Question

Suppose that a Ferrari and a Porsche begin a race with a moving start, and each moves with constant speed. One lap of the track is 2 km. The Ferrari laps the Porsche after the Porsche has completed 9 laps. If the speed of the Ferrari had been 10 km/h less, the Porsche would have traveled 18 laps before being overtaken. What were the speeds of the two cars?

Answer 1

Porsche = 180 kmph

Ferrari = 200 kmph

Step-by-step explanation:

Given that

The distance of one lap = 2 km

The distance the Ferrari laps the Porsche = 9 laps

The distance the Ferrari will lap the Porsche with 10 km/h less speed = 18 laps

So we say let,

the speed of the Ferrari = S(f),

the speed of the Porsche = S(p), and

the time that lapse before two cars lapped = t

From the question, we're told that

(S(f) * t1) - (S(p) * t1) = 2 km, also

S(p) * t1 = 9 * 2 = 18 km

Referring back to the initial equation

(S(f) * t1) = 2 + (S(p) * t1)

(S(f) * t1) = 2 + 18 = 20 km

Going back to the question again, we're told that

(S(f) - 10) * t₂ - S(p) * t₂ = 2 km

S(p) * t₂ = 18 * 2 = 36 km

So that,

(S(f) - 10) * t₂ = 2 km + S(p) * t₂

(S(f) - 10) * t₂ = 2 + 36 km = 38 km

(S(f) - 10) * t₂ = 38 km

S(p) * t₂ (36 km) = 2 * S(p) * t₁ (18 km),

S(p) * t₂ = 2 * S(p) * t₁, collect like terms

t₂ = 2t₁

(S(f) - 10) * 2t₁ - S(p) * 2t₁ = 2 km

(S(f) - 10) * 2t₁ = 38 km

(S(f) - 10) * 2t₁ = 38 km

2t₁ * S(f) - 2t₁ * 10 = 38 km

S(f) * t₁ = 20 km,

2 * 20 km - 2t₁ * 10 = 38 km

2t₁ * 10 = 2 * 20 km - 38 km

2t₁ * 10 = 2 km

20t₁ = 2 km

t₁ = 2/20 = 0.1 hour = 6 minutes

This means that

S(p) * t₁ = 18 km

S(p) * 0.1 h = 18 km

S(p) = 18 km/0.1 h = 180 km/h

The speed of the Porsche = S = 180 km/h

S(f) * t₁ = 20 km

S(f) × 0.1 h = 20 km

S(f) = 20 km/0.1 h = 200 km/h

The speed of the Ferrari = S = 200 km/h