okay, we have a boat that has a mass of 1000 kg and is traveling at 25
m/s along the x direction. Now the engine is turned off and the boat
slows down due to water drag. Lets say the force exerted by the water
is f = -70 * v, where v is the velocity of the boat along the x-dir
and f is the water drag in newtons.

how long will it take for the boat to slow down to 12.5 m/s?

now i solved this problem using numerical methods and get a time of
roughly 10 secs. But i was wondering how would you solve this problem
exactly?

i know W = F*x = -70*v*x.Also the total work done is -234,375 J. But
since v varies, how would you set this up to solve it exactly? i tired
using v = dx/dt and tried other differential methiods but to no avail.
mikecoon

(mikecoon) wrote in
om:

okay, we have a boat that has a mass of 1000 kg and is traveling at 25
m/s along the x direction. Now the engine is turned off and the boat
slows down due to water drag. Lets say the force exerted by the water
is f = -70 * v, where v is the velocity of the boat along the x-dir
and f is the water drag in newtons.

how long will it take for the boat to slow down to 12.5 m/s?

now i solved this problem using numerical methods and get a time of
roughly 10 secs. But i was wondering how would you solve this problem
exactly?

i know W = F*x = -70*v*x.Also the total work done is -234,375 J. But
since v varies, how would you set this up to solve it exactly? i tired
using v = dx/dt and tried other differential methiods but to no avail.
mikecoon

You know that F = ma and that F = -kv, where k = 70 newton-sec/meter. So:

ma = m*(dv/dt) = -k*v; so (dv/v) = -k*dt/m.

Integrating you get ln(v) = (-k/m)*t + constant. Solving you get:

v(t)/v(0) = exp(-k*t/m).

Set v(t)/v(0) to 1/2 and solve for t and you get 9.9 seconds.

--
Steve Gray

dear sir

thanks

i was so focused on the work formula that i forgot to start with f= ma.

thanks once again.
mike coon