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

Answer: (B)

$\frac{t_4}{2}−\frac{t_3}{3}+10t+4$

Answer 2

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
$∫_{4}^{θ}dθ=∫_{0}^{t}(2t^3−t^2+10)dt$
$θ=\frac{t_4}{2}-\frac{t^3}{3}+10t+4$