In one line
When acceleration depends on position or velocity, use \( a = v\,dv/dx \) or \( a = dv/dt \), separate the variables, and integrate.
Explained simply
A spring pushes harder the more it is compressed, so the acceleration it gives depends on position. Air drag grows with speed, so the deceleration depends on velocity. Time is not given directly, so we need a different form of the definition.
Technically, by the chain rule \( a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx} \).
How it works
- If \( a = f(x) \): write \( v\,dv = f(x)\,dx \) and integrate between limits: \( \tfrac12(v^2 - v_0^2) = \int_{x_0}^x f(x)\,dx \).
- If \( a = f(v) \) and time is needed: \( dt = \dfrac{dv}{f(v)} \).
- If \( a = f(v) \) and distance is needed: \( dx = \dfrac{v\,dv}{f(v)} \).
- Apply the given initial conditions as integration limits.
Key formula
\( a = v\dfrac{dv}{dx} \); for \( a = -kv \): \( v = v_0 e^{-kt} \) and \( x = \dfrac{v_0 - v}{k} \). k in s\(^{-1}\), v in m/s, x in m.
Quick example
(1) \( a = 4x \) m/s\(^2\), v = 2 m/s at x = 0: \( \tfrac12(v^2 - 4) = 2x^2 \), so \( v^2 = 4 + 4x^2 \); at x = 3 m, \( v = \sqrt{40} = 6.32 \) m/s. (2) A boat cuts its engine at 10 m/s and decelerates as \( a = -0.5v \): it coasts \( x = (10 - 0)/0.5 = 20 \) m before (theoretically) stopping, while its speed halves every \( \ln2/0.5 = 1.39 \) s.
Common mistakes
- Integrating \( a = f(x) \) with respect to time.
- Forgetting the factor \( \tfrac12 \) in \( \int v\,dv = v^2/2 \).
- Losing the minus sign for deceleration, which gives growing instead of decaying speed.
How MU asks it
Asked as 5 to 6-mark numericals: 'The acceleration of a particle is \( a = -kv^2 \) ... find the distance travelled when the speed halves'. Write the correct form of a, separate variables, integrate with limits, and give the final value with units.