Question

# A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in last two seconds is:

A
2v(n+1)n
B
v(n+1)n
C
v(n1)n
D
2v(n1)n

Solution

## The correct option is D $$\dfrac{2v(n-1)}{n}$$The distance traveled by the particle in the last 2 seconds = distance traveled by particle in nth second- distance traveled by particle in (n-2)th secondSo,$$s=\frac{1}{2}an^2-\frac{1}{2}a(n-2)^2$$$$s=\frac{1}{2}a[n^2-n^2+4n-4]$$$$s=\frac{a}{2}[4n-4]$$$$s=(2n-2)a=2a(n-1).........(1)$$Now,Velocity, $$v=acceleration\times time$$$$v=a\times n$$$$a=\frac{v}{n}...........................(2)$$Substitute value of 'a' from (2) to (1) we get,$$s=\dfrac{2v(n-1)}{n}$$Physics

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