A person walks at a speed of 6 km/h from point A to point B. If he improves his pace by 1.5 km/h, he will arrive 1 hour earlier. a) calculate the distance and time he intends to travel from a to b
b) the individual initially travels at v1=6 km/h for a distance s1, then pauses for 15 minutes to rest. So he goes at v2=7.5km/h for the rest of the route and arrives 30 minutes early than expected.

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

the answer would be a

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