Q: Kinematics: Use calculus in kinematics for motion in a straight line: v = dr/dt, a = dv/dt = d²r/dt², r = ∫v dt, v = ∫a dt; extend to 2 dimensions using vectors.

The idea

This topic connects calculus to kinematics. For straight-line motion, displacement r(t) gives velocity v = dr/dt and acceleration a = dv/dt = d²r/dt². Conversely, integrating acceleration gives velocity, and integrating velocity gives displacement, with constants found from initial conditions. In two dimensions, treat each component separately: r = xi + yj, v = ẋi + ẏj, a = ẍi + ÿj. For example, if a = 6ti + 2j, integrate to get v = 3t²i + 2tj + c; use v(0) to find c, then integrate again for r. Always check units and direction.

What earns the marks

• Correctly differentiate a given displacement vector to find velocity and acceleration, component by component.
• Integrate an acceleration vector to find velocity, including a constant vector determined from initial velocity.
• Integrate a velocity vector to find displacement, including a constant vector determined from initial position.
• Interpret scalar results in context, including direction and magnitude where required.
• Use calculus to solve problems involving variable acceleration in one or two dimensions.

Common mistakes and how to put them right

• Forgetting the constant of integration when finding velocity or displacement.

  Put it right: Use initial conditions to evaluate it.

• Differentiating or integrating only one component of a vector.

  Put it right: Apply the operation to each component separately.

• Confusing displacement, velocity and acceleration notation.

  Put it right: Ensure you know which function is being differentiated or integrated.

Exam tips

• Write down the given vector functions clearly and state what you are finding at each step.
• Check your answer by differentiating back or ensuring units are consistent.
• For two-dimensional problems, keep i and j components separate throughout your working.