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What is the acceleration if the given equation x=root(1+t^2)

What is the acceleration if the given equation x=root(1+t^2)

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
Acceleration would be the double differentiation of the x w.r.t for the given equation that is:
x = root(1+t^2)
differentiating on squaring we gt,
2x.dx = 2tdt
xdx = tdt
v = dx/dt = t/root(1+t^2)
sq. and differtiating again we get,
v^2 = t^2/(1+t^2)
2vdv = (2t(1+t^2) – 2t^3)dt/(1+t^2)^2 = (2t + 2t^3 – 2t^3)dt/(1+t^2)^2
vdv = 2tdt/(1+t^2)^2
a = dv/dt = (2t/(1+t^2))* t/root(1+t^2)
 

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