A particle moving in the +ve x direction has an acceleration, $a = 100 - 4v^2 m/s^2.$ Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s.

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A particle moving in the +ve x direction has an acceleration,

$a = 100 - 4v^2 m/s^2.$ Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s.

written 8.2 years ago by

teamques10 ★ 69k

modified 3.1 years ago by

sagarkolekar ★ 11k

engineering mechanics

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2 Answers

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written 8.2 years ago by

teamques10 ★ 69k

FBD of the lift

enter image description here

Using kinematic equation

$ v^2=u^2+2as$

Given that $v=3m/s, u=0, s=4m,$ we have,

$3^2=0+(2×4)a \\ a = \dfrac98 m/s^2$

By Newton’s Law,

$ma=T-mg \\ \therefore T=m(a+g)=750(\dfrac 98+9.81) \\ T=8201.25N \space or\space T=8.201kN -----Ans$

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written 8.2 years ago by

teamques10 ★ 69k

FBD of the lift

enter image description here

Using kinematic equation

$v^2=u^2+2as$

Given that $v=3m/s, u=0, s=4m,$ we have,

$3^2=0+(2×4)a \\ a = \dfrac98 m/s^2$

By Newton’s Law,

$ma=T-mg \\ \therefore T=m(a+g)=750(\dfrac 98+9.81) \\ T=8201.25N \space or\space T=8.201kN -----Ans$

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