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

Differentiating wrt t

$Vy = 18 t^2 – 5$

Differentiating wrt t

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

Integrating wrt t

$Vx = 7t^2 +k $

But, $Vx = 4$ at $ t=0$

So, $k = 4m/s$

$Vx = 7t^2 +4 \\ (Vx)_{t=1} = 7 + 4 = 11 m/s ..3 \\ (Ax)_{t=1} = 14 \times 1 = 14 m/s^2 ..4 \\ V= Vx(i) + Vy(j) \\ = 11i + 13j (\text { from 1 and 3}) $

Converting into polar form

$V = 17 m/s ↗ θ = 50 \\ A= Ax(i) + Ay(j) \\ = 14i + 36j (\text {from 2 and 4})$

Converting into polar form

$A = 38.6 m/s^2 ↗ θ = 68.75$

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