Vector applications to kinematics practice questions.

A particle moves so that its position vector at time t (seconds) is r = (3 + 2t, 5 − t). State the velocity vector of the particle.

• (3, 5)
• (2, −1)
• (2 + 3t, −1 + 5t)
• (2, 1)

Answer: (2, −1)

Velocity is the derivative of the position vector: dr/dt = (2, −1). The constants (3, 5) are the initial position, not the velocity.

A boat has velocity vector v = (4, 3) in km h⁻¹. Calculate its speed.

• 7 km h⁻¹
• 12 km h⁻¹
• 5 km h⁻¹
• 25 km h⁻¹

Answer: 5 km h⁻¹

Speed is the magnitude of the velocity: √(4² + 3²) = √25 = 5 km h⁻¹. Adding components (7) or multiplying them (12) are common errors.

The position of a particle is r = (t², 2t) metres. Determine its velocity vector at t = 3 s.

• (9, 6)
• (6, 2)
• (3, 2)
• (2, 0)

Answer: (6, 2)

v = dr/dt = (2t, 2). At t = 3: (2×3, 2) = (6, 2). Option (9, 6) is the position at t = 3, not the velocity.

Two ships start from the same point. Ship A moves with velocity (2, 5) km h⁻¹ and ship B with velocity (7, 1) km h⁻¹. Determine the velocity of ship B relative to ship A.

• (9, 6)
• (5, −4)
• (−5, 4)
• (14, 5)

Answer: (5, −4)

Relative velocity of B with respect to A is v_B − v_A = (7 − 2, 1 − 5) = (5, −4). Reversing the subtraction gives (−5, 4).

A particle has acceleration a = (0, −2) m s⁻² and initial velocity (4, 6) m s⁻¹. Determine its velocity vector at t = 3 s.

• (4, 0)
• (4, 12)
• (12, 0)
• (4, −6)

Answer: (4, 0)

Integrating acceleration: v = (4 + 0×3, 6 + (−2)×3) = (4, 0). Only the y-component changes because acceleration acts vertically.

The velocity of a particle is v = (2t, 3) m s⁻¹. Determine its acceleration vector.

• (2, 3)
• (2, 0)
• (t², 3t)
• (0, 0)

Answer: (2, 0)

Acceleration = dv/dt = (2, 0). The 3 is constant so its derivative is 0; (2, 3) wrongly keeps the constant.