• anonymous
Hi everyone. Anybody have an idea to point me in the right direction? A motorboat traveling at a speed of 2.2 m/s shuts off its engines at t=0. How far does it travel before coming to rest if it is noted that after 2.9 s its speed has dropped to half its original value? Assume that the drag force of the water is proportional to v.
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • schrodinger
I got my questions answered at in under 10 minutes. Go to now for free help!
  • matt101
The question gives us the following information: Initial velocity = vi = 2.2 m/s Final velocity = vf = 0.5vi = 1.1 m/s Time = t = 2.9 s We can also assume constant acceleration since the drag force is proportional to v. Using this information, you should be able to use equations of motion to find the distance travelled!
  • IrishBoy123
because drag \(\propto v\), we can say \(F = ma = - kv\) where k is some constant that creates the differential equation \({dv \over dt} = - {k \over m} v\) which we solve as \(\large \int\limits_{v=v_o}^{v(t)} {1 \over v} dv = -\alpha \int\limits_{t=0}^{t} dt \) where \(\alpha = {k \over m}\) [because we know neither the value of m or k, we may as well stuff them into a single constant called \(\alpha\)] your solution will be in the form of exponential decay, and then to solve for \(\alpha \), just plug in the various values we have been given for v and t. *however* please note that the boat in theory never actually stops. exponential decay is asymptotic....maybe you are supposed to spot this at the outset, i dunno, seems strange to me but this is the correct analysis :p
  • matt101
Good point, @IrishBoy123! To be honest I didn't think much of the assumption provided, but you've turned me around.

Looking for something else?

Not the answer you are looking for? Search for more explanations.