A ball is spun with angular acceleration $α = 6 t^{2} - 2 t$ where t is in second and $α$ is in rads^-2. At t = 0, the ball has angular velocity of 10 rads^-1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

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$\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

$2 t^{4} - \frac{t^{3}}{2} + 5 t + 4$

$\frac{3}{2} t^{4} - t^{2} + 10 t$

$\frac{2 t^{4}}{3} - \frac{t^{3}}{6} + 10 t + 12$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d ω}{dt} = 6 t^{2} - 2 t \\ \Rightarrow \int_{10}^{ω} d ω = 2 t^{3} - t^{2} \\ \Rightarrow ω = 10 + 2 t^{3} - t^{2} \\ \Rightarrow \frac{dθ}{dt} = 10 + 2 t^{3} - t^{2} \\ \Rightarrow \int_{4}^{θ} dθ = \int_{0}^{t} (10 + 2 t^{3} - t^{2}) d t \\ \Rightarrow \int_{4}^{θ} dθ = 10 t + \frac{t^{4}}{2} - \frac{t^{3}}{3} \\ \Rightarrow θ = 4 + 10 t + \frac{t^{4}}{2} - \frac{t^{3}}{3} \end{array}$