Question

A body starts from rest and travels with uniform acceleration on a straight line. If its velocity after making a displacement of 32 meter is 8 m/s then it's acceleration is:- (A) 1 m/s^2 [One meter per second square] (B) 2 m/s^2 [Two meter per second square] (C) 3 m/s^2 [Three meter per second square] (D) 4 m/s^2 [Four meter per second square]​

Answer 1

$\small{\underline{\sf{\red{Given:\rightarrow}}}}$

• Initial velocity [u]= 0m/s

• Final velocity [v]= 8m/s

• Displacement [s]= 32m

$\small{\underline{\sf{\blue{To\:Find\rightarrow}}}}$

• acceleration [a]

$\small{\underline{\sf{\green{Formula \ used:\rightarrow}}}}$

$\boxed{\bf{v^2-u^2= 2as }}$

$\small{\underline{\sf{\orange{SoLution:\rightarrow}}}}$

Putting the value in formula we get,

$\bf {\implies 8^2- 0^2= 2\times a \times 32}$

$\bf{\implies 64= a\times 64 }$

$\bf{{\implies a= 1}}$

$\boxed{\sf{\pink{Acceleration = (A) \ 1m/s^2\bigstar }}}$

_________________________________

Answer 2

Answer:

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=21an2−21a(n−2)2

s=21a[n2−n2+4n−4]

s=2a[4n−4]

s=(2n−2)a=2a(n−1).........(1)

Now,

Velocity, v=acceleration×time

v=a×n

a=nv...........................(2)

Substitute value of 'a' from (2) to (1) we get,

s=n2v(n−1)