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. \(\frac{3}{4}t^4−t^2+10t\)
• B. \(\frac{t_4}{2}−\frac{t_3}{3}+10t+4\)
• C. \(\frac{2t^4}{3}−\frac{t^3}{6}+10t+12\)
• D. \(2t^4−\frac{t^3}{2}+5t+4\)

Answer 1

The Correct Option is B

The correct answer is (B) : \(\frac{t_4}{2}−\frac{t_3}{3}+10t+4\)
\(α=\frac{d\omega}{dt}=6t^2−2t\)
\(∫_{0}^{\omega} d\omega=∫_{0}^{t}(6t^2−2t)dt\)
so ω = 2t³ – t² + 10
and
\(\frac{dθ}{dt}=2t^3−t^2+10\)
so
\(∫_{4}^{θ}dθ=∫_{0}^{t}(2t^3−t^2+10)dt\)
\(θ=\frac{t_4}{2}-\frac{t^3}{3}+10t+4\)