University Physics with Modern Physics (14th Edition)

We can find the magnitude of the acceleration when the engine is operating. $F = ma$ $a = \frac{F}{m} = \frac{8.0\times 10^4~N}{3.6\times 10^7~kg}$ $a = 2.22\times 10^{-3}~m/s^2$ We can find the time for the tanker to decelerate to rest. $t = \frac{0-1.5~m/s}{-2.22\times 10^{-3}~m/s^2}$ $t = 676~s$ We can find the distance required to decelerate to rest. $x = (1.5~m/s)(676~s)-\frac{1}{2}(2.22\times 10^{-3}~m/s^2)(676~s)^2$ $x = 507~m$ Therefore, since the distance to the reef is only 500 meters, the tanker will strike the reef. We can find the velocity of the tanker after it moves 500 meters. $v^2 = v_0^2+2ax$ $v = \sqrt{v_0^2+2ax}$ $v = \sqrt{(1.5~m/s^2)^2 - (2)(2.22\times 10^{-3}~m/s^2)(500~m)}$ $v = 0.173~m/s$ The tanker's speed when it hits the reef will be 0.173 m/s. Since the speed of the tanker is less than 0.2 m/s, the hull will be able to withstand the impact.