Questions: Rectilinear Kinematics of Particles

Question 1

A particle starts with velocity v(0) = 2 m/s and has acceleration a = 6t m/s². What is the velocity at t = 3 s?

• A. 20 m/s
• B. 29 m/s
• C. 56 m/s
• D. 18 m/s

Answer: (B)

Since a = 6t is a function of time, integrate: v(t) = ∫6t dt = 3t² + C. Applying the initial condition v(0) = 2 gives C = 2, so v(t) = 3t² + 2. At t = 3: v = 3(9) + 2 = 29 m/s. A common error is to use the constant-acceleration formula v = v₀ + at with a single value of a, which is only valid when a is constant.

Question 2

A particle travels 40 m to the right, then reverses and travels 10 m to the left. Its displacement is 30 m and its total distance traveled is also 30 m.

• T. True
• F. False

Answer: False

Displacement is the net change in position (signed): 40 - 10 = 30 m to the right. Total distance traveled is the total path length regardless of direction: 40 + 10 = 50 m. These are equal only when the particle never reverses direction. Confusing displacement and distance is one of the most common errors in kinematics.

Question 3

When should you use the identity a = v dv/dx instead of integrating a(t) with respect to time?

Model answer: Use a = v dv/dx when acceleration is given as a function of position, a(x). In that case, integrating a dt is not possible directly because t is not the independent variable in the expression for a. The chain rule identity a = v dv/dx converts the problem into a separable ODE in x, which can be integrated to find v(x).

The three-case framework is central to rectilinear kinematics: (1) a = constant → use kinematic equations; (2) a = f(t) → integrate with respect to time; (3) a = f(x) → apply a = v dv/dx to get v as a function of x. The identity comes from a = dv/dt = (dv/dx)(dx/dt) = v dv/dx by the chain rule.