Question

A ball is spun with angular acceleration α = 6t² – 2t, 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 :

• A. \(\frac{3}{4}\)t⁴-t²+10t
• B. \(\frac{t^4}{2}\)-\(\frac{t^3}{3}\)+10t+4
• C. \(\frac{2t^4}{3}\)-\(\frac{t^3}{6}\)+10t+12
• D. 2t⁴-\(\frac{t^3}{2}\)+5t+4

Answer 1

The Correct Option is B

\(\alpha=  \frac{dω}{dt}  =6t^2−2t\)

\(\int_{0}^{ω}dw=\int_{0}^{t} (6t^2-2t)dt\)

So, \(\omega = 2t^3 – t^2 + 10\)

\(\int_{4}^{θ} θ dθ = \int_{0}^{t}(2t^3-t^2+10)dt)\)

\(\theta =  \frac{t^4}{2} − \frac{t^3}{3}+10t+4\)

\(\therefore ,\) The correct option is (B): \(\frac{t^4}{2} − \frac{t^3}{3}+10t+4\)