Answer
$v_P=2.42~ft/s$
$a_P=34.4~ft/s^2$
Work Step by Step
The required velocity and acceleration can be determined as follows:
As $\alpha=400 t^3$
$\implies 400t^3=\frac{d\omega}{dt}$
$\implies d\omega=400t^3dt$
$\implies \int_0^{\omega_A} d\omega\implies \int_0^t 400 t^3 dt$
$\implies \omega|_0^{\omega_A}=(\frac{400t^4}{4})|_0^t$
$\implies \omega_A=100 t^4$
Similarly, $\omega_B=\frac{\omega_A r_A}{r_B}=\frac{(31.6406)(0.5)}{1.2}=13.1836rad/s$
and $\alpha_B=\frac{\alpha_A r_A}{r_B}=\frac{(168.75)(0.5)}{1.2}=70.3~rad/s^2$
Now the velocity at point B is given as
$v_P=\omega_B r_P$
$\implies v_P=13.1836\times 2.2=29in/s =29/12=2.4169ft/s$
We know that
$a_{Pt}=\alpha_B r_P$
$\implies a_{Pt}=70.3125\times \frac{2.2}{12}=12.8906ft/s^2$
and $a_{Pn}=\omega^2_B r$
$\implies a_{Pn}=(13.1836)^2(\frac{2.2}{12})=31.8646ft/s^2$
Now $a_P=\sqrt{(a_{Pt})^2+(a_{Pn})^2}$
We plug in the known values to obtain:
$a_P=\sqrt{(12.8906)^+(31.8646)^2}=34.4~ft/s^2$
