AP Physics 1 · Free-response skill

From a described launch to numbers and graphs

The move. Translate a description of motion into equations and numbers, then into velocity-time and acceleration-time graphs described in words.

How to do it

  1. Turn the words into a list of stages. Each stage with its own constant acceleration gets its own knowns: starting velocity, acceleration, and a time or a distance.
  2. The end of one stage is the start of the next, so carry the velocity and the position across the boundary.
  3. Choose one sign convention and keep it in every stage. With up as positive, gravity’s acceleration is negative whenever gravity acts alone.
  4. To describe a velocity-time graph in words, give each straight segment’s starting and ending values and its slope, which is the acceleration during that stage.
  5. For an acceleration-time graph, decide what the acceleration is during each stage and whether it changes within the stage, then show what happens at each boundary between stages.

Practice prompt Try it

A small model rocket is launched straight up from rest on a launch pad. While its engine fires, the rocket’s acceleration is a constant 6.0 m/s² upward (engine and gravity together); the engine fires for 2.0 s and then shuts off, after which the rocket moves under gravity alone. Ignore air resistance and the rocket’s change in mass, and take up as positive. Use g = 9.8 m/s².

Answer each part about the rocket’s flight from launch to its highest point.

  1. How fast is the rocket moving when the engine shuts off?
  2. How high above the pad is the rocket when the engine shuts off?
  3. How long after launch does the rocket reach its highest point?
  4. How high above the pad does the rocket go?
  5. Describe in words the rocket’s velocity-time graph from launch to its highest point. Give the slope of each part of the graph.
  6. Describe in words the rocket’s acceleration-time graph over the same time, giving the acceleration during each part of the flight.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 8 points

Model answer, annotated

(a) During the burn, v=v0+at=0+(6.0)(2.0)=12v = v_0 + at = 0 + (6.0)(2.0) = 12 m/s, moving upward.

  • Point 1 Uses the first stage alone: starts from rest, constant 6.0 m/s² for 2.0 s.

(b) Starting from rest, Δy=12at2=12(6.0)(2.0)2=12\Delta y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(6.0)(2.0)^2 = 12 m above the pad.

  • Point 2 Displacement during the burn from the constant-acceleration equation.

(c) After the engine shuts off, the acceleration is -9.8 m/s². The rocket is at its highest point when its velocity reaches zero: 0=12−9.8t0 = 12 - 9.8t, so t = 1.22 s after shutoff, or 2.0 + 1.22 ≈ 3.2 s after launch.

  • Point 3 Carries the 12 m/s into the second stage and adds the two times.

(d) While coasting it rises a further v2/(2g)=122/19.6=7.3v^2/(2g) = 12^2/19.6 = 7.3 m, so the greatest height is 12 + 7.3 = 19.3 m above the pad.

  • Point 4 Adds the coasting rise to the height at shutoff.

(e) From 0 to 2.0 s the velocity-time graph is a straight line rising from 0 to 12 m/s, with a slope of +6.0 m/s². From 2.0 s it becomes a steeper straight line falling with a slope of -9.8 m/s², reaching zero at about 3.2 s, the highest point.

  • Point 5 First segment: rising straight line, slope +6.0 m/s².
  • Point 6 Second segment: falling straight line, slope -9.8 m/s².

(f) The acceleration-time graph is a flat line at +6.0 m/s² from 0 to 2.0 s. At 2.0 s it jumps down to a flat line at -9.8 m/s², which continues to the highest point.

  • Point 7 Constant +6.0 m/s² during the burn.
  • Point 8 Constant -9.8 m/s² once gravity acts alone.