Plan the investigation
How does a constant acceleration change motion?
Constant acceleration means that velocity changes by the same amount during equal time intervals. The prepared Constant Acceleration scene starts a 1.00 kg block at x0 = 2.00 m with vi = 0.00 m/s. A constant +3.00 N horizontal applied force on a frictionless surface produces ax = +3.00 m/s²; gravity and the normal force balance vertically.
Change
Set the block’s initial x velocity, applied-force x component, and trial duration. Keep the mass, frictionless surface, gravity, and starting position fixed when comparing trials.
Measure
Record x, vx, and ax at equal times. Use the slope of x–t to find velocity, the slope of vx–t to find acceleration, and signed graph areas to connect velocity and displacement.
Choose a sign convention before you run: positive x points right. A positive acceleration points right even while an object is moving left; in that case the object slows, reaches vx = 0, and then reverses.
Procedure
A useful five-trial workflow
- Load and inspect. Launch the simulation and choose Constant acceleration from the experiment menu. Select the Constant-acceleration block in Objects, open Properties, and record x0, initial vx, mass, and the applied-force x component.
- Set one trial. Reset, edit the initial x velocity, and keep the y velocity at 0 m/s. For the prepared acceleration, leave the horizontal applied force at +3.00 N and keep the surface frictionless. Use the five initial velocities in the table below.
- Predict and run. Predict the final velocity and displacement with vf = vi + aΔt and Δx = viΔt + ½a(Δt)². Set a 2-second run duration beside Play, then run from t = 0.
- Analyze one graph at a time. Open Data → Graph and display x, vx, and ax. Keep one series visible while using Inspect or Slope. The x–t curve should be concave up for positive ax; vx–t should be a straight line with slope ax; and ax–t should be horizontal.
- Compare and preserve evidence. Reset before each velocity. Record the same time rows for every trial, compare measured values with the predictions, and use Export CSV if you need the complete sample history.
| vi (m/s) | vf (m/s) | xf (m) | Δx (m) |
|---|---|---|---|
| −3.00 | +3.00 | 2.00 | 0.00 |
| 0.00 | +6.00 | 8.00 | +6.00 |
| +3.00 | +9.00 | 14.00 | +12.00 |
| +6.00 | +12.00 | 20.00 | +18.00 |
The vi = −3.00 m/s trial is a useful reversal case: the block initially travels left, stops at t = 1.00 s, and finishes back at its starting position after 2.00 s.
Worked example
Read the prepared +3.00 m/s² trial
The preset begins at x0 = 2.00 m with vi = 0.00 m/s and ax = +3.00 m/s². After 2.00 s:
vf = vi + aΔt = 0.00 + 3.00(2.00) = +6.00 m/s
Δx = viΔt + ½a(Δt)² = 0.00 + ½(3.00)(2.00)² = +6.00 m
xf = x0 + Δx = 2.00 + 6.00 = 8.00 m
The x–t graph should curve upward because its slope grows with time. The vx–t graph should rise linearly from 0 to +6.00 m/s, and its slope should be +3.00 m/s². The ax–t graph should remain at +3.00 m/s².
Expected data pattern
Use equal-time rows as a check
For the unchanged preset, the ideal model gives these values. Use the displayed time in each Data row when comparing your own run; small rounding differences are expected.
| t (s) | x (m) | vx (m/s) | ax (m/s²) |
|---|---|---|---|
| 0.00 | 2.00 | 0.00 | +3.00 |
| 0.50 | 2.38 | +1.50 | +3.00 |
| 1.00 | 3.50 | +3.00 | +3.00 |
| 1.50 | 5.38 | +4.50 | +3.00 |
| 2.00 | 8.00 | +6.00 | +3.00 |
The area under vx–t from 0 to 2.00 s is +6.00 m, matching the displacement. The area under ax–t is +6.00 m/s, matching the change in velocity.
Common misconception
What does constant acceleration actually hold constant?
It holds the rate of change of velocity constant. It does not make speed, position, or displacement change by equal amounts during equal times.
Can an object move left while acceleration points right?
Yes. While vx is negative and ax is positive, the object slows. When vx reaches zero, it reverses and then speeds up to the right.
Does zero velocity mean zero acceleration?
No. At the instant of reversal, vx is zero, but ax can remain nonzero.
Does negative acceleration always mean slowing down?
No. Acceleration and velocity signs must be compared. Negative ax slows an object moving right, but speeds up an object already moving left.
For teachers
Connect motion, slope, area, and concavity
Have students sketch the three graphs before running. Ask them to explain why equal time intervals produce equal changes in velocity, why the vx–t area equals displacement, and why the position graph becomes curved when acceleration is nonzero.
Extend the investigation with a stopping trial: start at vi = +35 m/s and set ax = −15 m/s². The traveler reaches vx = 0 after about 2.33 s and travels about 40.83 m before reversing. Students can compare that result with the Constant Speed lab to see how a curved x–t graph differs from a straight one.
The Motion Graphing lab focuses on synchronized ticker-tape and graph patterns. The Acceleration Graphs guide develops slope, area, and concavity. Physics reference: OpenStax, University Physics Volume 1, §2.4.