Linear motion under a variable force

Further Maths · Further Mechanics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

When the force on a particle changes with time, position or speed, the acceleration is not constant and the suvat equations cannot be used. Newton's second law, F = ma, now gives a differential equation.

The skill is to write the acceleration in the form that fits the question. Acceleration is dv/dt, and by the chain rule it is also v dv/dx. Choose dv/dt when the question links speed and time. Choose v dv/dx when it links speed and distance.

Then separate the variables, integrate both sides and use the starting values to find the constant. Take one direction as positive and give a resistance a minus sign.

Rules to remember

Newton's second law: resultant force = mass × acceleration (F = ma)
Acceleration a = dv/dt = v dv/dx, and v = dx/dt
Force given in terms of t: use m dv/dt = F(t) and integrate with respect to t
Force given in terms of x: use mv dv/dx = F(x), which gives ½mv² = ∫F dx + C
Force given in terms of v: use m dv/dt = F(v) for time, or mv dv/dx = F(v) for distance, and separate the variables
Always add the constant of integration and find it from the initial conditions, or use limits on both integrals
Terminal (steady) speed is the speed at which the acceleration is zero

Common mistake

Students forget the v in v dv/dx and write m dv/dx = F. They also leave out the minus sign on a resistance. Write the equation of motion with signs first, then choose the form of the acceleration.

Worked example

A particle of mass 2 kg starts from rest at the origin and moves along the x-axis. When its displacement is x m, the only force on it is (3x² + 1) N in the positive x-direction. Find its speed when x = 2.

The force depends on x, so use mv dv/dx = F: 2v dv/dx = 3x² + 1.
Separate and integrate: ∫2v dv = ∫(3x² + 1) dx, so v² = x³ + x + C.
v = 0 when x = 0, so C = 0.
When x = 2: v² = 8 + 2 = 10, so v = √10 = 3.16.

Answer: 3.16 m s⁻¹ (3 significant figures)

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