IP AMaths Notes (Upper Sec, Year 3-4): 19) Kinematics

Kinematics connects calculus with motion. Derivatives give velocity and acceleration; integration recovers displacement.

Relationships

• Velocity: \( v = \dfrac{ds}{dt} \).
• Acceleration: \( a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2} \).
• Displacement from velocity: \( s = \int v , dt \) (plus constant).

Worked example 1 — From velocity to displacement

Given \( v(t) = 6t^2 - 4t + 3 \) in \( \pu{m.s-1} \), find displacement \( s(t) \) if \( s(0) = 2 \).

1. Integrate: \( s(t) = \int (6t^2 - 4t + 3) , dt = 2t^3 - 2t^2 + 3t + C \).
2. Apply initial condition: \( 2 = 0 - 0 + 0 + C \) ⇒ \( C = 2 \).
3. Hence \( s(t) = 2t^3 - 2t^2 + 3t + 2 \).

Worked example 2 — Maximum speed

A particle moves with \( s(t) = t^3 - 6t^2 + 9t \). Find the time when speed is greatest for \( t \geq 0 \).

1. Velocity: \( v(t) = 3t^2 - 12t + 9 \).
2. Acceleration: \( a(t) = 6t - 12 \).
3. Speed is maximum when \( \dfrac{dv}{dt} = 0 \) and acceleration switches sign → solve \( 6t - 12 = 0 \) giving \( t = 2 \).
4. Check endpoints: \( t = 0 \) gives \( v = 9 \); \( t = 2 \) gives \( v = -3 \) (speed 3); as \( t \to \infty \), \( v \to \infty \).
5. Since quadratic opens upward, minimum occurs at \( t = 2 \) but speed grows thereafter. For restricted interval, consider context; often question restricts to \( 0 \leq t \leq 3 \).
6. Within \( [0, 3] \), compare \( |v(0)| = 9 \), \( |v(2)| = 3 \), \( |v(3)| = 6 \). Greatest speed is \( 9 \pu{m.s-1} \) at \( t = 0 \).

Try this

Given \( a(t) = 12t - 8 \) with \( v(0) = 5 \) and \( s(0) = 0 \), find expressions for \( v(t) \) and \( s(t) \).