Answer
(a) $m = 1.25~kg$
(b)The new velocity of the boat is $0.90~m/s$ at an angle of $18.0^{\circ}$ south of west.
Work Step by Step
(a) $ma = F$
$m = \frac{F}{a}$
$m = \frac{0.375~N}{0.30~m/s^2}$
$m = 1.25~kg$
(b) We can find the west component of velocity after 2.0 seconds:
$v_f = v_0+at$
$v_f = 0.33~m/s+(0.30~m/s^2)~cos~28^{\circ}(2.0~s)$
$v_f = 0.86~m/s$
We can find the south component of velocity after 2.0 seconds:
$v_f = v_0+at$
$v_f = 0+(0.30~m/s^2)~sin~28^{\circ}(2.0~s)$
$v_f = 0.28~m/s$
We can find the magnitude of the final velocity:
$v = \sqrt{(0.86~m/s)^2+(0.28~m/s)^2} = 0.90~m/s$
We can find the direction south of west:
$tan~\theta = \frac{0.28~m/s}{0.86~m/s}$
$\theta = tan^{-1}(\frac{0.28~m/s}{0.86~m/s})$
$\theta = 18.0^{\circ}$
The new velocity of the boat is $0.90~m/s$ at an angle of $18.0^{\circ}$ south of west.
