Why does reducing air resistance by a factor of 4 give acceleration of 3F/4m?

[jerryez]

If F is the force of air resistance on an object with mass m moving at a constant velocity, which of the following best describes the acceleration of the object when the force of air resistance is reduced by a factor of 4?



F=ma


It says the answer is equal to (3/4)F/m = a

Can someone explain why its the 3/4?

 

[Coto]

It is an application of Newton's 2nd law.

Write down the sum of all forces in the case where an object is moving at constant velocity:

$$\sum F = F_{\mathrm{thrust}} - F_{\mathrm{res}} = 0$$,

i.e. the object is not accelerating or decelerating, thus the thrust must equal the wind resistance. Now write down the sum of all forces when the resistance force suddenly drops by 1/4:

$$\sum F = ma = F_{\mathrm{thrust}} - F_{\mathrm{res}}/4$$.

The thrust force remains the same as before which was equal to the old wind resistance, however the resistance has now dropped a quarter meaning the thrust is more powerful than the new wind resistance which results in an acceleration. From the first part you have that $$F_{\mathrm{thrust}} = F_{\mathrm{res}}$$, so plugging this into the second part yields the result.

 

[Aeroneer]

So, qualitatively speaking:
resisting force = driving force since its at constant velocity, right?
so if the resisting dropped from F to 1/4F... then the net force would be 3/4F in favour of the driving force. ...which means the object is accelerating.