40 km/h = 40,000 m/ 3600 s = 11.1 m/s
5 min = 300 s
time to accelerate = t
d1 = (1/2) a t^2
11.1 = a t
so
d1 = (1/2)(11.1) t
d1 = 5.56 t
runs at 11.1 m/s for t2 seconds
distance = d2 = 11.1 t2
deaccelerate time = 3 t
0 = 11.1 + 3 a2 t
a2 t = -11.1/3 = - 3.7 t
a2 = -3.7/t
d3 = 11.1 (3t) + (1/2) (-3.7 /t )(3t)^2
d3 = 33.3 t - 16.7 t
d3 = 16.7 t
so
5.56 t + 11.1 t2 + 16.7 t = 3000 meters
t + t2 + 3 t = 300 seconds
so t2 = (300 -4t)
solve for t
a train takes 5 minutes to cover a distance of 3 km between two stations p and q . Starting from rest at p , it accelerate at a constant rate to a speed 40 km/h and maintain this speed untill it is brought to rest at q . If train takes 3 times as long to deaccelerate as it does to accelerate , find time taken by the train to accelerate .
2 answers
T = 148
2 answers