QCEVault

Further integration — Question 323

Original QCE Vault practice · 8 marks

Q323 · Practice questionTechnology-freeComplex unfamiliar8 marks

QUESTION 323 (8 marks)

Visitors arrive at a night market at rate R(t)=120(e−t/2−e−t)R(t)=120(e^{-t/2}-e^{-t}) people per hour for t≥0t\geq0. Let N(t)N(t) be the modelled number who have arrived by time tt, with N(0)=0N(0)=0.
Arrival-rate graph R(t)=120(exp(-t/2)-exp(-t)), rising from zero to a peak and then approaching zero.
a)
Determine the time and value of the maximum arrival rate.
[3 marks]
b)
Show that N(t)=120(1−e−t/2)2N(t)=120(1-e^{-t/2})^2.
[3 marks]
c)
The model predicts a limiting total of 120120 visitors. Determine the exact time by which half have arrived.
[2 marks]
Question linkSyllabus coverage

Related questions

  1. Q6 · Original practice · 1 mark
    Evaluate ∫023x2 dx\displaystyle\int_0^2 3x^2\,dx.
    Further integration
  2. Q16 · Original practice · 1 mark
    An antiderivative of 2xx2+5\dfrac{2x}{x^2+5} is
    Further integration
  3. Q31 · Original practice · 3 marks
    Evaluate ∫014xx2+1 dx\displaystyle\int_0^1\frac{4x}{x^2+1}\,dx.
    Further integration
  4. Q41 · Original practice · 3 marks
    For 0≤x≤10\le x\le1, the graph y=kx(1−x)y=kx(1-x) lies above the xx-axis. The area between the graph and the xx-axis is 2 square units. Determine kk.
    Further integration