Kinematics Question

[Brophy7]

Hello folks, I have a query here on how to solve this. I am struggling to integrate the differential which will link a,v and x. Some help or a push in the right direction would be tremendous, last question on the paper and I have hit a wall..

"When, a = - kv2, k is a positive constant. Tests show that the constant k is 0.000 32 m-1 for a fully-laden oil tanker under calm sea-going conditions. If the ship loses all power when it is travelling at a speed of 12m/s, determine how far it will travel before its speed drops to 2m/s. How much time will have passed by then?”

 

[Snickster]

I took shot at it. I think I got the correct answer. I will give you a clue how to proceed. Try figuring it out and let me know your results.

First using the equation for "a" above (I assume V is squared and not multiplied by 2), Set up an integral that you can solve for "t". Solve the integral for "t" in terms of "v". Remember the resulting integral needs to include the constant of integration "c".

Use initial boundary conditions at t = 0 , v = 12 and solve for "c".

Rearrange resulting equation to put "v" on the left side as a function of "t" on the right side. Now you know "v" as a function of "t".

Integrate velocity with respect to "t" to find "x" as a function of "t". Remember the resulting integral needs to include the constant of integration "c".

Solve for "c" using the boundary condition at t=0 , x = 0. Now you have a function of "x" with respect to "t".

Find "t" by plugging into the equation derived for velocity previously at v = 2 m/s. This is time it takes for velocity to reach 2 m/s.

Find distance traveled by plugging this time into the equation derived for x(t).

Hint: I get an answer somewhere between 5000 and 6000 meters. Let me know what you get.

 

[GregLocock]

In the real world of engineering it is most unusual to find problems that are simple enough to solve with calculus, and yet have not been reduced to a plug and play equation.

As such I almost always solve problems numerically. Somewhere between 5001 and 5999m

 

[waross]

I once wrote a trail and error program to solve difficult problems.
The program would check if an equation balanced and if not, change the least significant digit by one and re-iterate.
When it found one digit too great,it would retreat by one and go after the next lower order of magnitude.
It would solve to 8 decimal places in a under a second. An average of about 50 iterations.

One problem was the 100 foot ladder leaning up against a vertical wall and just touching the corner of a 10 ft wide by 10 ft high addition.
How high up the building does the ladder reach?

 

[rb1957]

but that's "just" two similar triangles .. no?

but back to the question ... how are you getting on ? seriously, this is a high school type question

Spoiler: hint

S = 1/2*at^2+ut, a = (v-u)/t

 

[Snickster]

"a" is not a constant = -kv^2 so the double integral is not 1/2*at^2.

This is my first step:

a = dv/dt = -kv^2

Separate variables:

dt = dv/(-kv^2) = -(1/k)(v^-2) dv

Integrate and solve for "t" in terms of "v"

 

[rb1957]

ok, agreed ... and integrate v(t) to get the distance. Maybe that was the "trick" of the question, as posted it asked first for the distance and then the time taken; but it is much more direct to determine the time taken and then calc the distance.

 

[Snickster]

Yes the first step was the hardest part. It took me a while to realize the trick was to using a=dv/dt then solve the -kv^2 equation directly rather than indirectly such as change of variable method or other methods.