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AI HL · AHL 4

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AHL 4.21

Kinematics on a line: derivatives and integrals

AHL 4.21

Kinematics on a line: derivatives and integrals

Target: I can find velocity and acceleration from a displacement function by differentiating, using the chain rule. (+3 more)

By the end of this section

Key points

Differentiate and integrate

  • Velocity is the rate of change of displacement: \(v = \dfrac{ds}{dt}\).
  • Acceleration is the rate of change of velocity: \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\).
  • Differentiate to go from displacement to velocity to acceleration. Integrate, and add \(+c\), to go back.
  • Trig functions need radians. Check your calculator mode before every question with sin or cos in it.

Acceleration from displacement

  • When the velocity is given in terms of \(s\), use \(a = v\dfrac{dv}{ds} = \dfrac{d}{ds}\left(\tfrac12 v^2\right)\).
  • Differentiate with respect to \(s\), not \(t\), then multiply by \(v\).
  • The result is a function of \(s\). To find the acceleration at a given point, put that value of \(s\) in.

Integrals over a time interval

  • The change in velocity from \(t_1\) to \(t_2\) is \(\displaystyle\int_{t_1}^{t_2} a\,dt\).
  • The change in displacement from \(t_1\) to \(t_2\) is \(\displaystyle\int_{t_1}^{t_2} v\,dt\).
  • A change is not a final value. To find the final velocity, add the change to the velocity at \(t_1\).
  • Definite integrals of exponential and trig functions are quickest on the calculator.

On the calculator

Definite integral

  • Enter the function of \(t\) and the limits.
  • Evaluate the integral and read the value.
  • State what it is: a change in velocity if you integrated \(a\), a change in displacement if you integrated \(v\).

Screenshot and video to come

Radian mode

  • Check that the calculator is in radians before evaluating or integrating sin and cos of \(t\).
  • A wrong mode gives a plausible but wrong number.

Screenshot and video to come

Explore: displacement, velocity and acceleration

Choose a displacement function and move the time slider. The dot on the line is the object. The three graphs share one time axis: v is the gradient of s, and a is the gradient of v.

Displacement so far
Distance travelled so far
Where the object turns round

Open dots mark where v = 0. In the v graph, teal area is movement in the positive direction and gold area is movement in the negative direction. The calculator is in radians.

Move the time slider to move the object.

  1. Where does the v graph cross the axis? What is the object doing at those times, and what does the s graph look like there?
  2. Move the time past the first turn. Why is the distance travelled bigger than the displacement? Which areas in the v graph explain it?
  3. Find a time when v and a have the same sign and one when they have opposite signs. What is the object doing in each case?
  4. Choose the third function. Why are the distance and the displacement always equal?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Differentiating twice

The displacement of a particle from a fixed point is \(s = 3\sin 4t\) metres at time \(t\) seconds. Find expressions for the velocity and the acceleration, then find the acceleration when \(t = 0.5\).

Working
  1. Velocity

    \(v = \dfrac{ds}{dt} = 12\cos 4t\)

    Chain rule: the derivative of sin 4t is 4 cos 4t, so the 3 becomes 12.

  2. Acceleration

    \(a = \dfrac{dv}{dt} = −48\sin 4t\)

    Chain rule again: the derivative of cos 4t is −4 sin 4t, and 12 × (−4) = −48.

  3. Substitute

    From the calculator, in radians: \(a(0.5) = −48\sin 2 = −43.6\) m s⁻²

    Radians: the calculator must be in radian mode for time in seconds.

Video walkthrough coming
Acceleration from velocity in terms of s

A particle moves on a line with velocity \(v = 3 + 2s^2\) m s⁻¹ when its displacement is \(s\) metres. Find an expression for the acceleration in terms of \(s\), then find the acceleration when \(s = 2\).

Working
  1. The rule

    \(a = v\dfrac{dv}{ds}\)

    Use this when the velocity is given in terms of displacement, not time.

  2. Differentiate v

    \(\dfrac{dv}{ds} = 4s\)

    Differentiate with respect to s.

  3. Multiply

    \(a = (3+2s^2)(4s) = 8s^{3} + 12s\)

    Expand to give a in terms of s.

  4. Substitute

    \(a(2) = 12(2) + 8(2)^3 = 88\) m s⁻²

    Put s = 2 into the expression for a.

Video walkthrough coming
Change in velocity

A sprinter's acceleration is \(a = 5e^{-0.4t}\) m s⁻² for \(t \ge 0\), with \(t\) in seconds. Find the change in her velocity between \(t = 1\) and \(t = 4\).

Working
  1. The rule

    \(\text{change in } v = \displaystyle\int_1^4 a\,dt\)

    The change in velocity is the integral of the acceleration over the time interval.

  2. Write the integral

    \(\displaystyle\int_1^4 5e^{-0.4t}\,dt\)

    Enter this as a definite integral with limits 1 and 4.

  3. Evaluate

    From the calculator: \(\displaystyle\int_1^4 5e^{-0.4t}\,dt = 5.86\) m s⁻¹

    Positive: the sprinter is speeding up.

Video walkthrough coming
Change in displacement

The velocity of a buoy moving up and down a pole is \(v = 6\sin(0.5t)\) m s⁻¹, with \(t\) in seconds. Find the change in its displacement between \(t = 0\) and \(t = 5\).

Working
  1. The rule

    \(\text{change in } s = \displaystyle\int_0^5 v\,dt\)

    The change in displacement is the integral of the velocity.

  2. Evaluate

    From the calculator, in radians: \(\displaystyle\int_0^5 6\sin(0.5t)\,dt = 21.6\) m

    The buoy ends 21.6 m higher than it started, if the positive direction is up.

Video walkthrough coming