How Do You Calculate the Distance a Motorboat Travels After Engine Shutdown?

[physicsss]

Help!

A motorboat traveling at a speed of 2.4 m/s shuts off its engines at t = 0. How far does it travel before coming to rest if it is noted that after 3.0 s its speed has dropped to half its original value? Assume that the drag force of the water is proportional to v.

 

[physicsss]

...anyone?

 

[Tide]

The equation of motion would be
$$\frac {dv}{dt} = -k v$$
Can you integrate that?

 

[physicsss]

Do I get lnv=-kt ? What do I need to do after I find what k is?

 

[Hypercase]

Whats the answer given in your reference book?
-Cheers.

 

[physicsss]

The answer is 10m, and i have no idea how they got it

 

[poolwin2001]

When you solve the differential eqn you will have 2 variables(1 from integration & other 'k')Use the initial conditions given to find them.
At t=0,v=? and one more.
Else,if you did definite integration ,you have to figure out k by the 2nd condition given.

 

[Tide]

$$v = v_0 e^{-kt}$$

 

[HallsofIvy]

Do I get lnv=-kt ? What do I need to do after I find what k is?

Since ln v=-kt+ C (you forgot to add the constant), v= Ce^-kt which has two unknown parameters, C and k. Now use the information you were given: "traveling at a speed of 2.4 m/s shuts off its engines at t = 0". Okay, when t=0, v= Ce^-k(0)= C= 2.4. "it is noted that after 3.0 s its speed has dropped to half its original value" Okay, when t= 0, v= "half its original value" which is 2.4/2= 1.2 m/s. v= 2.4e^-k(3)= 1.2 . Solve that for k.