Recovering Motion by Integrating Acceleration · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

From acceleration back to motion

Science · Forces & Motion · ages 18-19
Name ______________________   Date ____________

  1. Each integration constant needs its own starting value.

    Circle one:   True   False

  2. You integrate an acceleration function once. What do you get?

    • Position, with both constants already fixed
    • Velocity plus one unknown constant
    • Force plus two unknown constants

  3. Acceleration stays constant at value a. After two integrals, the position is what?

    • Starting position plus starting velocity times t plus a times t squared
    • Starting velocity plus starting position times t plus half a times t squared
    • Starting position plus starting velocity times t plus half a times t squared

  4. What does the area under the acceleration curve between two times tell you?

    • The distance travelled over that stretch of time
    • The starting position of the trip
    • The change in velocity over that stretch

  5. You know a(t) but never wrote down the starting velocity. What happens?

    • You get a family of possible velocities
    • You still get one exact velocity, just as before
    • You get the position function directly

  6. A sled coasts with velocity v(t) = 4.0 - t squared/4 meters per second. The math gives negative velocity after 4 seconds. What does that mean?

    • The sled stays stopped at that spot for good
    • The sled slides back toward where it started
    • The slowing force suddenly turns positive

  7. A particle starts from rest at the origin while its acceleration falls with time. What fixes the two constants?

    • The acceleration function fixes both by itself
    • Zero starting position fixes the first, and zero starting velocity fixes the second
    • Zero starting velocity fixes the first, and zero starting position fixes the second

  8. A trolley moves with v(t) = 9 - t squared meters per second from the origin. It stops after 3 seconds, 18 meters along. A student says the model proves the trolley rests there forever. What is the mistake?

    • Past 3 seconds the equations give negative velocity
    • The stopping distance should have come out as zero meters
    • The velocity function never quite reaches zero in this model
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Answer key

For grown-ups. Fold this page away before handing over the rest.

From acceleration back to motion W1-mt_CXo7wKOA46-s1

  1. True · Velocity at time zero fixes the first constant, and position at time zero fixes the second.
  2. Velocity plus one unknown constant · One integral undoes one derivative, so you land on velocity with one constant to fix.
  3. Starting position plus starting velocity times t plus half a times t squared · A constant integrand integrates to the old equation: x0 plus v0 t plus half a t squared.
  4. The change in velocity over that stretch · Integrating acceleration over time accumulates velocity change, which is the area beneath.
  5. You get a family of possible velocities · Without the starting velocity the first constant stays free, so many motions still fit.
  6. The sled slides back toward where it started · Negative velocity points opposite the trip, so the sled reverses direction.
  7. Zero starting velocity fixes the first, and zero starting position fixes the second · Rest at the origin hands you both starting values, one per integral.
  8. Past 3 seconds the equations give negative velocity · The solution holds past the stopping moment, and there it describes rolling back.
Worksheet · LightMySky