Acceleration of falling object with drag force proportional to v^n

[LukeEvans]

A body of mass m (kg) is falling vertically under gravity g m/s² in a medium whose resistance to the speed of the body, v m/s, is proportional to vⁿ (n positive).

If the body was released from rest and has terminal velocity, v_t m/s, use Newton's 2^nd law of motion to show that its acceleration, a m/s², may be expressed as

a = g(1-(v/v_t)ⁿ)

(4 Marks)

I would obviously like to attempt to solve this problem, but am unsure what the first step should be. Perhaps the resistive force will need to be deducted from the velocity... many thanks to anybody who looks.

 

[tiny-tim]

Hi LUke!

start by writing the first sentence as an equation …

A body of mass m (kg) is falling vertically under gravity g m/s² in a medium whose resistance to the speed of the body, v m/s, is proportional to vⁿ (n positive).

 

[LukeEvans]

A body of mass m (kg) is falling vertically under gravity g m/s2 in a medium whose resistance to the speed of the body, v m/s, is proportional to vn (n positive).

m = g - vⁿ

would be an uneducated guess

 

[tiny-tim]

what about "proportional to" ?

 

[LukeEvans]

m = g - (v*vⁿ)

possibly?

 

[tiny-tim]

m = g - (v*vⁿ)

possibly?

That's g - vⁿ⁺¹ …

how is vⁿ⁺¹ proportional to vⁿ ? ​

 

[LukeEvans]

how about:

m = g * vⁿ

or

m = g*(v+vⁿ)

 

[tiny-tim]

Suppose I told you that the medium's resistance was 3vⁿ, what would the equation be then?

 

[LukeEvans]

m = g - 3vⁿ

I'll be honest I don't know where the 3 has come from.

 

[tiny-tim]

the resistive force is proportional to vⁿ.

3 is proportional

it has to be something

 

[LukeEvans]

Right.

m = g - vvⁿ

 

[LukeEvans]

How does that look?

 

[tiny-tim]

" proportional to vⁿ " means kvⁿ for some constant k