Kinematics Question 169
Question: A body starts from rest with uniform acceleration. If it’s velocity after n second is v, then it’s displacement in the last two seconds is
Options:
A) $ \frac{2v(n+1)}{n} $
B) $ \frac{v(n+1)}{n} $
C) $ \frac{v(n-1)}{n} $
D) $ \frac{2v(n-1)}{n} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d]
$ \therefore v=0+na\Rightarrow a=v/n $
Now, distance travelled I n sec $ \alpha \Rightarrow S _{n}=\frac{1}{2}an^{2} $ and distance travelled in $ (n-2) $
$ \sec \Rightarrow {S _{n-2}}=\frac{1}{2}a{{(n-2)}^{2}} $
$ \therefore $ distance travelled in last two seconds, $ =S _{n}-{S _{n-2}}=\frac{1}{2}an^{2}-\frac{1}{2}a{{(n-2)}^{2}} $
$ =\frac{a}{2}[n^{2}-{{(n-2)}^{2}}]=\frac{a}{2}[n+(n-2)][n-(n-2)] $
$ =a(2n-2)=\frac{v}{n}(2n-2)=\frac{2v(n-1)}{n} $
BETA
