Answers by visitors named: Kalin

4,3,7,11,10,9,13

I know this is quite old but for any future souls seeking help, I struggled a lot on a similar problem but it's actually very simple. For starters, your expression is correct. v=vo*e^kt Plugging in 2.1 m/s for vo, (2.1)/2 (or 1.05 m/s) for v, and 2.7 s for t, we can solve for k. k=(ln(2.1/1.05))/(2.7) = -0.2567 1/s Alternatively, because this time is the amount necessary for velocity to reduce to half its initial value, we can utilize the half-life expression, where k=-ln(2)/t Because we want an expression for the distance traveled, we want to integrate our first velocity expression to find an expression for distance. INT (vo*e^kt) = (vo/k)*e^kt Because it took 2.7 seconds for the velocity to reduce to half of its original value, we can assume that this is the half-life at every moment. That is, it takes 2.7 seconds to reduce from 2.1 m/s to 1.05 m/s, 2.7 seconds to reduce from 1.05 m/s to 0.525 m/s, 2.7 seconds to reduce from .525 m/s to 0.2625 m/s, etc. After about 10 half-lives, any system is going to achieve 99.9% of its goal. In this case, after 10 half-lives, the boat is going to cover 99.9% of the distance it is going to travel. It will technically keep moving forever, but for any practical calculation you can use this approach. Using 10 half-lives, or 10*2.7 s, as our upper bound and 0 as our lower bound in the integral, 2.1 m/s for vo, and -0.2567 1/s for k, we get distance = ((2.1/-0.2567)*e^(-0.2567)(27)) - ((2.1/-0.2567)*e^(-0.2567)(0)) =(-0.00799 m) - (-8.18076 m) = 8.17276 m So the distance the boat travels before coming to rest, with two significant digits, is 8.2 meters. Hope this helps.

Its 16 yall the real anwser is 16