In one line
When acceleration is a function of time, integrate it once to get velocity and again to get position, using the given initial conditions.
Explained simply
The equations \( v = u + at \) and \( s = ut + \tfrac12 at^2 \) work only when acceleration is constant, like a ball in free fall. A car pressing the accelerator harder and harder has changing acceleration, so we go back to the definitions and use calculus.
Technically, \( v = \dfrac{dx}{dt} \) and \( a = \dfrac{dv}{dt} \). If \( a = f(t) \), then \( v = v_0 + \int_0^t f(t)\,dt \) and \( x = x_0 + \int_0^t v\,dt \).
How it works
- If x(t) is given: differentiate once for v(t), again for a(t).
- If a(t) is given: integrate to get v(t), add the constant from the initial velocity.
- Integrate v(t) to get x(t), add the constant from the initial position.
- To find when the particle stops or reverses, set v = 0.
- Distance vs displacement: if v changes sign during the interval, add the magnitudes of the separate displacements to get total distance travelled.
Key formula
\( v = \dfrac{dx}{dt} \), \( a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2} \). x in m, v in m/s, a in m/s\(^2\), t in s.
Quick example
\( a = 6t \) m/s\(^2\), with \( v = 2 \) m/s and \( x = 0 \) at t = 0. Then \( v = 2 + 3t^2 \) and \( x = 2t + t^3 \). At t = 2 s: v = 14 m/s, x = 12 m. Second case: \( x = t^3 - 6t^2 + 9t \) gives \( v = 3t^2 - 12t + 9 = 0 \) at t = 1 s and t = 3 s, so the particle reverses at those instants.
Common mistakes
- Using \( v = u + at \) when a is not constant.
- Forgetting the constant of integration (initial conditions).
- Reporting displacement when the question asks for total distance, after the particle has reversed.
How MU asks it
Typical 5 to 6-mark numerical: 'The motion of a particle is defined by \( x = \ldots \); find the time when velocity is zero, the position and the total distance travelled in the first 4 s'. Show differentiation/integration, initial conditions, and a short table of x at the turning points for the distance.