Predict constant-acceleration displacement

Supplied materials

• All measurements and observations below are supplied classroom examples. They are not data from actual learners or a claim about a particular real site.
• Paper or browser document; an ordinary calculator is optional.
• Rule: Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. Formula: a*c+0.5*b*c*c. Variable key: a = initial velocity (m/s), b = signed acceleration (m/s²), c = elapsed time (s)
• Distinct worked inputs and result: A distinct model has initial 1 m/s, acceleration 2 m/s² and time 2 s: 1×2+0.5×2×4=6 m.
• Assessment case 1: Supplied case 1: initial velocity (m/s)=2; signed acceleration (m/s²)=1; elapsed time (s)=4, using the explicitly named quantities. Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. What result follows?
• Assessment case 2: Supplied case 2: initial velocity (m/s)=3; signed acceleration (m/s²)=2; elapsed time (s)=3, using the explicitly named quantities. Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. What result follows?
• Assessment case 3: Supplied case 3: initial velocity (m/s)=4; signed acceleration (m/s²)=-1; elapsed time (s)=2, using the explicitly named quantities. Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. What result follows?
• Assessment case 4: Supplied case 4: initial velocity (m/s)=0; signed acceleration (m/s²)=3; elapsed time (s)=4, using the explicitly named quantities. Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. What result follows?
• Assessment case 5: Supplied case 5: initial velocity (m/s)=5; signed acceleration (m/s²)=1; elapsed time (s)=2, using the explicitly named quantities. Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. What result follows?
• Assessment case 6: Supplied case 6: initial velocity (m/s)=-2; signed acceleration (m/s²)=-1; elapsed time (s)=3, using the explicitly named quantities. Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. What result follows?

Learn the idea

For a constant-acceleration straight-line model, displacement is initial velocity times time plus one-half acceleration times time squared. This differs from total path length if direction reverses.

Displacement equals initial velocity times seconds plus one-half acceleration times seconds squared. The formula in the supplied model is a*c+0.5*b*c*c. Variable key: a = initial velocity (m/s), b = signed acceleration (m/s²), c = elapsed time (s). Square the time in the acceleration term. Use displacement, not automatically path length.

A distinct model has initial 1 m/s, acceleration 2 m/s² and time 2 s: 1×2+0.5×2×4=6 m. Constant acceleration is an explicit assumption. Track the quantity and unit before calculating. Explain the meaning of a positive, negative or fractional result when relevant. A numerical answer supports the stated model calculation; broader claims require the additional measurements and assumptions described here.

1. Name the measured quantities and their units before substituting values.
2. Write the stated rule and use the variable key to match each symbol to its explicitly named quantity.
3. Calculate each supplied case separately, keeping intermediate arithmetic and the result unit.
4. Explain the result in context and identify the assumption that would invalidate direct use of the rule.

Check your reasoning

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Take it further

Changed-input task: use the same stated model with initial velocity (m/s)=3; signed acceleration (m/s²)=1; elapsed time (s)=4. Expected result: 20 m. Explain what changes and keep this assumption: Constant acceleration is an explicit assumption.

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