Kinematics with Calculus · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Kinematics with Calculus

Mathematics · Mechanics · ages 17-18
Name ______________________   Date ____________

  1. You know the acceleration and want the velocity. What do you do?

    • differentiate the acceleration
    • integrate the acceleration
    • divide the acceleration by the time
    • square the acceleration

  2. A particle being "at rest" means its velocity is zero.

    Circle one:   True   False

  3. Integrating acceleration to get velocity needs an initial condition to fix the constant.

    Circle one:   True   False

  4. A particle moves with s(t) = t squared + 3t metres, t in seconds. What is its velocity at t = 2 s?

    • 7 m/s
    • 5 m/s
    • 10 m/s
    • 4 m/s

  5. Lia is asked when a particle is at rest and solves a = 0. Where is the slip?

    • she should have solved s = 0 instead
    • a = 0 has no solutions for a moving particle
    • a = 0 is where the velocity stops changing, which is not the same question
    • she should have differentiated once more before solving

  6. A particle has a = 6t m/s² and its velocity at t = 0 is 2 m/s. What is its velocity at t = 2?

    Answer: ______________

  7. A particle has v = t² - 9 m/s. At t = 2 it has stopped.

    Circle one:   True   False

  8. Acceleration is a(t) = 4t m/s squared and v(0) = 1 m/s. What is v(3) in m/s?

    • 19 m/s
    • 18 m/s
    • 13 m/s
    • 37 m/s

  9. A particle has s = t³ - 6t² + 9t metres. What is its acceleration at t = 1?

    • 0 m/s²
    • 6 m/s²
    • 9 m/s²
    • -6 m/s²

  10. Sam integrates a = 8t and reports the velocity as v = 4t². Where is the slip?

    • the constant is missing, and it is the starting velocity, so this forces v(0) = 0
    • the answer should have been 8t²/2 without simplifying
    • he should have differentiated rather than integrated
    • the power should have gone down to zero, giving v = 8
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Answer key

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

Kinematics with Calculus W1-mt_2kNspOOoDw-s1

  1. integrate the acceleration · Acceleration is the derivative of velocity, so going the other way means integrating. That step brings a constant with it.
  2. True · At rest means not moving, and velocity is what measures the movement. So the equation to solve is v = 0.
  3. True · Integration always leaves a constant, and here that constant is the velocity at t = 0. Without it the velocity is only known up to an unknown starting value.
  4. 7 m/s · Differentiate term by term: v(t) = 2t + 3, so v(2) = 4 + 3 = 7 m/s.
  5. a = 0 is where the velocity stops changing, which is not the same question · A particle can have zero acceleration while travelling at a steady speed, which is the opposite of at rest. The equation for at rest is v = 0.
  6. 14 · Integrating gives v = 3t² + c, and v(0) = 2 gives c = 2. At t = 2 that is 12 + 2 = 14 m/s.
  7. False · At t = 2 the velocity is 4 - 9 = -5 m/s. That is not zero, so the particle is moving, and the minus sign says it is going backwards.
  8. 19 m/s · v(t) = 2t squared + 1, so v(3) = 18 + 1 = 19 m/s.
  9. -6 m/s² · v = 3t² - 12t + 9, so a = 6t - 12. At t = 1 that is 6 - 12, which is -6 m/s².
  10. the constant is missing, and it is the starting velocity, so this forces v(0) = 0 · His integration is right as far as it goes, but v = 4t² + c is the general answer. Dropping the c silently claims the particle started from rest.
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