Question

A train moving with a uniform speed 90 km/h . If we increase its speed byb15km/h then it takes 30 min less time than before. Calculate the original speed

Answer 1

distance covered = 90 km

let the speed of the train = x km/hr

time (t1 ) = distance /speed

t1 = 90/x -----(1)

2) distance = 90 km

speed = (x+15) km/hr

t2 = 90 /(x+15) ----(2)

given

t1-t2 = 30 minutes

90/x - 90 / (x+15) = 1/2 hr

[90(x+15) -90x]/x(x+15) = 1/2

[90x + 1350 -90x]/ (x^2+15x) =1/2

1350 *2= x^2+15x

2700 = x^2+15x

x^2+15x-2700=0

x*x +60x -45x - 45*60=0

x(x+60) - 45(x+60)=0

(x+60)(x-45)=0

x+60=0 or x-45=0

x= -60 0r x= 45

x should not be negative

x= 45

therfore speed of the train = 45km/hr

Answer 2

Answer:

heya..

Velocity at the time of applying break = 90 km/h = 90 km/h ... and the speed of car reduces to half its origi

i hope its help u