Velocity and Acceleration from Displacement

Kinematics links displacement, velocity and acceleration by differentiation: $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt}$. From a displacement function, differentiate once for velocity, twice for acceleration. (To reverse, you integrate.)

A particle moves in a straight line so that its displacement from a fixed point after $t$ seconds is $s = t^3 - 6t^2 + 9t$ metres. (i) Find expressions for the velocity and acceleration. [3] (ii) Find the velocity at the instant when the acceleration is zero. [2]

The working

(i) Velocity is $\frac{ds}{dt}$: $v = \frac{ds}{dt} = 3t^2 - 12t + 9 \quad \text{(M1, A1)}$

Acceleration is $\frac{dv}{dt}$ (differentiate again): $a = \frac{dv}{dt} = 6t - 12 \quad \text{(A1)}$

(ii) Set $a = 0$ to find the time: $6t - 12 = 0 \;\Rightarrow\; t = 2 \quad \text{(M1)}$

Substitute $t = 2$ into the velocity expression: $v = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3\ \text{m s}^{-1} \quad \text{(A1)}$

So the velocity is $-3\ \text{m s}^{-1}$ when the acceleration is zero (the negative sign means it’s moving in the negative direction).

Where the marks are won and lost

• Differentiate in the right direction: displacement → velocity → acceleration. Integrating by mistake goes the wrong way.
• In (ii), set the acceleration to zero to find $t$, then substitute into the velocity. Substituting into the wrong expression is the usual slip.
• Keep the sign: $v = -3$ is a valid answer; the negative direction is physically meaningful, don’t drop it.

Common mistakes

• Differentiating once and calling it acceleration (that’s velocity).
• Setting velocity to zero instead of acceleration in (ii).
• Dropping the negative sign on the final velocity.

Full method: Kinematics notes. Topic home: Calculus pillar.