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Boss lesson · The River Crossing
Exam-hall rules, last run of the day
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Exam-hall rules, last run of the day

This is the boss. No calculator, and every value is pinned so it fits in your head. One ferry, one river, and every answer feeds the next: the speed over the ground, then the heading of the track, then the crossing time, then the drift — and finally the verdict the pilot actually cares about, which is where the boat touches the far bank. Write each line down as you go; one early slip floats all the way downstream.

Two things worth carrying in: every track here is a 3-4-5 triangle in disguise, so its length is Pythagoras done in your head. And that triangle's own angles are 36.9∘36.9^\circ and 53.1∘53.1^\circ — the pair every Grade 12 student should recognise on sight, the way you recognise 2≈1.414\sqrt{2} \approx 1.414. One of them will be waiting for you. Nothing else in here needs a keypad.

Everything from the chapter rides along. 5★ under par is the chapter crown. Cast off.

θ|v|vx

vx=∣v⃗∣cos⁡θv_x = |\vec{v}| \cos\theta

  • vxv_x= x-component
  • ∣v⃗∣|\vec{v}|= Vector magnitude
  • θ\theta= Angle from +x axis (angle)
x-Component from Magnitude and Angle solver →
θvy|v|

vy=∣v⃗∣sin⁡θv_y = |\vec{v}| \sin\theta

  • vyv_y= y-component
  • ∣v⃗∣|\vec{v}|= Vector magnitude
  • θ\theta= Angle from +x axis (angle)
y-Component from Magnitude and Angle solver →
|v|vyvx

∣v⃗∣=vx2+vy2|\vec{v}| = \sqrt{v_x^2 + v_y^2}

  • ∣v⃗∣|\vec{v}|= Vector magnitude
  • vxv_x= x-component
  • vyv_y= y-component
Magnitude of a 2D Vector solver →
θvyvx

θ=atan2⁡(vy,  vx)\theta = \operatorname{atan2}(v_y,\; v_x)

  • θ\theta= Direction angle from +x axis (angle)
  • vxv_x= x-component
  • vyv_y= y-component
Direction Angle of a 2D Vector solver →
no calculator in the boss room — the numbers are chosen to fit in your head