A ball is spun with angular acceleration =6 t ^ 2 -2 t where t is… - Vidyadip

MCQ

A ball is spun with angular acceleration $\alpha=6 t ^{2}-2 t$ where $t$ is in second and $\alpha$ 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
$\frac{3}{2} t^{4}-t^{2}+10 t$
✓
$\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4$
C
$\frac{2 t ^{4}}{3}-\frac{ t ^{3}}{6}+10 t +12$
D
$2 t^{4}-\frac{t^{3}}{2}+5 t+4$

✓

Answer

Correct option: B.

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

b
$\frac{d \omega}{d t}=6 t^{2}-2 t$

$\int \limits_{10}^{m} d \omega=2 t^{5}-t^{2}$

$\omega=10+2 t^{5}-t^{2}$

$\frac{d \theta}{d t}=10+2 t^{5}-t^{2}$

$\int \limits_{4}^{\theta} d \theta=10+2 t^{3}-t^{2}$

$\int \limits_{4}^{\theta} d \theta=10 t+\frac{t^{4}}{2}-\frac{t^{3}}{3}$

$\theta=4+10 t+\frac{t^{4}}{2}-\frac{t^{5}}{3}$

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