Question

derivation for the distance travelled in nth second by calculus

Answer 1

more details?

Answer 2

its just a derivation

Answer 3

k d equation is Sn=u+a/2(2n-1)

Answer 4

ok gotcha..wait a min...

Answer 5

k :)

Answer 6

You have to find the distance traveled in n seconds and distance traveled in (n−1) seconds and then subtract them to find the dist. travelled in n seconds:

Therefore:
\[S _{n th} = S_{n} - S_{n-1}\]
We know that distance traveled in n seconds:

\[S = ut + \frac{ 1 }{ 2}at^2\]

Here for n seconds the distance covered will be:

(we know that here t = n)

Therefore:

\[S= un+\frac{ 1 }{ 2 }un^2\]

SO the EQ for the n-1 will be:

\[S_{n-1} = u(n-1) + \frac{ 1 }{ 2}a(n-1)^2\]

Therefore S_nth will be:

\[S_{nth} = S_{n} - S_{n-1}\]
\[=> S _{nth}= un + \frac{ 1 }{ 2 }un^2 - un + u -\frac{ 1 }{ 2 }an^2 -\frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }a(2n)\]

\[=>S_{nth}= u + an -\frac{ 1 }{ 2 }a = u + a(n -\frac{ 1 }{ 2 })\]

Therefore:
\[S_{nth} = u + a (n - \frac{ 1 }{ 2 })\]

@thirty

Answer 7

@thrity

Answer 8

i no dis but i want it using calculus

Answer 9

SO you want to find derivative of it

Answer 10

s

Answer 11

is the answer coming is u+a/2(2n-1)

Answer 12

sorry typor error:is the answer is

u(t +1)?

Answer 13

yup!!

Answer 14

no

Answer 15

u(t +1) isn't answer to that..?

Answer 16

at + u:

Answer 17

its derivation so thr is no ans

Answer 18

so what's yar book's answer

Answer 19

its just proving sum and they dont hav answers

Answer 20

So is my first answer apropriate by your perspective

Answer 21

s

Answer 22

and that's the right method too..

check this out if you don't believe:
(you confused me ..Lol):

http://openstudy.com/study#/updates/5026314ee4b07eea166b278d

Answer 23

@thrity my answer and working is correct

Answer 24

\[dx \div dt=v\] ...{integrating b/s} {taking limits as n & (n-1)}
\[\int\limits_{x _{0}}^{x} dx=\int\limits_{n-1}^{n}vdt\] { put v=u+at}
\[[x]_{x0}^{x}=\int\limits_{n-1}^{n}(u+at)dt\]
\[[x]_{x0}^{x}=u \int\limits\limits_{n-1}^{n} dt+a \int\limits\limits_{n-1}^{n}tdt\]
\[[x-x_{0}]=u[t]_{n-1}^{n}+a[t ^{2}/2]_{n-1}^{n}\]
\[s _{nth}=u[n-n+1]+a/2[n ^{2}-(n-1)^{2}]\]
\[s _{nth}=u+a/2[n ^{2}-n ^{2}-1+2n]\]
\[s _{nth}=u+a/2[2n-1]\]