Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : MAY 2015

$$Y = 6t^3 – 5t$$

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 wrt t

$Vx = 7t^2 +k $

But, $Vx = 4 \space at \space t=0 \\ So, \space k = 4m/s \\ Vx = 7t^2 +4 \\ (Vx)_{t=1} = 7 + 4 = 11 m/s ..3 \\ (Ax)_{t=1} = 14 \times 1 = 14 m/s^2 ..4 \\ V= Vx(i) + Vy(j) \\ = 11i + 13j$ (from 1 and 3)

Converting into polar form

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

Converting into polar form

$A = 38.6 m/s^2 ↗ θ = 68.75$