Question

a small blok slides, without friction, down an inclined plane starting from rest. let sn be the distance travelled from t=(n-1) to t=(n).
then (sn)/(sn+1)???

Answer 1

@ParthKohli

Answer 2

Let a be the constant acceleration of the motion.
Let the block be released at t=0.
Speed after (n-1) seconds is given by
v=a(n-1)

sn=vt+0.5at^2
t=1second (since sn is distance travelled from t=n-1 to t=n)
v=a(n-1) , the initial speed at t=n-1 seconds
So, sn=a(n-1) +a/2 =an-a/2 .....................................(1)

Speed after time n seconds will be ,v= an
So sn+1 = an +a/2 .....................................(2)
[you may get it from the same procedure as above]

sn/sn+1 = (n-0.5) / (n+0.5)

Answer 3

so we can say it is equal to (2n-1)/(2n+1).... jeee thnks a bunch

Answer 4

Remember n has to be greater than or equal to 1 for this result to hold good. If suppose you take n=0.1sec. Then s1 is distance travelled from t=-0.9 sec to t=0.1 sec. But motion was not present before t=0 and we get meaningless result.

Answer 5

ok... thnq.... btw.... u a lecturer?? :)

Answer 6

Nope, a student!

Answer 7

lol.... which grade?

Answer 8

currently enjoying holidays!

Answer 9

I am in first year of graduation.

Answer 10

what about u?

Answer 11

cool....thnx a bunch dude.... me in +1... nd ya i hav anoder question... me taggin u on it.... plz do help me wid it

Answer 12

no problem. Always glad to help if i can