A passenger train takes 3 hours less for a journey of 360km . if its speed increased by 10km/hr from its usual speed . what is its usual speed?
360 = v t
360 = (v+10)(t-3) = vt -3v + 10t -30
so
vt = vt -3v + 10 t - 30
10 t = 3 (v+10)
but t = 360/v
3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30
360 = v t
360 = (v+10)(t-3) = vt -3v + 10t -30
so
vt = vt -3v + 10 t - 30
10 t = 3 (v+10)
but t = 360/v
3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30
To find the usual speed of the passenger train, we need to set up an equation based on the given information.
Let's assume the usual speed of the train is 'x' km/hr.
According to the problem, when the train's speed is increased by 10 km/hr, its new speed becomes (x + 10) km/hr.
We know that the duration of the journey is 3 hours less when the speed is increased. So, if the usual speed of the train is 'x' km/hr, the time taken for the journey at this speed is:
Time taken = Distance / Speed = 360 km / x km/hr
When the speed is increased by 10 km/hr, the time taken for the journey is:
Time taken = Distance / Speed = 360 km / (x + 10) km/hr
According to the problem, the time taken for the journey at increased speed is 3 hours less than the time taken at the usual speed. So, we can set up the following equation:
360 / x = 360 / (x + 10) + 3
To solve this equation, we can multiply both sides by the common denominator of (x)(x + 10):
360(x)(x + 10) / x = 360(x)(x + 10) / (x + 10) + 3(x)(x + 10)
This simplifies to:
360(x + 10) = 360(x) + 3(x)(x + 10)
Expanding and simplifying:
360x + 3600 = 360x + 3x^2 + 3600
Simplifying further:
3x^2 = 0
Dividing both sides by 3:
x^2 = 0
Taking the square root of both sides:
x = 0
However, the usual speed of the train cannot be zero. Hence, there must be an error in the problem or information provided.
Please double-check the question or provide any additional relevant information to rework the solution.