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

    Circle one:   True   False

  2. A particle has s = 4t² metres after t seconds. What is its velocity at t = 3, in metres per second?

    Answer: ______________

  3. A particle has v = 3t² - 6t m/s. What is its acceleration at t = 4?

    • 18 m/s²
    • 24 m/s²
    • 6 m/s²
    • -6 m/s²

  4. A journey is measured with displacement in kilometres and time in hours. What are the units of acceleration?

    • kilometres per hour
    • kilometres
    • kilometres per hour squared
    • hours per kilometre

  5. A particle has s = t² - 8t + 12 metres. At what time t is it at rest?

    Answer: ______________

  6. A particle has a = 10 - 2t m/s² and starts from rest, so v(0) = 0. What is v?

    • v = 10 - t²
    • v = 10t - 2t²
    • v = 10t - t²
    • v = 10t - t² + 10

  7. A particle has v = t² - 5t + 6 m/s. At what times is it at rest?

    • t = 2 and t = 3
    • t = 5 and t = 6
    • t = -2 and t = -3
    • t = 1 and t = 6

  8. A particle has v = 12 - 4t m/s and its displacement at t = 0 is 0. What is s?

    • s = 12 - 4t²
    • s = 12t - 4t²
    • s = -4
    • s = 12t - 2t²

  9. A particle has a = 12t - 6 m/s², with v(0) = 4 m/s and s(0) = 1 m. What is its displacement at t = 1, in metres?

    Answer: ______________

  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. 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.
  2. 24 · Differentiating gives v = 8t. At t = 3 that is 24 m/s.
  3. 18 m/s² · Differentiating gives a = 6t - 6. At t = 4 that is 24 - 6, which is 18 m/s².
  4. kilometres per hour squared · The rule does not depend on which units you started with. Each differentiation divides by another hour, so velocity is km per hour and acceleration is km per hour squared.
  5. 4 · v = 2t - 8, and at rest means v = 0, so 2t = 8 and t = 4 seconds.
  6. v = 10t - t² · Integrating gives v = 10t - t² + c. Starting from rest means v(0) = 0, so c = 0 and v = 10t - t².
  7. t = 2 and t = 3 · At rest means v = 0, so factorise: (t - 2)(t - 3) = 0, giving t = 2 and t = 3 seconds.
  8. s = 12t - 2t² · Integrating gives s = 12t - 2t² + c, and s(0) = 0 puts c = 0. So s = 12t - 2t².
  9. 4 · Integrating gives v = 6t² - 6t + 4, then s = 2t³ - 3t² + 4t + 1. At t = 1 that is 2 - 3 + 4 + 1 = 4 m.
  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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