A particle is moving in a straight line under constant acceleration Travels a distance 3d in 5th and 5d in 6th seconds .the displacement in 8 seconds will be
Answers
Given : A particle is moving in a straight line under constant acceleration Travels a distance 3d in 5th and 5d in 6th seconds .
To Find : the displacement in 8 seconds will be
Solution:
S = ut + (1/2)at²
Distance covered in n seconds
= un + (1/2)an²
Distance covered in n-1 seconds
= u(n - 1) + (1/2)a(n - 1)²
Distance covered in nth second
= un + (1/2)an² - (u(n - 1) + (1/2)a(n - 1)²)
= u + (1/2)a (2n - 1)
Distance covered in 5th second
3d = u + (1/2)a (2 * 5 - 1)
3d = u + 9a/2
Distance covered in 6th second
5d = u + (1/2)a (2 * 6 - 1)
5d = u +11a/2
=> 2d = 2a/2
=> a = 2d
3d = u + 9a/2
=> 3d = u + 9(2d)/2
=> u = -6d
displacement in 8 seconds
S = 8u + (1/2)a8²
= 8(-6d) + (1/2)(2d)8²
= -48d + 64d
= 16d
displacement in 8 seconds will be 16d
and distance in 8th second would be 9d
Additional Info:
u = -6d , a = 2d
ut + (1/2)at²
-6dt + dt²
t d in that second
0 0 0
1 -5d -5d
2 -8d -3d
3 -9d -d
4 -8d d
5 -5d 3d
6 0 5d
7 7d 7d
8 16d 9d
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