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New YorkPhysicsSyllabus dot point

How do the constant-acceleration equations link displacement, velocity, acceleration and time, and how do you choose the right one?

Apply the constant-acceleration kinematic equations to solve problems for displacement, initial and final velocity, acceleration and time, selecting the equation that omits the unknown not asked for.

A Regents Physics answer on the constant-acceleration kinematic equations: the four printed on the Reference Tables, what each one omits, how to choose the right equation, and how to solve one-dimensional motion problems, with worked examples.

Generated by Claude Opus 4.812 min answer

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  1. What this topic is asking
  2. The constant-acceleration equations
  3. Choosing the right equation
  4. Sign conventions
  5. Multi-stage motion
  6. Reference Tables note
  7. Try this

What this topic is asking

This is the calculation core of Regents kinematics. The Physical Setting/Physics course gives you a set of constant-acceleration (uniform-acceleration) equations that link the five quantities of straight-line motion, displacement dd, initial velocity viv_i, final velocity vfv_f, acceleration aa and time tt, and the exam asks you to pick the right one and solve. The four equations are printed on the Reference Tables, so the skill is selection and substitution, not memorizing.

The constant-acceleration equations

Each of the first three equations leaves out exactly one of the five quantities:

  • vf=vi+atv_f = v_i + at contains no displacement.
  • d=vit+12at2d = v_i t + \tfrac{1}{2}at^2 contains no final velocity.
  • vf2=vi2+2adv_f^2 = v_i^2 + 2ad contains no time.

That structure is the key to choosing quickly.

Choosing the right equation

For example, if a problem gives viv_i, aa and tt and asks for dd, the final velocity vfv_f is not involved, so use d=vit+12at2d = v_i t + \tfrac{1}{2}at^2. If it gives viv_i, vfv_f and dd and asks for aa, time is not involved, so use vf2=vi2+2adv_f^2 = v_i^2 + 2ad. Reading off which quantity is missing is faster than trying equations at random.

Sign conventions

Because velocity, displacement and acceleration are vectors in one dimension, choose a positive direction and apply it consistently. An object slowing down has an acceleration opposite to its motion, so aa is negative if motion is positive. Starting from rest means vi=0v_i = 0; coming to a stop means vf=0v_f = 0. Writing the knowns with their signs before substituting prevents most errors.

Multi-stage motion

If the acceleration changes (a car accelerates, then cruises at constant speed, then brakes), the equations apply to each stage separately, because each requires constant acceleration. Solve one stage, carry the final velocity and position forward as the initial values of the next, and add the displacements. A constant-velocity stage uses d=vtd = vt (zero acceleration).

Reference Tables note

All the kinematic equations you need are printed in the Mechanics section of the Reference Tables, so you should never have to recall them. What you bring is the method: identify the missing quantity, choose the matching equation, and substitute with correct signs and units. The averaging relation vˉ=12(vi+vf)\bar{v} = \tfrac{1}{2}(v_i + v_f) is a quick check, since dd should equal the average velocity times the time.

Try this

Q1. An object accelerates from 4.04.0 m/s at 2.02.0 m/s squared for 3.03.0 s. Calculate its final velocity. [2 points]

  • Cue. vf=vi+at=4.0+(2.0)(3.0)=10.v_f = v_i + at = 4.0 + (2.0)(3.0) = 10. m/s.

Q2. State which kinematic equation you would use to find displacement when given viv_i, aa and tt but not vfv_f. [1 point]

  • Cue. d=vit+12at2d = v_i t + \tfrac{1}{2}at^2 (it omits the final velocity).

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart B-2 (constructed response). A car starts from rest and accelerates uniformly at 3.03.0 m/s squared for 5.05.0 s. Calculate the distance it travels. Show the equation, substitution and answer.
Show worked answer →

A 2-point constructed-response calculation using a Reference-Table kinematic equation.

Equation: with vi=0v_i = 0, use d=vit+12at2d = v_i t + \tfrac{1}{2}at^2 (printed on the tables).
Substitution: d=(0)(5.0)+12(3.0)(5.0)2=12(3.0)(25)d = (0)(5.0) + \tfrac{1}{2}(3.0)(5.0)^2 = \tfrac{1}{2}(3.0)(25).
Answer: d=37.5d = 37.5 m, which rounds to 3838 m.

Markers reward the equation from the tables, correct substitution with units, and the final answer. A common error is using d=vtd = vt with the final velocity, which is wrong because the velocity is not constant.

Regents (style)3 marksPart C (extended response). A motorcycle travelling at 2020 m/s brakes uniformly and stops in a distance of 5050 m. (a) Calculate the acceleration. (b) Calculate the time taken to stop. Show all work.
Show worked answer →

A 3-point Part C item combining two kinematic equations.

(a) Acceleration (2 points): the unknown not given is time, so use vf2=vi2+2adv_f^2 = v_i^2 + 2ad. With vf=0v_f = 0: 0=(20)2+2a(50)0 = (20)^2 + 2a(50), so a=−400100=−4.0a = \dfrac{-400}{100} = -4.0 m/s squared (the negative sign shows deceleration).
(b) Time (1 point): use vf=vi+atv_f = v_i + at: 0=20+(−4.0)t0 = 20 + (-4.0)t, so t=204.0=5.0t = \dfrac{20}{4.0} = 5.0 s.

Markers reward choosing the equation that omits the unknown, correct substitution including the sign of aa, and the time from a second equation.

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