Question

train starting from rest attains velocity of 108km/h in 20 min. find the acceleration and distance travelled by the train for attaining this velocity assuming the acceleration to be uniform​

Answer 1

Answer:

Acceleration = 324 km/hr^2

Distance travelled = 18 km

Explanation:

given:- u = 0 (initial velocity)

,v = 108km/hr (final velocity)

and ,t = 20min. = (1/3)hr. (time taken)

So, by applying v = u + a x t

where, a = acceleration

=> 108 = 0 + a x (1/3)

=> a = 108 x 3

=> a = 324 km/hr^2

Now, by applying

S = u x t + (1/2) x a x t^2

where , S = distance travelled

=> S = 0 + (1/2) x 324 x (1/3)^2

=> S = 324/18

=> S = 18 km ....

Answer 2

initial velocity(u)=o (starts from rest)

final velocity(v) = 108km/hr = 30m/s

time taken = 20min = 1200s

using equation of motion

v = u + at

30 = 0 + a*1200

a= 1/40 ms^-2

distance traveled(s)

s= ut + 1/2 at^2

= 1800m or 1.8km

hope it helps