Question

a particle starts from rest with uniform acceleration a.its velocity after n seconds is v.the displacement of the body in last two seconds is:

Answer 1

A particle starts from rest so, initial velocity of particle , u = 0
particle acceleration with uniform acceleration a
So, particle must follow all kinematical equations .
I mean, S = ut + 1/2at²
At u = 0, S = 1/2at²
Here S is displacement during t time .

∴ displacement in last 2 sec = distance travelled in n sec- distance travelled in (n -2) sec
= $\bold{S_n-S_{n-1}}$ = 1/2 an² - 1/2a(n -2)²
= 1/2 a (n² - n² + 4n - 4)
= a (2n -2)
So, displacement in last 2 sec = a(2n - 2)

A/C to question,
Final velocity = v
Time = t
Acceleration = a
Then, use formula,
v = u + at
initial rest so, u = 0
v = a × n ⇒a = v/n
Now, displacement in last 2 sec = v/n(2n - 2) = 2v - 2v/n

Answer 2

the dist travelled by the particle in last 2 seconds=distance travelled by particle in nth second - distance travelled by particle in (n-2)th second so,

s=(1/2)an² - (1/2)a(n-2)²

 =(1/2)a.[n²-n²+4n-4]

 =a/2.[4n-4]

so,

velocity = acceleration*n(time)

a=v/n

s=2(v/n).(n-1)