derive the expression for displacement in the nth second.
Asked by abhilipsa satpathy
| 21st May, 2013,
04:54: PM
Expert Answer:
a=accelleration
v=velocity, u=initial velocity
s=distance
s=ut + 1/2at^2
To calculate the distance traveled during the nth second, you calculate the distance covered in n seconds and subtract the distance covered in n-1 seconds and get
s = un-u(n-1) + 1/2an^2-1/2a(n-1)^2
simplifying gives us
u term = u(n-(n-1)) = u*1 = u
(n-1)^2 = n^2-2n+1
a term = 1/2a(n^2-(n-1)^2) = 1/2a(n^2-n^2+2n-1) the n^2's cancel and give us 1/2a(2n-1) .
Final equation for displacement in the nth second is
s = u + 1/2a(2n-1)
v=velocity, u=initial velocity
s=distance
s=ut + 1/2at^2
To calculate the distance traveled during the nth second, you calculate the distance covered in n seconds and subtract the distance covered in n-1 seconds and get
s = un-u(n-1) + 1/2an^2-1/2a(n-1)^2
simplifying gives us
u term = u(n-(n-1)) = u*1 = u
(n-1)^2 = n^2-2n+1
a term = 1/2a(n^2-(n-1)^2) = 1/2a(n^2-n^2+2n-1) the n^2's cancel and give us 1/2a(2n-1) .
Final equation for displacement in the nth second is
s = u + 1/2a(2n-1)
Answered by
| 22nd May, 2013,
08:15: AM
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