Skip to content

9231 · 3.5

Linear motion under a variable force — FAQ

Frequently asked questions for 9231 Linear motion under a variable force. Direct answers first, then deeper explanation — then practise with marking.

How do I know which form of acceleration to use?

Look at the variable in the force function. If force is a function of time tt, use a=dvdta = \frac{dv}{dt}. If force is a function of displacement xx, use a=vdvdxa = v\frac{dv}{dx}. If force is a function of velocity vv, you can use either, but the question will usually guide you towards finding either v(t)v(t) or v(x)v(x).

Can I ever use the suvat equations in this topic?

Only if you can prove that the net force on the particle is constant. If the force is given as a function like F(t)=2tF(t) = 2t or F(x)=3/x2F(x) = 3/x^2, then the force is variable, acceleration is variable, and the suvat equations are not valid.

What's the difference between work done and the work-energy principle in these problems?

They are directly related. The work done by the net force on a particle equals the change in its kinetic energy. When you solve the equation F(x)=mvdvdxF(x) = mv\frac{dv}{dx} by integrating to get F(x)dx=mvdv\int F(x)\,dx = \int mv\,dv, you are actually using the work-energy principle. The left side is the work done, and the right side integrates to 12mv2\frac{1}{2}mv^2, representing kinetic energy.