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How can the equations of constant acceleration predict where a moving object will be and how fast it will be going?

Use the constant-acceleration (kinematic) equations from the reference sheet to solve for an unknown displacement, velocity, acceleration, or time in straight-line motion (MA STE Introductory Physics, Motion and Forces).

A standard-level answer on the kinematic equations for the Massachusetts High School Introductory Physics MCAS: the constant-acceleration relationships on the reference sheet, how to pick the right one, and how to solve for displacement, velocity, acceleration, or time.

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  1. What this topic is asking
  2. The constant-acceleration equations
  3. A method that always works
  4. Signs and direction
  5. Reference-sheet note
  6. Try this

What this topic is asking

Once an object has constant acceleration, its whole future motion is fixed, and the kinematic equations let you calculate it. The Massachusetts Introductory Physics MCAS gives you the constant-acceleration relationships on the reference sheet, so the skill being tested is choosing the right equation for the quantities you have and substituting correctly. This is the practice of using mathematics and computational thinking at the heart of mechanics.

The constant-acceleration equations

The kinematic equations describe an object whose acceleration does not change. The two printed on the Introductory Physics reference sheet are:

vf=vi+atv_f = v_i + at

d=vit+12at2d = v_i t + \tfrac{1}{2}at^2

where viv_i is the initial velocity, vfv_f is the final velocity, aa is the acceleration, tt is the time, and dd is the displacement. The first relates velocity, acceleration, and time; the second relates displacement, velocity, acceleration, and time. Many problems are solved with one of these, or with both used in turn.

A method that always works

The reliable approach for any kinematics problem is the same:

  1. List the quantities. Write down viv_i, vfv_f, aa, tt, and dd, filling in the values you are given. Convert to SI units first.
  2. Spot the hidden values. "Starts from rest" means vi=0v_i = 0. "Comes to a stop" means vf=0v_f = 0. "Constant velocity" means a=0a = 0.
  3. Identify the unknown. What is the question asking for?
  4. Choose the equation that contains your known quantities and the unknown, and no other unknown.
  5. Substitute and solve, then check the unit and whether the size is sensible.

This method turns a wordy problem into a short calculation, and it is exactly what the MCAS rewards in a constructed response: the equation, the substitution, and the answer.

Signs and direction

Because velocity and acceleration are vectors, signs matter. Choose a positive direction (usually the direction of initial motion). Then:

  • A velocity in the positive direction is positive; in the opposite direction, negative.
  • An acceleration that speeds the object up (same direction as motion) is positive; one that slows it down (opposite to motion), negative.

A braking car moving in the positive direction has a negative acceleration. Putting the minus sign in keeps the algebra honest and gives a sensible positive time when you solve for tt.

Reference-sheet note

The reference sheet prints vf=vi+atv_f = v_i + at and d=vit+12at2d = v_i t + \tfrac{1}{2}at^2 in the motion section. It does not print the time-free equation vf2=vi2+2adv_f^2 = v_i^2 + 2ad that some textbooks give, so when neither time is known you find the time from the two given equations first, as in the worked example. Knowing which relationships are on the sheet, and which you must work around, is part of the skill.

Try this

Q1. An object at rest accelerates at 4.04.0 m/s squared for 3.03.0 s. Find its final velocity. [2]

  • Cue. vf=vi+at=0+(4.0)(3.0)=12v_f = v_i + at = 0 + (4.0)(3.0) = 12 m/s.

Q2. A car travels at constant 2020 m/s for 8.08.0 s. How far does it go? [2]

  • Cue. With a=0a = 0, d=vit=(20)(8.0)=160d = v_i t = (20)(8.0) = 160 m.

Exam-style practice questions

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

MA Physics MCAS (style)3 marksA car starts from rest and accelerates at 3.03.0 m/s squared for 5.05.0 s. (a) Calculate the final velocity. (b) Calculate the distance traveled. Show the equation, substitution, and answer for each.
Show worked answer →

A 3-point constructed-response item using two reference-sheet kinematic equations.

(a) Final velocity (up to 2 points): use vf=vi+atv_f = v_i + at. Since it starts from rest, vi=0v_i = 0, so vf=0+(3.0)(5.0)=15v_f = 0 + (3.0)(5.0) = 15 m/s. (1 point equation, 1 point answer with unit.)
(b) Distance (1 point): use d=vit+12at2=0+12(3.0)(5.0)2=12(3.0)(25)=37.5d = v_i t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(3.0)(5.0)^2 = \tfrac{1}{2}(3.0)(25) = 37.5 m, which rounds to 3838 m.

Markers reward choosing equations that fit the given quantities and recognizing vi=0v_i = 0 for "starts from rest."

MA Physics MCAS (style)2 marksA cyclist moving at 1212 m/s brakes at a constant 2.02.0 m/s squared until stopping. Calculate the time taken to stop.
Show worked answer →

A 2-point calculation selecting the kinematic equation that links velocity, acceleration, and time.

Equation (1 point): rearrange vf=vi+atv_f = v_i + at to t=vf−viat = \dfrac{v_f - v_i}{a}.
Substitution and answer (1 point): the braking acceleration opposes the motion, so a=−2.0a = -2.0 m/s squared. t=0−12−2.0=−12−2.0=6.0t = \dfrac{0 - 12}{-2.0} = \dfrac{-12}{-2.0} = 6.0 s.

Markers reward a positive time and the recognition that "stopping" means vf=0v_f = 0. Treating the deceleration as negative gives the correct positive time.

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