Kinematics (A2 Unit 4: Applied Mathematics B) — WJEC A-Level Mathematics
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Kinematics (A2 Unit 4: Applied Mathematics B) explained
Straight-line suvat formulas generalize directly to two dimensions under constant vector acceleration a.
Read the full explanation
The vector velocity equation is v = u + at. The vector displacement equation is r = ut + ½at², with position vector r = r₀ + ut + ½at² from initial position r₀. These equations can be derived by integrating constant vector acceleration a with respect to time t. Because horizontal and vertical vector components operate independently, motion can be analysed component-wise using i and j: v_x = u_x + a_x t and v_y = u_y + a_y t. Candidates find particle speed as the vector magnitude |v| and direction of motion using trigonometry.
Your focus
- Derive two-dimensional vector equations of motion under constant acceleration by integration.
- Apply vector suvat equations to compute position and velocity vectors at specified times.
- Determine speed and direction of travel from two-dimensional velocity vector components.
Kinematics (A2 Unit 4: Applied Mathematics B) exam tips
Marking Points
- applying vector suvat formulas v = u + at or r = r₀ + ut + ½at² in two dimensions
- deriving the vector equations by integrating constant acceleration a with respect to time
- calculating the speed of the particle by evaluating vector magnitude |v| = √(v_x² + v_y²)
- calculating the direction of motion or bearing from velocity components
Examiner Tips
- 💡Work in column vectors to keep i and j components organized and avoid sign errors.
- 💡Remember: 'speed' is the scalar magnitude of velocity √(v_x² + v_y²), while velocity is the vector itself.
Common Mistakes
- omitting the initial position vector r₀ when calculating position vectors from displacement
- attempting to use scalar suvat equations with vector quantities directly without component resolution