Answer:
A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km
Explanation:
Given that,
A train travelled a distance of 240km
Let the initial speed be
S_1 = x km/hr
Let assume the time spent on the first journey is
t_1 = a
Now if he increase the speed to
S_2 = (x + 20) km/hr
Then, he would have take 2hrs less time
Then, time t_2 = a - 2
The common data fore the two journey is the distance
Speed = distance / time
For the first stage
S_1 = d / t_1
d = S_1 × a
d = x × a
240 = x•a
x = 240 / a Equation 1
For stage two
d = S_2 × t_2
d = (x+20) × (a - 2)
240 = (x+20) × (a - 2). Equation 2
Substitute equation 1 into 2
240 = (240/a + 20) × (a -2)
240 = 240 - 480/a + 20a - 40
240 - 240 + 40 = - 480/a + 20a
240 - 240 + 40 = (-480 + 20a²) / a
40 = (-480 + 20a²) / a
40a = -480 + 20a²
20a² - 40a -480 = 0
Divided through by 20
a² - 2a - 24 = 0
a² + 4a - 6a - 24 = 0
a(a+4) -6(a+4) = 0
(a-6)(a+4) = 0
(a-6) = 0 or (a+4) = 0
So, a = 6 or a = -4
The time cannot be negative, then, the time is a = 6hours
So, t_1 = a = 6hours,
So, the time used in the first journey is 6hours
So, in the second journey the time use is 2hours less than the first journey
Then, t_2 = 6 - 2 = 4 hours
t_1 = 6 hours
t_2 = 4 hours
Spanish
Un tren recorrió una distancia de 240 km.
Deje que la velocidad inicial sea
S_1 = x km / h
Supongamos que el tiempo dedicado al primer viaje es
t_1 = a
Ahora si aumenta la velocidad a
S_2 = (x + 20) km / h
Entonces, habría tomado 2 horas menos de tiempo
Entonces, el tiempo t_2 = a - 2
Los datos comunes para los dos viajes son la distancia.
Velocidad = distancia / tiempo
Para la primera etapa
S_1 = d / t_1
d = S_1 × a
d = x × a
240 = x • a
x = 240 / a Ecuación 1
Para la etapa dos
d = S_2 × t_2
d = (x + 20) × (a - 2)
240 = (x + 20) × (a - 2). Ecuación 2
Sustituye la ecuación 1 en 2
240 = (240 / a + 20) × (a -2)
240 = 240 - 480 / a + 20a - 40
240 - 240 + 40 = - 480 / a + 20a
240 - 240 + 40 = (-480 + 20a²) / a
40 = (-480 + 20a²) / a
40a = -480 + 20a²
20a² - 40a -480 = 0
Dividido entre 20
a² - 2a - 24 = 0
a² + 4a - 6a - 24 = 0
a (a + 4) -6 (a + 4) = 0
(a-6) (a + 4) = 0
(a-6) = 0 o (a + 4) = 0
Entonces, a = 6 o a = -4
El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas
Entonces, t_1 = a = 6 horas,
Entonces, el tiempo utilizado en el primer viaje es de 6 horas
Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje
Entonces, t_2 = 6 - 2 = 4 horas
t_1 = 6 horas
t_2 = 4 horas
