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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
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B
v(n+1)n
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C
v(n1)n
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D
2v(n1)n
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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 second
So,
$$s=\frac{1}{2}an^2-\frac{1}{2}a(n-2)^2$$
$$s=\frac{1}{2}a[n^2-n^2+4n-4]$$
$$s=\frac{a}{2}[4n-4]$$
$$s=(2n-2)a=2a(n-1).........(1)$$
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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