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AI HL · AHL 2

Further Derivatives

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AHL 2.9

Kinematics: displacement, velocity and acceleration

AHL 2.9

Kinematics: displacement, velocity and acceleration

Target: I can find velocity and acceleration from a displacement function, and say what their signs mean. (+5 more)

By the end of this section

Key points

Displacement, velocity and acceleration

  • Displacement \(s\) is the position of the particle measured from a fixed point O. It can be negative: the particle is on the other side of O.
  • Velocity is the rate of change of displacement: \(v=\frac{ds}{dt}\). Acceleration is the rate of change of velocity: \(a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}\).
  • The sign of \(v\) gives the direction of motion. Speed is the size of the velocity, \(|v|\), so speed is never negative.
  • Units: displacement in m, velocity in m s\(^{-1}\), acceleration in m s\(^{-2}\).

Reading the signs

  • At rest means \(v=0\). If \(v\) changes sign there, the particle turns round.
  • Speeding up: \(v\) and \(a\) have the same sign. Slowing down: \(v\) and \(a\) have opposite signs. Check both signs before you answer.
  • The velocity is greatest or least where \(a=0\) and the sign of \(a\) changes. Use the second derivative test, as for any other function.
  • A trig model such as \(s=A\sin(kt)\) has velocity \(Ak\cos(kt)\): the factor \(k\) comes from the chain rule. Use radian mode.

Going back, and distance travelled

  • To go from \(a\) to \(v\) to \(s\), integrate. Each integration adds a constant, which the starting values (such as \(v(0)\) and \(s(0)\)) fix.
  • Displacement over an interval is \(\int v\,dt\). The forwards and backwards parts cancel.
  • Total distance travelled is \(\int|v|\,dt\). Every part counts as positive. Use the calculator.
  • Warning: a velocity of zero is not the same as an acceleration of zero. A particle thrown upwards is at rest at the top and still accelerating.

On the calculator

Radian mode and solving a trig equation

  • Check the calculator is in radian mode before you use \(\sin(kt)\) or \(\cos(kt)\) with a time t.
  • Solve \(v=0\) or \(v=\) a value over the interval you are given, and list every solution in it.

Screenshot and video to come

Distance travelled

  • Evaluate the definite integral of the absolute value of v over the interval.
  • Evaluate the integral of v on its own for the displacement, and compare.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Velocity, acceleration and rest

A particle moves in a straight line. Its displacement from a fixed point O is \(s\) metres after \(t\) seconds, where \(s=t^{3}-9t^{2}+15t\) for \(t\ge 0\). (a) Find the velocity and the acceleration when \(t=2\). (b) Find the times when the particle is at rest.

Working
  1. Link the quantities

    \(v=\frac{ds}{dt},\quad a=\frac{dv}{dt}\)

    Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity.

  2. Differentiate twice

    \(v=3t^{2}-18t+15,\quad a=6t-18\)

    Use the power rule on each term.

  3. Substitute t = 2

    \(v(2)=12-36+15=-9,\quad a(2)=12-18=-6\)

    The negative velocity means the particle is moving in the negative direction at 9 m s⁻¹.

  4. At rest

    \(3t^{2}-18t+15=0\Rightarrow 3(t-1)(t-5)=0\)

    At rest means v = 0.

  5. State

    \(t=1\) s and \(t=5\) s

    Between t = 1 and t = 5 the velocity is negative, so the particle changes direction at both times.

Video walkthrough coming
A buoy on the waves

A buoy rises and falls on the sea. Its height above the seabed is \(h\) metres after \(t\) seconds, where \(h=6+1.2\sin(1.5t)\) and \(t\ge0\). (a) Find an expression for the vertical velocity of the buoy. (b) Find the first time the buoy is momentarily at rest. (c) Find the greatest speed of the buoy.

Working
  1. Differentiate

    \(v=\frac{dh}{dt}=1.2\times1.5\cos(1.5t)=1.8\cos(1.5t)\)

    The chain rule brings out the factor 1.5.

  2. At rest

    \(1.8\cos(1.5t)=0\Rightarrow 1.5t=\frac{\pi}{2}\)

    Radian mode. The first positive solution of cos = 0 is π/2.

  3. Solve for t

    \(t=\frac{\pi}{3}=1.05\) s

    Divide both sides by 1.5.

  4. Greatest speed

    \(|v|=1.8|\cos(1.5t)|\le1.8\), so the greatest speed is 1.8 m s⁻¹

    Cosine never goes beyond 1 or −1. The buoy is fastest as it passes the middle height.

Video walkthrough coming
Speeding up or slowing down?

A particle moves in a straight line. Its displacement from O is \(s=4te^{-0.5t}\) metres after \(t\) seconds, \(t\ge0\). (a) Find expressions for the velocity and the acceleration. (b) Find the time when the particle is at rest. (c) When \(t=1\), is the particle speeding up or slowing down?

Working
  1. Product rule

    \(u=4t,\ w=e^{-0.5t}\Rightarrow v=4e^{-0.5t}+4t(-0.5)e^{-0.5t}=(4-2t)e^{-0.5t}\)

    The chain rule on \(e^{-0.5t}\) gives the factor −0.5.

  2. Acceleration

    \(a=-2e^{-0.5t}-0.5(4-2t)e^{-0.5t}=(t-4)e^{-0.5t}\)

    Differentiate v with the product rule again, then collect.

  3. At rest

    \((4-2t)e^{-0.5t}=0\Rightarrow t=2\)

    The exponential is never 0, so the bracket must be 0.

  4. Signs at t = 1

    \(v(1)=2e^{-0.5}=1.21>0,\quad a(1)=-3e^{-0.5}=−1.82<0\)

    Compare the signs of v and a.

  5. State

    v and a have opposite signs, so the particle is slowing down.

    Same sign: speeding up. Opposite signs: slowing down.

Video walkthrough coming
A boat coming to a stop

A boat is slowing down in a canal. Its distance from a lock is \(s=40\left(1-e^{-0.25t}\right)\) metres after \(t\) seconds. (a) Find the velocity at \(t=0\). (b) Find the time when the velocity is 2 m s\(^{-1}\). (c) Describe what happens to \(s\) and \(v\) as \(t\) gets very large.

Working
  1. Differentiate

    \(v=40\times0.25e^{-0.25t}=10e^{-0.25t},\quad a=-2.5e^{-0.25t}\)

    The derivative of \(-e^{-0.25t}\) is \(+0.25e^{-0.25t}\).

  2. Start

    \(v(0)=10\) m s⁻¹

    The boat starts at 10 m s⁻¹.

  3. Solve v = 2

    \(10e^{-0.25t}=2\Rightarrow e^{-0.25t}=0.2\Rightarrow t=-4\ln0.2=6.44\)

    Take natural logs of both sides.

  4. Long run

    \(s\to40,\quad v\to0\)

    The boat stops 40 m from the lock. v is never exactly 0, and a is always negative, so it slows down for ever.

Video walkthrough coming
From acceleration back to displacement

A particle moves in a straight line. Its acceleration is \(a=6t-8\) m s\(^{-2}\). When \(t=0\) its velocity is 5 m s\(^{-1}\) and its displacement is 2 m. Find its displacement when \(t=3\).

Working
  1. Integrate a

    \(v=\int(6t-8)\,dt=3t^{2}-8t+c\)

    Going from a to v is the reverse of differentiating.

  2. Use v(0) = 5

    \(c=5\Rightarrow v=3t^{2}-8t+5\)

    The initial velocity fixes the constant.

  3. Integrate v

    \(s=\int(3t^{2}-8t+5)\,dt=t^{3}-4t^{2}+5t+k\)

    A new constant, k.

  4. Use s(0) = 2

    \(k=2\Rightarrow s=t^{3}-4t^{2}+5t+2\)

    The initial displacement fixes k.

  5. Substitute t = 3

    \(s(3)=27-36+15+2=8\) m

    Check the units: metres.

Video walkthrough coming
Displacement and distance travelled

A robot moves along a straight rail with velocity \(v=t^{2}-5t+4\) m s\(^{-1}\) for \(0\le t\le5\). Find the total distance the robot travels in the 5 seconds.

Working
  1. Find where v = 0

    \(t^{2}-5t+4=0\Rightarrow t=1\text{ and }t=4\)

    The robot turns round at these times, so the sign of v changes.

  2. Distance, not displacement

    \(\text{distance}=\int_{0}^{5}|v|\,dt\)

    Distance adds every part as a positive amount. \(\int v\,dt\) would let the backwards part cancel.

  3. From the calculator

    \(\int_{0}^{5}|t^{2}-5t+4|\,dt=8.17\)

    Type the absolute value inside the integral.

  4. Compare

    \(\int_{0}^{5}(t^{2}-5t+4)\,dt=-0.833\) m is the displacement

    The distance is bigger than the displacement because the robot goes backwards between t = 1 and t = 4.

Video walkthrough coming