Question

Dos motos, A y B, toman la salida en una carrera de 90 km. La moto A realiza el recorrido con una velocidad media inferior en 20 km/h a la de la moto B, con lo que llega a la meta 3 minutos después que B. Calcula la velocidad de cada moto.

Answer 1

motorcycle A: 180 km/h

motorcycle B: 200 km/h

Explanation:

To solve this question we need to write a system of equations for A and B, using the equation:

distance = speed * time

The speed of A is 20 less than the speed of B, so:

speedA = speedB - 20

And the time A traveled is 3 minutes (0.05 hours) more than B's time, so:

timeA = timeB + 0.05

Then, using the distance equation, we have that:

distanceB = speedB * timeB

90 = speedB * timeB

distanceA = speedA * timeA

90 = (speedB - 20) * (timeB + 0.05)

90 = speedB * timeB + 0.05 * speedB - 20*timeB - 1

90 = 90 + 0.05 * speedB - 20*timeB - 1

20*timeB = 0.05 * speedB - 1

timeB = (speedB - 20)/400

using this timeB in the distance equation, we have:

90 = speedB * (speedB - 20)/400

90 * 400 = speedB^2 - 20*speedB

speedB^2 - 20*speedB - 36000 = 0

Solving this quadratic equation, we have speedB = 200 km/h

And the speedA is:

speedA = speedB - 20 = 200 - 20 = 180 km/h