Lesson 1.4 · Unit 1 · CED topic 1.3 (equations)
The constant-acceleration equations
Graphs show motion; equations let you predict it. If you know where an object starts, how fast it is moving and how it is accelerating, three short equations tell you where it will be and how fast it will be moving at any later time. They apply whenever the acceleration is constant, which covers a car braking steadily, a rock falling near the ground, a sled sliding down a smooth slope, and many more situations you will meet this year. The skill is not remembering the equations, which are printed on the AP reference sheet, but choosing the right one and keeping the signs straight. This lesson builds the equations from the graphs of lesson 1.3 so that none of them is a mystery.
Describe the position, velocity, and acceleration of an object using representations of that object’s motion.
- For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension: \(v_x = v_{x0} + a_x t\), \(x = x_0 + v_{x0}t + \tfrac12 a_x t^2\), and \(v_x^2 = v_{x0}^2 + 2a_x(x - x_0)\).
- The equations above are written to indicate motion in the \(x\)-direction, but these equations can be used in any single dimension as appropriate.
Where the equations come from
Start with a velocity graph for constant acceleration (Figure 1.4a). The graph is a straight line that begins at \(v_{x0}\) and ends at \(v_x\) after a time \(t\). Its slope is the acceleration, so the rise, \(v_x - v_{x0}\), equals \(a_x t\). Rearranging that one statement gives the first equation: \[ v_x = v_{x0} + a_x t. \] The displacement is the area under the line. That area is a rectangle of height \(v_{x0}\) and width \(t\), plus a triangle of base \(t\) and height \(a_x t\), so \(\Delta x = v_{x0}t + \tfrac12 (a_x t)\,t\). This is the second equation: \[ x = x_0 + v_{x0} t + \tfrac12 a_x t^2. \]
The third equation removes time. Solve the first equation for \(t = (v_x - v_{x0})/a_x\) and substitute it into the second, or equivalently use the trapezoid: the displacement is the average velocity times the time, \(\Delta x = \tfrac12 (v_{x0} + v_x)\,t\), and eliminating \(t\) gives the third equation: \[ v_x^2 = v_{x0}^2 + 2a_x (x - x_0). \] Whenever the time is neither given nor wanted, this equation is the shortest route.
The average-velocity statement \(\Delta x = \tfrac12 (v_{x0} + v_x) t\) is useful in its own right, but be careful: it holds only when the acceleration is constant. For any motion, the average velocity is simply \(\Delta x/\Delta t\); the version with the arithmetic mean of the initial and final velocities is a property of straight-line velocity graphs.
A method for choosing the equation
Five quantities appear in the three equations: the displacement \(\Delta x\), the time \(t\), the initial velocity \(v_{x0}\), the final velocity \(v_x\) and the acceleration \(a_x\). Each equation contains four of them and leaves one out (Figure 1.4b). If you know three and want a fourth, exactly one equation contains your four quantities and omits the fifth, so there is never a long search.
- Draw a sketch and choose an axis. Write “right (or up) is positive.”
- List the knowns with signs and units, including hidden ones: “starts from rest” means \(v_{x0} = 0\); “comes to a stop” means \(v_x = 0\); a dropped object has \(v_{x0} = 0\).
- Name the unknown and find the quantity that is neither known nor wanted.
- Choose the equation that omits that quantity, solve symbolically, then substitute.
- Check the answer for sign, units and reasonableness.
The AP exam values this symbolic habit. Free-response questions often ask for a derived expression in terms of given variables, with the numbers coming later or not at all.
A car moving at \(+20\ \text{m/s}\) brakes with a constant acceleration of \(-5.0\ \text{m/s}^2\) until it stops. How long does the stop take, and how far does the car travel?
Knowns. \(v_{x0} = +20\ \text{m/s}\), \(v_x = 0\), \(a_x = -5.0\ \text{m/s}^2\). Unknowns: \(t\) and \(\Delta x\).
Time. The distance is not wanted, so use the first equation: \(t = \dfrac{v_x - v_{x0}}{a_x} = \dfrac{0 - 20}{-5.0} = 4.0\ \text{s}\).
Distance. Time is already known, but the third equation avoids relying on it: \(0 = (20)^2 + 2(-5.0)\Delta x\), so \(\Delta x = \dfrac{400}{10} = 40\ \text{m}\).
Check. The average velocity during the stop was \(\tfrac12(20 + 0) = 10\ \text{m/s}\); \(10\ \text{m/s} \times 4.0\ \text{s} = 40\ \text{m}\). The two methods agree.
A car starts from rest and accelerates at \(+4.0\ \text{m/s}^2\) for 10 s. Find its final velocity and displacement. How long does it need to cover the first 100 m?
Velocity. \(v_x = 0 + (4.0)(10) = +40\ \text{m/s}\).
Displacement. \(\Delta x = 0 + \tfrac12(4.0)(10)^2 = 200\ \text{m}\).
First 100 m. From rest, \(\Delta x = \tfrac12 a_x t^2\), so \(t = \sqrt{\dfrac{2\Delta x}{a_x}} = \sqrt{\dfrac{2(100)}{4.0}} = \sqrt{50} = 7.1\ \text{s}\).
Interpretation. The car covers the first 100 m in 7.1 s and the second 100 m in only 2.9 s more, because it is moving faster. Starting from rest at constant acceleration, distance grows as \(t^2\).
A car on an entrance ramp speeds up from \(+10\ \text{m/s}\) to \(+25\ \text{m/s}\) with a constant acceleration of \(+3.0\ \text{m/s}^2\). How far does it travel, and how long does it take?
Distance. Time is neither given nor wanted first, so \(v_x^2 = v_{x0}^2 + 2a_x\Delta x\) gives \(\Delta x = \dfrac{625 - 100}{2(3.0)} = \dfrac{525}{6.0} = 87.5\ \text{m}\).
Time. \(t = \dfrac{25 - 10}{3.0} = 5.0\ \text{s}\).
Check. \(\Delta x = v_{x0}t + \tfrac12 a_x t^2 = (10)(5.0) + \tfrac12(3.0)(25) = 50 + 37.5 = 87.5\ \text{m}\). ✓
Why braking distance grows so fast
The third equation explains one of the most important facts in everyday physics. If a car brakes to a stop with deceleration magnitude \(|a|\), then \(0 = v_0^2 - 2|a|d\) and \[ d = \frac{v_0^2}{2|a|}. \] The stopping distance depends on the square of the initial speed. With \(|a| = 5.0\ \text{m/s}^2\), a car at 10 m/s stops in 10 m, at 20 m/s in 40 m, and at 30 m/s in 90 m (Figure 1.4c). Doubling the speed quadruples the distance. On a real road the driver also covers distance during the reaction time, at the full speed: at 20 m/s and a 1.0 s reaction time that adds another 20 m, for a total of 60 m.
- Constant acceleration only. Before using any of the three equations, confirm the acceleration is constant. Say so in your solution.
- Hidden zeros. “From rest,” “dropped” and “stops” each supply a known velocity of zero. Read the wording for them.
- Signs on \(a_x\). A braking car moving in the positive direction has a negative acceleration. Entering a positive value gives a stop that takes negative time, which is your cue that a sign is wrong.
- Symbolic first. Solve for the unknown before substituting numbers; graders can then follow your reasoning even if the arithmetic slips.
- Squares. If the speed doubles and the acceleration stays the same, the distance covered in changing the speed is four times as large. Use this ratio reasoning on qualitative questions.
- A cyclist starts from rest and accelerates at \(0.50\ \text{m/s}^2\) for 12 s. Find the final speed and the distance covered.
Show answer
\(v = 0 + (0.50)(12) = 6.0\ \text{m/s}\). \(\Delta x = \tfrac12(0.50)(12)^2 = 36\ \text{m}\).
- A train slows from 30 m/s to 10 m/s while traveling 400 m at constant acceleration. Find the acceleration and the time.
Show answer
\(a = \dfrac{v^2 - v_0^2}{2\Delta x} = \dfrac{100 - 900}{800} = -1.0\ \text{m/s}^2\). \(t = \dfrac{10 - 30}{-1.0} = 20\ \text{s}\). Check: \(\tfrac12(30 + 10)(20) = 400\ \text{m}\).
- A car traveling at 15 m/s stops in 25 m. Find the magnitude of its acceleration and the stopping time.
Show answer
\(|a| = \dfrac{v_0^2}{2d} = \dfrac{225}{50} = 4.5\ \text{m/s}^2\). \(t = \dfrac{2d}{v_0} = \dfrac{50}{15} = 3.3\ \text{s}\).
- A rocket sled goes from rest to 100 m/s in 5.0 s with constant acceleration. Find the acceleration and the distance.
Show answer
\(a = 100/5.0 = 20\ \text{m/s}^2\). \(\Delta x = \tfrac12(20)(5.0)^2 = 250\ \text{m}\) (or average velocity \(50\ \text{m/s} \times 5.0\ \text{s}\)).
- A puck slides on ice with an initial velocity of \(+8.0\ \text{m/s}\) and a constant acceleration of \(-0.50\ \text{m/s}^2\). How far does it slide before stopping?
Show answer
\(0 = 8.0^2 + 2(-0.50)\Delta x\), so \(\Delta x = 64/1.0 = 64\ \text{m}\). The slide lasts \(8.0/0.50 = 16\ \text{s}\).
- Car A and car B brake with the same deceleration. A is going twice as fast as B when the brakes are applied. Compare their stopping distances and stopping times. Justify your answer with equations.
Show answer
From \(d = v_0^2/(2|a|)\), A’s stopping distance is four times B’s. From \(t = v_0/|a|\), A’s stopping time is twice B’s. Speed enters the distance as a square and the time only to the first power.