DE with mechanics

[Sonal7]

I tried to solve this by using the equation F = m x a
I also did the following:
F/m = dV/dt and F =-KV, therefore K=-F/V

K is 2, as when F is 10 the a is 0, and velocity is stated as 2 m/s

I dont know what else I might need to get the equation for dv/dt

 

[MarkFL]

We are given two points of the form \((v,F)\):

[MATH](0,10),\,(5,0)[/MATH]
Hence:

[MATH]F=-2v+10[/MATH]
Now, if \(x(t)\) represents the position of the particle at time \(t\), then using Newton's 2nd Law:

[MATH]m\frac{d^2x}{dt^2}=-2\frac{dx}{dt}+10[/MATH]
It would have been nice if you had included the choices from which you are to select.

 

[Sonal7]

Thanks for the reply

I cant figure out where I am going wrong but you have brought me closer!

F=m x a

a is dv/dt

therefore m x dv.dt = -2V+10?
Then dv/dt = (-2V+10)/mass

This is rearranged to dv/dt =10-2v/m This is incorrect

Here are the options, i guessed until i got the correct answer.

 

[Sonal7]

I got it with some help, I think I need to used the resistive forces which means working out a constant of proportionality.
R is proportional to V, meaning R=K.V
Given that when a = 0, V =2

(-2V+10)-kv=ma (1)
K works out be 3.

substituting k in (1) gives the equation that is correct.

 

[topsquark]

There's just one little problem: According to the graph MarkFL's derivation is correct. The initial data (a = 0, V = 2) is not what the graph provided says!

-Dan

 

[MarkFL]

I got it with some help, I think I need to used the resistive forces which means working out a constant of proportionality.
R is proportional to V, meaning R=K.V
Given that when a = 0, V =2

(-2V+10)-kv=ma (1)
K works out be 3.

substituting k in (1) gives the equation that is correct.

I started to add in a resistive force, but there was one already present, and as Dan stated, it matches the given graph. I did fail to incorporate the given initial values though. Not a very well constructed problem, in my opinion.

 

[Sonal7]

I agree Lets hope the exam doesnt ask questions in similar manner!

 

[Sonal7]

There's just one little problem: According to the graph MarkFL's derivation is correct. The initial data (a = 0, V = 2) is not what the graph provided says!

-Dan

I thought so too that the graph doesnt fit the description but I thought I am just a student, the expert has gotta be right? wrong!

 

[topsquark]

I thought so too that the graph doesnt fit the description but I thought I am just a student, the expert has gotta be right? wrong!

Pffl. I once wrote an exam that had two long answer problems. One of the problems ended up being so badly written I just dopped it from the test instead of trying to fix it.

-Dan

 

[Deleted member 4993]

The other quibble I have:

The graph does not show the origin explicitly!