The y coordinate of a particle is given by $y = 6t^3 -5t.$ If $a_x = 14t m/sec^2 \&\space v_x = 4m/sec$ at $ t = 0,$ determine the velocity & acceleration of particle when t=1 second.

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written 8.9 years ago by

teamques10 ★ 70k

modified 3.8 years ago by

pedsangini276 • 4.8k

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : MAY 2015

engineering mechanics

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written 8.9 years ago by

teamques10 ★ 70k

$$Y = 6t^3 – 5t$$

Differentiating wrt t

$Vy = 18 t^2 – 5$

Differentiating wrt t

$Ay = 36t \\ At\space t = 1 s \\ (Vy)_{t=1} = 18 – 5 = 13 m/s .. 1 \\ (Ay)_{t=1} = 36 m/s^2 .. 2 \\ Ax = 14t $

Integrating …

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