Answer:
a) The distance is 30 km
The time duration is 5 hours
b) s₁ is approximately 28.142 km or s₁ is approximately 1.505 km
Explanation:
The initial speed with of the person, v₁ = 6 km/h
The distance the walked by the person, d = From point A to point B
The rate at which the person increases the speed, Δv = 1.5 km/h
The time it takes for the person to arrive at point B from point A at the new speed, t₂ = 1 hour earlier than when walking at 6 km/h
a) Let t₁ represent the time it takes the person walking from point A to point B at 6 km/h, we have;
t₂ = t₁ - 1...(1)
d/t₁ = 6...(2)
d/t₂ = 6 + 1.5 = 7.5
∴ d/t₂ = 7.5...(3)
From equation (2), we have;
d = 6 × t₁ = 6·t₁
Plugging in d = 6·t₁, and t₂ = t₁ - 1 in equation (3) gives;
d/t₂ = 7.5
∴ 6·t₁/t₁ - 1 = 7.5
6·t₁ = 7.5 × (t₁ - 1) = 7.5·t₁ - 7.5
7.5·t₁ - 6·t₁ = 7.5
1.5·t₁ = 7.5
t₁ = 7.5/1.5 = 5
t₁ = 5
The time it takes the person walking from point A to point B at 6 km/h, t₁ = 5 hours
The distance from point A to point B, d = 6 km/h × 5 hours = 30 km
b) The distance the person travels at the initial speed, v₁ (6 km/h) = s₁
The duration the person pauses for a rest = 15 minutes = 1/4 hours
The speed with which he walks the rest of the journey, v₂ = 7.5 km/h
The time earlier than expected that he arrives, Δt = 30 minutes = 0.5 hours
We note that the total distance, d = 30 km
The expected time, t₁ = 5 hours
Therefore, we have;
s₁ + s₂ = 30 km
s₂ = 30 - s₁
v₁/s₁ + 1/4 + v₂/s₂ = t₁ - 0.5
Therefore;
6/s₁ + 1/4 + 7.5/(30 - s₁) = 5 - 0.5 = 4.5
6/28.142+ 1/4 + 7.5/(30 - 28.142) = 5 - 0.5 = 4.5
6/s₁ + 7.5/(30 - s₁) = 4.5 - 1/4 = 4.25
-(3·s₁ + 360)/(2·s₁²- 60·s₁) = 4.25
2·s₁²- 60·s₁) × 4.25 + 3·s₁ + 360 = 0
17·s₁²- 504·s₁ + 720 = 0
s₁ = (504 ± √((-504)² - 4 × 17 × 720))/(2 × 17)
s₁ ≈ 28.142 or s₁ = 1.505
The distance the individual travels at v₁ = 6 km/h, s₁ ≈ 28.142 km or 1.505 km
