A particle moving in the +ve x direction has an acceleration, $a = 100

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

$a = 100 - 4v^2 m/s^2.$

written 8.7 years ago by

teamques10 ★ 69k

• modified 4.9 years ago

Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s.

engineering mechanics

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written 8.7 years ago by

teamques10 ★ 69k

FBD of the lift

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Using kinematic equation

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

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

$3^2=0+(2×4)a$

$\therefore a=\dfrac 98 m/s^2$

By Newton’s Law,

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

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