Acceleration as a Function of Position
Where v(x) is differentiable along a one-dimensional trajectory, a = dv/dt = (dv/dx)(dx/dt) = v dv/dx. Equivalently, a = (1/2)d(v²)/dx.
Why this shows up in the exam
Velocity-displacement graphs · Power-law v(x) motion · Time-free acceleration calculations
Learn the idea
When velocity is given versus position, use a = v dv/dx instead of forcing time into the problem. Velocity can change because the particle reaches a new place. The chain rule converts change per metre into change per second by multiplying by the current velocity.
🧠 Memory hook: If v knows x, acceleration is v times its x-slope.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- a = v dv/dx — chain-rule acceleration for a known v(x)
- a = (1/2)d(v²)/dx — useful form for v-squared versus position
How to approach it
- 1Write v as a function of x
- 2Differentiate with respect to x
- 3Multiply by v and check acceleration units
Common slip-ups that cost marks
- •Using dv/dx alone as acceleration
- •Differentiating with respect to time without x(t)
- •Dropping the factor one-half in the v² form
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?
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