course.

CB