Lesson 1.5 · Unit 1 · HS-PS2-1
Four equations, none of them new information
For motion with constant acceleration there are four standard equations. They are usually presented as a list to memorize, which is the worst way to hold them: four similar-looking formulas are four things to confuse. They all follow from the definitions in lessons 1.3 and 1.4, so this lesson derives them, and after that they can be rebuilt rather than recalled.
- All four assume constant acceleration. Applying them to changing acceleration is the error that invalidates everything downstream.
- \( v = v_0 + at \) comes straight from the definition of acceleration.
- \( \Delta x = \dfrac{v_0 + v}{2}t \) is the trapezoid area from lesson 1.4.
- \( \Delta x = v_0t + \tfrac{1}{2}at^2 \) comes from substituting the first into the second.
- \( v^2 = v_0^2 + 2a\Delta x \) comes from eliminating \( t \) between them, and is the one to use when time is neither known nor wanted.
- Each equation omits one of the five quantities. Choose by finding which quantity the problem neither gives nor asks for.
- The five quantities are \( v_0 \), \( v \), \( a \), \( t \) and \( \Delta x \). Listing what you have before choosing an equation is most of the work.
Where students lose marks: using these equations across a change in acceleration. A car that accelerates, cruises, then brakes needs the equations applied to each phase separately. One application spanning all three describes a motion that never happened.
The problem. (a) Derive all four equations from the definitions. (b) A car accelerates uniformly from rest to 28 m/s in 7.0 s. Find its acceleration and the distance covered, using two different equations as a check. (c) Explain how to choose which equation to use.
Step one: derive the first equation. Acceleration is defined as \( a = \dfrac{v - v_0}{t} \). Multiply both sides by \( t \) and add \( v_0 \): \[ v = v_0 + at \] That is the definition rearranged, not a new fact.
Step two: derive the second. When acceleration is constant, the velocity-time graph is a straight line, so the average velocity over the interval is the ordinary average of the endpoints, \( \dfrac{v_0 + v}{2} \). Since displacement is average velocity times time, \[ \Delta x = \frac{v_0 + v}{2}\,t \] This is the trapezoid area from lesson 1.4 written algebraically. Note the condition. The midpoint average is only the true average because the graph is a straight line. That is where "constant acceleration" enters.
Step three: derive the third. Substitute \( v = v_0 + at \) into the second equation: \[ \Delta x = \frac{v_0 + (v_0 + at)}{2}\,t = \frac{2v_0 + at}{2}\,t \] Expanding gives \[ \Delta x = v_0t + \tfrac{1}{2}at^2 \]
Step four: derive the fourth. Solve the first equation for time, \( t = \dfrac{v - v_0}{a} \), and substitute into the second: \[ \Delta x = \frac{v_0 + v}{2} \cdot \frac{v - v_0}{a} = \frac{v^2 - v_0^2}{2a} \] using the difference of squares. Multiplying by \( 2a \) and rearranging: \[ v^2 = v_0^2 + 2a\Delta x \] All four came from two definitions and some algebra. Nothing was assumed about the physical situation beyond constant acceleration.
Step five: find the acceleration in (b). List the quantities: \( v_0 = 0 \), \( v = 28\ \text{m/s} \), \( t = 7.0\ \text{s} \), \( a \) unknown, \( \Delta x \) unknown. Use the first equation: \( 28 = 0 + a(7.0) \), so \( a = \dfrac{28}{7.0} = 4.0\ \text{m}/\text{s}^2 \).
Step six: find the distance two ways. Using the third: \( \Delta x = 0(7.0) + \tfrac{1}{2}(4.0)(7.0)^2 = \tfrac{1}{2}(4.0)(49) = 98\ \text{m} \). Using the second: \( \Delta x = \dfrac{0 + 28}{2}(7.0) = 14 \times 7.0 = 98\ \text{m} \) ✓ A third check with the fourth equation: \( v^2 = 0 + 2(4.0)(98) = 784 \), and \( \sqrt{784} = 28\ \text{m/s} \) ✓ matching the given final speed.
Step seven: set out the selection method for (c). Write down the five quantities, mark the three you know and the one you want. The one left over is the one to avoid, and each equation is identified by what it leaves out.
| Equation | Leaves out |
|---|---|
| \( v = v_0 + at \) | \( \Delta x \) |
| \( \Delta x = \dfrac{v_0+v}{2}t \) | \( a \) |
| \( \Delta x = v_0t + \tfrac{1}{2}at^2 \) | \( v \) |
| \( v^2 = v_0^2 + 2a\Delta x \) | \( t \) |
Step eight: work an example of the choice. A stone is dropped and lands at 25 m/s; how far did it fall? Known: \( v_0 = 0 \), \( v = 25\ \text{m/s} \), \( a = 9.8\ \text{m}/\text{s}^2 \) taking down positive. Wanted: \( \Delta x \). Not given and not wanted: \( t \). So use the equation that omits \( t \): \( 25^2 = 0 + 2(9.8)\Delta x \), giving \( \Delta x = \dfrac{625}{19.6} = 31.9\ \text{m} \). The alternative route works but costs more. Finding \( t \) first from the first equation, then \( \Delta x \) from the third, gives the same answer through two steps and two chances to slip. Choosing well is not elegance; it is error reduction.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Take \( g = 9.8\ \text{m}/\text{s}^2 \) where needed.
- What condition must hold for the kinematic equations to apply?
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Constant acceleration
- Which equation omits displacement?
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\( v = v_0 + at \)
- Which equation omits time?
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\( v^2 = v_0^2 + 2a\Delta x \)
- An object starts at rest with \( a = 3.0\ \text{m}/\text{s}^2 \). Find its speed after
5.0 s.
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\( v = 0 + 3.0(5.0) \). 15 m/s
- How many quantities appear across the four equations?
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Five
- A cyclist accelerates from 4.0 m/s at
\( 1.2\ \text{m}/\text{s}^2 \) over 30 m. Find the final speed.
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Time is neither given nor wanted, so use the fourth equation. \( v^2 = 4.0^2 + 2(1.2)(30) = 16 + 72 = 88 \). \( v = \sqrt{88} = 9.4\ \text{m/s} \). Check the size. Starting at 4.0 and accelerating over 30 m should give a moderate increase, and 9.4 is plausible ✓ 9.4 m/s
- A stone is dropped from rest and falls for 2.0 s. Find its speed and the distance
fallen.
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Take down positive, so \( a = +9.8\ \text{m}/\text{s}^2 \) and \( v_0 = 0 \). Speed: \( v = 0 + 9.8(2.0) = 19.6\ \text{m/s} \). Distance: \( \Delta x = \tfrac{1}{2}(9.8)(2.0)^2 = \tfrac{1}{2}(9.8)(4.0) = 19.6\ \text{m} \). The two numbers matching at 19.6 is a coincidence of this particular time, not a rule. They have different units and different meanings. 19.6 m/s and 19.6 m
- A car braking at \( -5.0\ \text{m}/\text{s}^2 \) stops in 10 m. Find its initial speed.
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Time is not given, so use the fourth equation with \( v = 0 \). \( 0 = v_0^2 + 2(-5.0)(10) \), so \( v_0^2 = 100 \) and \( v_0 = 10\ \text{m/s} \). Watch the sign of \( a \). Using \( +5.0 \) would give \( v_0^2 = -100 \), which has no real solution, and that impossibility is itself the signal that the sign was wrong. 10 m/s
- Explain why doubling a car's speed more than doubles its stopping distance.
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Because the relationship is quadratic, not linear, and the fourth equation shows it directly. The derivation. Setting \( v = 0 \) in \( v^2 = v_0^2 + 2a\Delta x \) and solving: \( \Delta x = \dfrac{v_0^2}{2|a|} \). The stopping distance depends on the square of the initial speed. What that means numerically, at a braking deceleration of \( 5.0\ \text{m}/\text{s}^2 \): at 10 m/s, \( \dfrac{100}{10} = 10\ \text{m} \); at 20 m/s, \( \dfrac{400}{10} = 40\ \text{m} \); at 30 m/s, \( \dfrac{900}{10} = 90\ \text{m} \). Doubling the speed quadruples the distance, and tripling it multiplies the distance by nine. Why it is squared rather than linear. Two things worsen together at higher speed. The car takes longer to stop, and during that longer time it is covering ground faster. Each effect scales with the speed, so the product scales with the square. The energy reading, which unit 5 will formalize. Kinetic energy goes as \( v^2 \), and the brakes remove energy at a roughly constant rate per meter, so four times the energy needs four times the distance. Why this matters outside the exam. The difference between 30 and 35 mph in a residential street is a much larger difference in stopping distance than the speeds suggest, which is the reasoning behind speed limits near schools. The quadratic is doing the work that intuition does not. Stopping distance goes as the square of the speed
- A rocket accelerates upward from rest at
\( 4.0\ \text{m}/\text{s}^2 \) for 15 s, then its engine cuts out. Find its speed and
height at cutout, and then the maximum height it reaches.
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This needs two phases, because the acceleration changes when the engine stops. Applying one equation across both would be the error this lesson warns about. Phase 1, engine on. Take up positive, \( a = +4.0\ \text{m}/\text{s}^2 \), \( v_0 = 0 \), \( t = 15\ \text{s} \). Speed at cutout: \( v = 0 + 4.0(15) = 60\ \text{m/s} \). Height at cutout: \( \Delta x = \tfrac{1}{2}(4.0)(15)^2 = \tfrac{1}{2}(4.0)(225) = 450\ \text{m} \). Phase 2, engine off. The rocket is still moving up at 60 m/s, but now the only acceleration is gravity, \( a = -9.8\ \text{m}/\text{s}^2 \). It rises until its velocity reaches zero. Time is not wanted, so use the fourth equation: \( 0 = 60^2 + 2(-9.8)\Delta x \), giving \( \Delta x = \dfrac{3600}{19.6} = 183.7\ \text{m} \). Total height. \( 450 + 183.7 = 633.7\ \text{m} \), about \( 6.3 \times 10^2\ \text{m} \) to two significant figures. Check the coasting phase a second way. Time to stop: \( t = \dfrac{60}{9.8} = 6.12\ \text{s} \). Distance: \( \dfrac{60 + 0}{2}(6.12) = 30(6.12) = 183.7\ \text{m} \) ✓ Why the phases cannot be merged. The acceleration is \( +4.0 \) for 15 s and then \( -9.8 \) afterward. No single constant value describes the motion, so no single application of the equations can. A check on plausibility. The coasting phase adds about 40 percent to the powered height, which is reasonable: the rocket is moving fast at cutout, and gravity takes six seconds to stop it. 60 m/s and 450 m at cutout; maximum height about 634 m