Speed of Faster Car?
The average speed of two cars are in ratio of 4:5. What is the speed of the faster one, if it takes half an hour less to complete a 50-km distance than the other?
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answers:
Captain Matticus, LandPiratesInc say: The faster one completes a 50 km distance in t hours
The slower one completes a 50 km distance in t + 0.5 hours

Distance = Speed * Time
50 = F * t
50 = S * (t + 0.5)

F = (5/4) * S

F * t = S * (t + 0.5)
(5/4) * S * t = S * (t + 0.5)
(5/4) * t = t + 0.5
1.25 * t = t + 0.5
0.25 * t = 0.5
t = 2

50 = F * 2
25 = F

Faster one moves at 25 km/hr
Slower one moves at 20 km/hr
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Krishnamurthy say: The average speed of two cars are in ratio of 4 : 5.
What is the speed of the faster one,
if it takes half an hour less to complete a 50-km distance than the other?
Distance traveled equals Speed times Time taken to travel.
d1 = s1 × t1, 50 = 4x × t1------> (1)
d2 = s2 × t2, 50 = 5x × (t1 - 0.5) -----> (2)
Subtracting equation (2) by equation (1)
x(t1- (0.5) = 0
Dividing equation (2) by equation (1)
t1 ≈ 0.5
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la console say: Recall: s = d/t → where s is the speed, d is the distance, t is the time


For the first (no.1) car, it gives us: → s₁ = d₁/t₁ → d₁ = s₁.t₁

For the second (no.2) car, it gives us: → s₂ = d₂/t₂ → d₂ = s₂.t₂


You know that the distance is 50 km for the 2 cars:

d₁ = d₂

s₁.t₁ = s₂.t₂

s₁/s₂ = t₂/t₁ → we supose here that the car no.1 is the faster car

s₁/s₂ = t₂/t₁ → it takes half an hour less to complete a 50 km distance than the other → t₂ = t₁ + (1/2)

s₁/s₂ = [t₁ + (1/2)] / t₁ → the average speed of the 2 cars are in ratio of 4:5 → s₁/s₂ = 5/4

5/4 = [t₁ + (1/2)] / t₁

5.t₁ = 4.[t₁ + (1/2)]

5.t₁ = 4.t₁ + 2

t₁ = 2


Recall:

s₁ = d₁/t₁ → where: d₁ = 50 km

s₁ = 50/t₁ → we've just seen that: t₁ = 2 h

s₁ = 50/2

s₁ = 25 km/h ← this is the speed of the faster car


Recall:

s₁/s₂ = 5/4

s₂/s₁ = 4/5

s₂ = s₁.(4/5) → you know that: s₁ = 25 km/h

s₂ = 25 * (4/5)

s₂ = 20 km/h
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