Kinematics: Finding Velocity from Acceleration

Displacement, velocity and acceleration are linked by calculus: differentiating goes $s \to v \to a$, so integrating reverses it, $a \to v \to s$. Each integration brings a constant, fixed by the initial conditions.

A particle moves in a straight line with acceleration $a = (6t - 2)\ \text{m/s}^2$ at time $t$ seconds. When $t = 0$ the velocity is $3\ \text{m/s}$ and the displacement is $0$. Find the displacement when $t = 2$. [6]

The working

Step 1, integrate acceleration to get velocity: $v = \int (6t - 2)\,dt = 3t^2 - 2t + C \quad \text{(M1)}$

Step 2, use $v = 3$ when $t = 0$: $3 = 0 - 0 + C \implies C = 3, \quad \text{so} \quad v = 3t^2 - 2t + 3 \quad \text{(A1)}$

Step 3, integrate velocity to get displacement: $s = \int (3t^2 - 2t + 3)\,dt = t^3 - t^2 + 3t + D \quad \text{(M1)}$

Step 4, use $s = 0$ when $t = 0$: $0 = 0 - 0 + 0 + D \implies D = 0, \quad \text{so} \quad s = t^3 - t^2 + 3t \quad \text{(A1)}$

Step 5, evaluate at $t = 2$: $s = 8 - 4 + 6 = 10\ \text{m} \quad \text{(M1, A1)}$

Where the marks are won and lost

• Both constants matter. $C$ comes from the velocity condition, $D$ from the displacement condition. Dropping either loses the accuracy marks downstream.
• Integrate in the right order: acceleration to velocity first, then velocity to displacement. You cannot jump straight from $a$ to $s$.
• Apply each initial condition to the matching expression: $v(0)$ to the velocity, $s(0)$ to the displacement.

Common mistakes

• Omitting the constant of integration (the single biggest error in kinematics).
• Differentiating instead of integrating.
• Using the velocity condition on the displacement expression.

Topic home: Calculus pillar. More: Worked examples.