## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Problems - Page 115: 68

#### 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.

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