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?
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Answer 1

Answer:

The speed of the Porsche =  180 km/h

The speed of the Ferrari =  200 km/h

Explanation:

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

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

we have;

S$_F$ × t₁ - S$_P$ × t = 2 km

S$_P$ × t₁ = 9 laps × 2 km/lap = 18 km

S$_P$ × t₁ = 18 km

S$_F$ × t₁ =  2 km + S$_P$ × t = 2 km + 18 km = 20 km

S$_F$ × t₁ =  20 km

∴

Similarly, we are given that the following relation;

(S$_F$ - 10) × t₂ - S$_P$ × t₂ = 2 km...(2)

From which we have;

S$_P$ × t₂ = 18 laps × 2 km/lap = 36 km

S$_P$ × t₂ = 36 km

(S$_F$ - 10) × t₂ = 2 km + 36 km = 38 km

(S$_F$ - 10) × t₂ = 38 km

Therefore, given that S$_P$ × t₂ (36 km) = 2 × S$_P$ × t₁ (18 km), we have;

S$_P$ × t₂ = 2 × S$_P$ × t₁

t₂ = 2×t₁

Equation (2) becomes;

(S$_F$ - 10) × 2×t₁ - S$_P$ × 2×t₁ = 2 km...(2)

From which we have;

(S$_F$ - 10) × 2×t₁  = 38 km

(S$_F$ - 10) × 2×t₁  = 38 km

2 × t₁ × S$_F$ - 2 × t₁ × 10 = 38 km

∵ S$_F$ × t₁ =  20 km, we have;

2 × 20 km - 2 × t₁ × 10 = 38 km

2 × t₁ × 10 = 2 × 20 km - 38 km = 2 km

20 × t₁ = 2 km

t₁ = 2/20 = 0.1 hour = 6 minutes

Therefore, we have;

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$_P$ = 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$_F$ = 200 km/h