uniform acceleration

[markosheehan]

A particle travels srating with a initial speed u, with uniform acceleration a. Show that the distance travelled during the nth second is u+an-.5a.
I tried working this out by putting this information into the formula
S=ut+.5at^2 but it did not work

[Janus]

In the formula

 

S = ut+.5at^2

 

u is the velocity at the beginning of the time interval you are interested in, and t is the duration of the time interval you are interested in.

 

So, with the information given, What would be the new value for u at the start of the nth second, and what is the duration of the nth second?

[swansont]

 

A particle travels srating with a initial speed u, with uniform acceleration a. Show that the distance travelled during the nth second is u+an-.5a.

I tried working this out by putting this information into the formula

S=ut+.5at^2 but it did not work

 

 

 

 

It should. You need the speed at time t to go into the formula.

[J.C.MacSwell]

If you want to use u as described, and use t as n seconds, you need to subtract the distance traveled up to n-1 seconds.

[Country Boy]

Yes, S=ut+.5at^2 is the distance traveled in time t- from 0 to t seconds if u is in units of "distance per second" and a is in units of "distance per second per second". The distance traveled "in the nth second" is the distance traveled from 0 up to n seconds minus the distance traveled in n-1 seconds. That is,

(un+ .5an^2)- (u(n-1)+ .5a(n- 1)^2)= un+ .5an^2- (un- u+ .5an^2- an+ .5a)= u+ an- .5a since the "un" and ".5an^2" terms cancel.