The relationship of distance-speed-time
distance = velocity × time
d = vt
Given:
v₁ = 35 mph
v₂ = 50 mph
the hypotenuse = 366
for example h is the time in hour
Solution:
d₁² + d₂² = (hypotenuse)²
(v₁t)² + (v₂t)² = 366²
(35h)² + (50h)² = 366²
1,225h² + 2,500h² = 133,956
3,725h² = 133,956
h² = 133,956 / 3,725
h² = 35.9613
h = 5.99
to the nearest hour ⇒ 6
The buses will be 366 miles apart in 6 hours