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Mathematics 15 Online
OpenStudy (anonymous):

A particle starts from rest with uniform acceleration a. Its velocity after n seconds is v. The displacement of the particle in the last two seconds is

OpenStudy (nincompoop):

so what is the question?

OpenStudy (anonymous):

I forgot to add "is" _-_

OpenStudy (nincompoop):

can you phrase it in a question? I mean even 3-year-olds can ask proper questions

OpenStudy (kainui):

What, this is asked appropriately. Clean your eyes nincompoop

OpenStudy (nincompoop):

haha he forgot the "is"

OpenStudy (anonymous):

wear some eye lenses

OpenStudy (nincompoop):

eyes have lens

OpenStudy (anonymous):

artificial

OpenStudy (nincompoop):

so do you know what is being asked?

OpenStudy (kainui):

Ok no.name where are you stuck? In these kinds of equations we are generally looking for what is known and we can find that by simply listing what is known. So list your known variables and equations. If you have two variables and two unknowns, you're probably on the right track.

OpenStudy (kainui):

Keep in mind, displacement is technically two unknowns combined into one slightly vague unknown.

OpenStudy (tylerd):

average velocity in the last 2 seconds times 2 seconds?

OpenStudy (anonymous):

that last two seconds is bothering me somehow

OpenStudy (nincompoop):

why?

OpenStudy (perl):

We have x(t) = 1/2*a*t^2 + vi*t

OpenStudy (perl):

dx/dt = at + vi , when t = 2 we have a(2) + vi = v ,

OpenStudy (nincompoop):

if you can solve it without it being 2 seconds, then you can solve with it

Parth (parthkohli):

\[s_{n-2}=\dfrac{1}{2}a (n-2)^2 = \dfrac{1}{2}a(n^2 - 4n + 4)\]\[s_{n}=\dfrac{1}{2}an^2\]\[s_n - s_{n-2} = \dfrac{1}{2}a(4n - 4) = 2a(n-1)\]\[v = an \Rightarrow 2an - 2a =\boxed{ 2(v - a)}\]

OpenStudy (aum):

If the answer is 2(v - a) then I can explain.

OpenStudy (aum):

oh cool!

OpenStudy (anonymous):

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