2.4 Constant Acceleration and the 5 Kinematic Equations

Constant acceleration is the engine of every kinematics problem you will meet in high school and university physics. Once acceleration is constant, position becomes a parabola, velocity becomes a straight line, and a small set of five equations connects everything you know about displacement, time, initial velocity, final velocity, and acceleration. Mastering how those equations are derived, when they are valid, and which one to choose when a variable is missing is what separates students who plug numbers from students who solve AP, A-Level, IB HL, and Olympiad problems quickly and correctly. In this definitive guide we rebuild the entire constant-acceleration framework from its graphical meaning to its calculus proof, and we work through the complete autonomous-car passing problem as it is developed in Fundamentals of Physics.

Key Insights

• Constant acceleration makes average and instantaneous acceleration equal.
• Five kinematic equations describe motion when acceleration is constant.
• Each equation is missing one variable, so choose by what is not given.
• Position versus time is quadratic, velocity versus time is linear, acceleration versus time is flat.
• Integration of constant acceleration gives \(v\) = \(v_0\) + \(at\) and \(x\) - \(x_0\) = \(v_0 t\) + 1/2\(at^2\).

Constant Acceleration

Constant acceleration is a special but extremely common case. In many types of motion the acceleration is either constant or approximately so. When a traffic light turns green you might accelerate a car at an approximately constant rate. Later when you brake the car to a stop the acceleration, or deceleration in common language, might also be approximately constant.

The entire approach rests on a warning you must carry into every homework and exam: the equations in this module are valid only for constant acceleration, or for situations in which you can approximate the acceleration as being constant. If acceleration varies with time, these equations are not valid.

Constant Acceleration: A Special Case

Three stacked 3D graphs for constant acceleration - position quadratic curve labeled slope varies, velocity linear line labeled slope = a, acceleration horizontal constant line labeled slope = 0, on clean white background

When acceleration is constant, the graphical family looks like this:

• The position curve \(x\)(\(t\)) has a slope that varies with time. Because the slope of position is velocity and velocity is changing, the position graph is curved, specifically a parabola.
• The velocity curve \(v\)(\(t\)), plotted with velocity on the vertical axis and time on the horizontal, has a constant slope because acceleration is the slope of velocity. The fact that \(a\)(\(t\)) is constant requires that \(v\)(\(t\)) be linear. Its slope is equal to a.
• The acceleration curve \(a\)(\(t\)) is a horizontal line. Its slope is zero because \(a\) does not change.

This is the visual signature you must recognize instantly in multiple-choice questions. Curved \(x\)- \(t\), straight \(v\)- \(t\), flat \(a\)- \(t\) means constant \(a\).

Callout Box:The Validity Check: Before you write down any of the five equations below, ask: Is \(a\) constant? If \(x\)(\(t\)) is given as \(x\) = -5\(t^3\) + 4\(t^2\) + 6, the \(t^3\) term means acceleration depends on time, so constant-acceleration equations do not apply. If \(x\)(\(t\)) = 3\(t\) - 4 or \(x\)(\(t\)) = 5\(t^2\) - 3, the highest power is \(t^1\) or \(t^2\), which is consistent with constant acceleration.

First Basic Equation

When acceleration is constant, the average acceleration and the instantaneous acceleration are equal. We can start from the definition of average acceleration

\(a_{avg}\) = (\(v_2\) - \(v_1\)) / (\(t_2\) - \(t_1\))

and set \(t_1\) = 0, \(v_1\) = \(v_0\) as the velocity at time zero, and \(t_2\) = \(t\), \(v_2\) = \(v\) as the velocity at any later time. Because \(a_{avg}\) = \(a\) for constant acceleration,

\(a\) = (\(v\) - \(v_0\)) / (\(t\) - 0)

Rearranged, this gives the first basic equation:

\(v\) = \(v_0\) + \(at\)

As a check, this reduces to \(v\) = \(v_0\) when \(t\) = 0, as it must. As a further check, take the derivative of this equation. Doing so yields \(dv\) / \(dt\) = \(a\), which is the definition of acceleration. The \(v\)(\(t\)) function is linear and thus the plot of velocity versus time is a straight line.

Exam Tip:\(v\) = \(v_0\) + \(at\) is your time-to-velocity converter. If you know \(v_0\), \(a\), and \(t\), you get \(v\) immediately without needing displacement. It is also the only equation that contains no \(x\) - \(x_0\) term.

Second Basic Equation

We can do the same for average velocity. The definition of average velocity is

\(v_{avg}\) = (\(x\) - \(x_0\)) / (\(t\) - 0)

where \(x_0\) is the position of the particle at \(t\) = 0 and \(v_{avg}\) is the average velocity between \(t\) = 0 and a later time \(t\). Rearranged,

\(x\) = \(x_0\) + \(v_{avg}t\)

For the linear velocity function \(v\) = \(v_0\) + \(at\), the average velocity over any time interval from \(t\) = 0 to a later time \(t\) is simply the average of the velocity at the beginning of the interval and the velocity at the end of the interval. This is a special property of a straight line.

For the interval from 0 to \(t\), the average velocity is

\(v_{avg}\) = 1/2(\(v_0\) + \(v\))

Substituting the right side of \(v\) = \(v_0\) + \(at\) for \(v\) yields, after a little rearrangement,

\(v_{avg}\) = \(v_0\) + 1/2\(at\)

Finally, substituting this expression into \(x\) = \(x_0\) + \(v_{avg}t\) yields the second basic equation:

\(x\) - \(x_0\) = \(v_0t\) + 1/2\(at^2\)

As a check, putting \(t\) = 0 yields \(x\) = \(x_0\), as it must. As a further check, taking the derivative of this equation yields \(v_0\) + \(at\), which is the first basic equation, as it must. The function is quadratic and thus the plot of position versus time is curved, a parabola.

Callout Box:Why Average of Initial and Final Works Only Here: \(v_{avg}\) = 1/2(\(v_0\) + \(v\)) is true only when acceleration is constant and therefore velocity versus time is a straight line. If acceleration varies, the average is not the simple arithmetic mean of endpoints. This is a favorite AP and IB trap.

Three Other Equations

Equations \(v\) = \(v_0\) + \(at\) and \(x\) - \(x_0\) = \(v_0t\) + 1/2\(at^2\) are the basic equations for constant acceleration. They can be used to solve any constant acceleration problem. However, we can derive other equations that might prove useful in certain specific situations.

First, note that as many as five quantities can possibly be involved in any problem about constant acceleration: displacement \(x\) - \(x_0\), final velocity \(v\), elapsed time \(t\), acceleration \(a\), and initial velocity \(v_0\). Usually, one of these quantities is not involved in the problem, either as a given or as an unknown. We are then presented with three of the remaining quantities and asked to find the fourth.

The two basic equations each contain four of these quantities, but not the same four. In the first basic equation, the missing ingredient is the displacement \(x\) - \(x_0\). In the second basic equation, the missing ingredient is the final velocity \(v\). These two equations can also be combined in three ways to yield three additional equations, each of which involves a different missing variable.

Elimination of time \(t\) gives:

\(v^2\) = \(v_0^2\) + 2\(a\)(\(x\) - \(x_0\))

This equation is useful if we do not know \(t\) and are not required to find it. It directly connects velocity and displacement.

Elimination of acceleration \(a\) between the two basic equations gives an equation in which \(a\) does not appear:

\(x\) - \(x_0\) = 1/2(\(v_0\) + \(v\))\(t\)

This is especially clean when you know both velocities and time but not acceleration.

Elimination of initial velocity \(v_0\) gives:

\(x\) - \(x_0\) = \(vt\) - 1/2\(at^2\)

Note the subtle difference between this and \(x\) - \(x_0\) = \(v_0t\) + 1/2\(at^2\). One involves the initial velocity \(v_0\); the other involves the velocity \(v\) at time \(t\).

Five glossy 3D cards floating on white background showing the five constant acceleration equations each labeled with missing variable

Equation	Form	Missing Quantity	When to Use
\(v\) = \(v_0\) + \(at\)	velocity-time	\(x\)-\(x_0\)	No displacement given or asked
\(x\) - \(x_0\) = \(v_0t\) + 1/2\(at^2\)	displacement-time with \(v_0\)	\(v\)	No final velocity given or asked
\(v^2\) = \(v_0^2\) + 2\(a\)(\(x\) - \(x_0\))	velocity-displacement	\(t\)	No time given or asked
\(x\) - \(x_0\) = 1/2(\(v_0\) + \(v\))\(t\)	displacement-time with average	\(a\)	No acceleration given or asked
\(x\) - \(x_0\) = \(vt\) - 1/2\(at^2\)	displacement-time with \(v\)	\(v_0\)	No initial velocity given or asked

Exam tip:Do not memorize all five separately. Remember only the first two. The other three are algebra. If you can remember \(v\) = \(v_0\) + \(at\) and \(x\) - \(x_0\) = \(v_0t\) + 1/2\(at^2\), you can derive the rest in 30 seconds, which is safer under pressure.

Exam Trap:Using these equations when \(a\) is not constant. Checkpoint: \(x\) = 3\(t\) - 4 is linear, so \(v\) constant, \(a\) = 0, constant acceleration equations apply (with \(a\) = 0). \(x\) = 5\(t^2\) - 3 is quadratic, so a constant, applies. \(x\) = -5\(t^3\) + 4\(t^2\) + 6 has \(t^3\), so \(a\) depends on time, does NOT apply. \(x\) = 2/\(t^2\) - 4/\(t\) is not polynomial, does NOT apply.

Sample Problem: Autonomous Car Passing Slower Car

You are riding in a car controlled by an autonomous driving system and trail a slower car that you want to pass. Both cars have length \(L\) = 4.50 m, speed \(v_0\) = 22.0 m/s (49 mi/h, slower than the speed limit), and travel on a straight road with one lane in each direction. Your car initially trails \(A\) by distance 3.00\(L\) when you ask it to pass the slow car. That would require you to move into the other lane where there could be an oncoming vehicle. Your system must determine the time required for passing \(A\), to see if passing would be safe. So, the first step in the system's control is to calculate that passing time.

We want \(B\) to pull into the other lane, accelerate at a constant \(a\) = 3.50 m/s² until it reaches a speed of \(v\) = 27.0 m/s (60 mi/h, the speed limit) and then, when it is 3.00\(L\) ahead of \(A\), pull back into the initial lane. It will then maintain 27.0 m/s. Assume the lane changing takes negligible time. The situation at the onset of the acceleration, with the rear of car \(B\) at \(x_{B1}\) = 0 and the rear of car \(A\) at \(x_{A1}\) = 4\(L\), shows the initial situation.

We define \(t_1\) and \(d_1\) be the time required for the acceleration and the distance traveled during the acceleration. Let \(t_2\) be the time from the end of the acceleration to when B is ahead of A by 3\(L\) and ready to pull back. We want the total time \(t_{tot}\) = \(t_1\) + \(t_2\).

(a) What are the values of \(t_1\) and \(d_1\)? (b) In terms of \(L\), \(v_0\), \(t_1\), and \(t_2\), what is the coordinate \(x_{B2}\) of the rear of car \(B\) when \(B\) is ready to pull back? (c) In terms of \(L\), \(v_0\), \(t_1\), and \(t_2\), what is the coordinate \(x_{A2}\) of the rear of car \(A\) just then? (d) What is \(x_{B2}\) in terms of \(x_{A2}\) and \(L\)? Putting the pieces together, find the values of (f) \(t_2\) and (g) \(t_{tot}\).

We can apply the equations of constant acceleration to both stages of passing: when car \(B\) has acceleration \(a\) = 3.50 m/s² and when it travels at constant speed and thus with constant \(a\) = 0.

Calculations: In the passing lane, \(B\) accelerates at constant rate \(a\) = 3.50 m/s² from initial speed \(v_0\) = 22.0 m/s to final speed \(v\) = 27.0 m/s. From \(v\) = \(v_0\) + \(at\),

\(t_1\)	= (\(v\) - \(v_0\)) / \(a\)
	= [(27.0 m/s) - (22.0 m/s)] / 3.50m/s2
	= 1.4285 s ≈ 1.43 s

The distance \(d_1\) traveled by \(B\) during acceleration from \(v^2\) = \(v_0^2\) + 2\(a d_1\),

\(d_1\)	= (\(v^2\) - \(v_0^2\)) / \(2a\)
	= [(27.0)2 - (22.0)2] / 2(3.50)
	= 35.0 m

After acceleration through displacement \(d_1\) from \(x_{B1}\) = 0, the rear of car \(B\) moves at constant speed \(v\) for the unknown time \(t_2\). Its position is then \(x_{B2}\) = \(d_1\) + \(v t_2\).

From its initial position \(x_{A1}\) = 4\(L\), the rear of car \(A\) moves at constant speed \(v_0\) for total time \(t_1\) + \(t_2\), so \(x_{A2}\) = 4\(L\) + \(v_0\)(\(t_1\) + \(t_2\)).

The rear of \(B\) is then 3\(L\) from the front of \(A\) and thus 4\(L\) from the rear of \(A\), so \(x_{B2}\) = \(x_{A2}\) + 4\(L\).

Putting pieces together, \(d_1\) + \(v t_2\) = 4\(L\) + \(v_0\)(\(t_1\) + \(t_2\)) + 4\(L\), which gives

\(t_2\)(\(v\) - \(v_0\)) = 8\(L\) + \(v_0 t_1\) - \(d_1\)

\(t_2\)	= (8\(L\) + \(v_0 t_1\) - \(d_1\)) / (\(v\) - \(v_0\))
	= [8(4.50) + (22.0)(1.4285) - 35.0] / (27.0 - 22.0)
	= 6.4854 s ≈ 6.49 s

The total time is \(t_{tot}\) = \(t_1\) + \(t_2\) = 1.4285 + 6.4854 = 7.91 s. As explored in one of the end-of-chapter problems, the next step for your car's control system is to detect the speed and distance of any oncoming car, to see if this much time is safe.

Realistic top-view 3D cars on road showing autonomous passing scenario with distances L, 3L, 4L and velocities v0 and v and acceleration a

Another Look at Constant Acceleration

Two 3D graphs side by side showing integration of acceleration - a(t) constant area under curve equals change in velocity, and v(t) linear area under curve equals displacement, with equations dv = a dt and dx = v dt

The first two equations in the table of five are the basic equations from which the others are derived. Those two can be obtained by integration of the acceleration with the condition that \(a\) is constant.

To find \(v\) = \(v_0\) + \(at\), we rewrite the definition of acceleration \(a\) = \(dv\) / \(dt\) as

\(dv\) = \(a\,dt\)

We next write the indefinite integral, or antiderivative, of both sides:

∫\(dv\) = ∫\(a\,dt\)

Since acceleration \(a\) is a constant, it can be taken outside the integration:

∫\(dv\) = \(a\) ∫\(dt\)

or

\(v\) = \(at\) + \(C\)

where \(C\) is a constant of integration. To evaluate \(C\), we let \(t\) = 0, at which time \(v\) = \(v_0\). Substituting these values into \(v\) = \(at\) + \(C\) yields \(v_0\) = \(a\)(0) + \(C\) = \(C\). Substituting this into \(v\) = \(at\) + \(C\) gives us the first basic equation.

To derive \(x\) - \(x_0\) = \(v_0t\) + 1/2\(at^2\), we rewrite the definition of velocity \(v\) = \(dx\) / \(dt\) as

\(dx\) = \(v\,dt\)

and then take the indefinite integral of both sides:

∫\(dx\) = ∫\(v\,dt\)

Next, we substitute for \(v\) with \(v_0\) + \(at\):

∫\(dx\) = ∫(\(v_0\) + \(at\)) \(dt\)

Since \(v_0\) is a constant, as is the acceleration \(a\), this can be rewritten as

∫\(dx\) = \(v_0\) ∫\(dt\) + \(a\) ∫\(t\,dt\)

Integration now yields

\(x\) = \(v_0t\) + 1/2\(at^2\) + \(C'\)

where \(C'\) is another constant of integration. At time \(t\) = 0, we have \(x\) = \(x_0\). Substituting these values yields \(x_0\) = \(C'\). Replacing \(C'\) with \(x_0\) gives us the second basic equation.

Concept	Algebraic Derivation	Integral Derivation	Meaning
Change in velocity	\(a\) = (\(v\) - \(v_0\))/\(t\)	∫\(dv\) = ∫\(a\,dt\) → \(v\) = \(at\) + \(C\)	Area under \(a\)- \(t\) graph
Change in position	\(x\) = \(x_0\) + \(v_{avg}t\)	∫\(dx\) = ∫\(v\,dt\) → \(x\) = \(v_0t\) + 1 / 2\(at^2\) + \(C'\)	Area under \(v\)- \(t\) graph
Constant \(C\) evaluation	Use initial condition \(t\) = 0, \(v\) = \(v_0\)	\(C\) = \(v_0\)	Initial velocity
Constant \(C'\) evaluation	Use initial condition \(t\) = 0, \(x\) = \(x_0\)	\(C'\) = \(x_0\)	Initial position

Exam Tip:Integration view gives you two fast checks: The change in velocity equals the area under \(a\) versus \(t\). The displacement equals the area under \(v\) versus \(t\). If you forget a formula, sketch the graph and compute the area.

Exam Trap:Forgetting to evaluate constants of integration. Many students write \(v\) = \(at\) and drop \(v_0\). The constant \(C\) is not zero unless \(v_0\) = 0. Always apply \(t\) = 0 condition to find \(C\) and \(C'\).

Practice Questions

MCQs (Multiple Choice Questions)

01. Which condition is required for the five kinematic equations to be valid?

1. Velocity is constant
2. Acceleration is constant
3. Displacement is zero
4. Time is zero

Correct: B
They are derived assuming \(a\) constant. If \(a\) varies, they fail.

02. For constant acceleration, \(x\)(\(t\)) is:

1. Linear
2. Quadratic
3. Cubic
4. Constant

Correct: B
\(x\) - \(x_0\) = \(v_0t\) + 0.5 \(at^2\), parabola.

03. For constant acceleration, \(v\)(\(t\)) is:

1. Quadratic
2. Linear
3. Flat
4. Cubic

Correct: B
\(v\) = \(v_0\) + \(at\), straight line with slope \(a\).

04. Which equation has no \(t\)?

1. \(v\) = \(v_0\) + \(at\)
2. \(v^2\) = \(v_0^2\) + 2\(a\)(\(x\) - \(x_0\))
3. \(x\) - \(x_0\) = \(v_0t\) + 0.5\(at^2\)
4. \(x\) - \(x_0\) = 0.5(\(v_0\) + \(v\))\(t\)

Correct: B
Missing \(t\), useful when time not given.

05. A car accelerates from 10 m/s to 20 m/s at 2 m/s². Time taken:

1. 5 s
2. 10 s
3. 2 s
4. 20 s

Correct: A
\(t\) = (20 - 10)/2 = 5 s.

06. Using \(v^2\) = \(v_0^2\) + \(2aΔx\), if \(v_0\) = 0, \(a\) = 3, \(Δx\) = 6, \(v\) is:

1. 6 m/s
2. 9 m/s
3. 36 m/s
4. 18 m/s

Correct: A
\(v^2\) = 0 + 2 × 3 × 6 = 36, v = 6.

07. Which \(x\)(\(t\)) does NOT allow constant acceleration equations?

1. \(x\) = 3\(t\) - 4
2. \(x\) = 5\(t^2\) - 3
3. \(x\) = -5\(t^3\) + 4\(t^2\) + 6
4. \(x\) = 2\(t^2\) + 5\(t\)

Correct: C
\(t^3\) term makes \(a\) time-dependent.

08. In the autonomous car example, \(d_1\) = 35.0 m is found via:

1. \(v\) = \(v_0\) + \(at\)
2. \(v^2\) = \(v_0^2\) + 2\(ad_1\)
3. \(x\) = \(vt\) - 0.5\(at^2\)
4. Area under \(a\)- \(t\)

Correct: B
Distance during acceleration without \(t_2\).

09. ∫\(a\,dt\) with constant \(a\) equals:

1. \(a\)
2. \(at\) + \(C\)
3. 0.5\(at^2\)
4. \(v_0\)

Correct: B
Integration gives \(v\) = \(at\) + \(C\), \(C\) = \(v_0\).

10. The total passing time in the sample was approximately:

1. 1.43 s
2. 6.49 s
3. 7.91 s
4. 35 s

Correct: C
\(t_{tot}\) = \(t_1\) + \(t_2\) = 1.4285 + 6.4854 = 7.91 s.

FRQs (Free Response Questions)

FRQ 1: (AP Style)
A particle starts at \(x_0\) = 0 with \(v_0\) = 4 m/s and \(a\) = 2 m/s² constant.
(a) Find \(v\) at \(t\) = 3 s.
(b) Find \(x\) at \(t\) = 3 s.
(c) Which equation would you use if you did not know \(t\)?

Rubric:

• (a) \(v\) = \(v_0\) + \(at\) = 4 + 6 = 10 m/s (2 pts)
• (b) \(x\) = \(v_0t\) + 0.5\(at^2\) = 12 + 9 = 21 m (2 pts)
• (c) \(v^2\) = \(v_0^2\) + 2\(a\)(\(x\) - \(x_0\)) (1 pt)

FRQ 2: (A-Level Style)
Show that \(x\) - \(x_0\) = 1/2(\(v_0\) + \(v\))\(t\) is valid only for constant acceleration. A car traveling 22 m/s accelerates to 27 m/s in 1.43 s, find distance covered during acceleration using this equation.

Rubric:

• Derivation requires \(v_{avg}\) = 0.5(\(v_0\) + \(v\)) which holds only for linear \(v\)(\(t\)) i.e. constant \(a\) (2 pts)
• Distance: 0.5(22 + 27)*1.4285 = 35.0 m (2 pts)
• Units correct (1 pt)

FRQ 3: (IB HL Style)
Derive \(v\) = \(v_0\) + \(at\) by integration starting from \(dv\) = \(a\,dt\). State the initial condition used.

Rubric:

• ∫\(dv\) = ∫\(a\,dt\), \(a\) constant outside integral (2 pts)
• \(v\) = \(at\) + \(C\) (1 pt)
• At \(t\) = 0, \(v\) = \(v_0\) so \(C\) = \(v_0\) (2 pts)

FRQ 4: (College Physics)
In the passing problem, explain why \(t_2\) formula is \(t_2\) = (8\(L\) + \(v_0t_1\) - \(d_1\)) / (\(v\) - \(v_0\)). What does 8\(L\) represent?

Rubric:

• 8\(L\) = 4\(L\) initial separation plus 4\(L\) final separation requirement, \(v_0t_1\) distance A moves during B's acceleration (2 pts)
• \(d_1\) distance B already covered, denominator \(v\) - \(v_0\) relative speed after acceleration (2 pts)
• Shows understanding of relative motion (1 pt)

FRQ 5: (USAPhO Style)
A particle has \(a\) = 3 m/s² constant. Its velocity changes from 2 m/s to 8 m/s. Find \(Δx\) without finding time first, then find time via integration reasoning.

Rubric:

• Use \(v^2\) = \(v_0^2\) + 2\(aΔx\) → 64 = 4 + 6 \(Δx\) → \(Δx\) = 10 m (2 pts)
• Time: \(v\) = \(v_0\) + \(at\) → 6 = 3\(t\) → \(t\) = 2 s (1 pt)
• Mention area under \(a\)- \(t\) equals \(Δv\) = 6 m/s, so rectangle 3*2 (2 pts)

Frequently Asked Questions

What are the five constant acceleration equations?

They are \(v\) = \(v_0\) + \(at\), \(x\) - \(x_0\) = \(v_0t\) + 1 / 2 \(at^2\), \(v^2\) = \(v_0^2\) + 2\(a\)(\(x\) - \(x_0\)), \(x\) - \(x_0\) = 1 / 2 (\(v_0\) + \(v\))\(t\), and \(x\) - \(x_0\) = \(vt\) - 1 / 2 \(at^2\). Each is valid only when \(a\) is constant.

When can I use v_avg = 1 / 2 (v₀ + v)?

Only when acceleration is constant and velocity versus time is a straight line. Then the average over any interval is the arithmetic mean of initial and final velocities. If acceleration varies, this fails.

How do I choose which kinematic equation to use?

List what you know: \(x\) - \(x_0\), \(v_0\), \(v\), \(a\), \(t\). Identify which one you are not given and not asked for. Pick the equation that is missing that variable. For example, if \(t\) is missing, use \(v^2\) = \(v_0^2\) + 2\(aΔx\).

Why is position quadratic when acceleration is constant?

Because \(v\) grows linearly as \(v_0\) + \(at\). Integrating \(v\) with respect to time gives \(x\) = \(x_0\) + \(v_0t\) + 1 / 2 \(at^2\). The \(t^2\) term makes \(x\) versus \(t\) a parabola.

How do you derive v = v₀ + at by integration?

Start with \(dv\) = \(a\,dt\). Integrate: ∫\(dv\) = \(a\,∫dt\), so \(v\) = \(at\) + \(C\). Apply initial condition at \(t\) = 0, \(v\) = \(v_0\), so \(C\) = \(v_0\), giving \(v\) = \(v_0\) + \(at\).

What does the area under a velocity-time graph give?

Displacement \(x\) - \(x_0\). For constant acceleration, the area is a trapezoid with average height 1 / 2 (\(v_0\) + \(v\)) times base \(t\), which yields \(x\) - \(x_0\) = 1 / 2 (\(v_0\) + \(v\))\(t\).

Can I use these equations if acceleration is zero?

Yes, zero is a constant. Then \(v\) = \(v_0\), and \(x\) - \(x_0\) = \(v_0t\). This is uniform motion, a special case of constant acceleration.

Why does braking distance depend on v²?

From \(v^2\) = \(v_0^2\) + 2\(aΔx\), if final \(v\) = 0, then Δx = - \(v_0^2\) / (2\(a\)). Since \(a\) is negative for braking, distance scales with initial speed squared, explaining why faster speeds need much longer stopping distance.

Exam Mapping

Concept	A-Level Physics	AP Physics 1 & C	IB Physics HL	College Physics 1	USAPhO / IPhO
Five equations, choose by missing variable	Kinematics in straight line, SUVAT equations, selection technique	Unit 1 Kinematics, selecting equation by knowns/unknowns, derivation not required but application critical	Topic 2.1 Kinematics, use of \(v\) = \(v_0\) + \(at\) and \(s\) = \(v_0t\) + 1 / 2 \(at^2\)	Chapter 2, algebraic and graphical methods, constant \(a\) only	Rapid selection, combined with relative motion
\(x\)(\(t\)) quadratic, \(v\)(\(t\)) linear, \(a\)(\(t\)) constant graphs	Graphical interpretation, gradient and area	Interpretation of \(x\)- \(t\), \(v\)- \(t\), \(a\)- \(t\) graphs, slope = \(a\)	Graphing and data analysis, IA	Lab: motion sensor graphs	Graphical reasoning under time pressure
Integration derivation \(dv\) = \(a\,dt\), \(dx\) = \(v\,dt\)	A-Level Maths link, optional calculus	AP C required: integration of \(a\) to get \(v\), \(v\) to get \(x\), initial conditions	HL calculus extension, indefinite integrals with constants	Calculus-based physics, derivation of kinematics	Fundamental calculus tool for variable acceleration extensions
Autonomous car passing with two phases (accelerated + constant velocity)	Relative motion, two-stage problems, safe overtaking distance	Multi-stage kinematics, relative motion, system analysis	Complex kinematics, overtaking problems	Word problems with multiple intervals	Relative motion plus safety time analysis, combined stages
Checkpoint: power of \(t\) tells if \(a\) constant	Recognizing uniform acceleration from \(x\)(\(t\)) polynomial degree	Conceptual check: which \(x\)(\(t\)) allows SUVAT	Nature of science: model limits	Identifying when model applies	Model applicability check

Conclusion

Constant acceleration gives you a complete algebraic and graphical toolkit. You now recognize the signature of curved position, straight velocity, and flat acceleration graphs, you can derive the first two equations from average velocity and acceleration and recover all five by eliminating the variable you do not need, and you can re-derive the same pair by integration with the constants \(C\) = \(v_0\) and \(C'\) = \(x_0\). You have seen how the missing-variable table guides equation choice and how a real two-stage problem, the autonomous car overtaking a slower car, splits into an accelerated phase \(t_1\) = (\(v\) - \(v_0\))/\(a\) and \(d_1\) = (\(v^2\) - \(v_0^2\))/2\(a\) and a constant-speed phase that yields \(t_2\) = (8\(L\) + \(v_0t_1\) - \(d_1\))/(\(v\) - \(v_0\)) and a total of about 7.91 seconds. With this foundation, you can handle free fall, projectile motion, and any exam problem that asks you to pick the right equation rather than remember all five.

References

Fundamentals of Physics by Halliday, Resnick, and Walker, Latest Edition, Module 2.4 Constant Acceleration: A Special Case and Another Look at Constant Acceleration - Primary source for five kinematic equations, graphical interpretation, Table of Equations for Motion with Constant Acceleration, checkpoint on \(x\)(\(t\)) forms, and sample problem autonomous car passing slower car with complete calculations.