Question

A train accelerates from 10 kilometre per hour to 40 kilometre per hour in 2 minute how much distance does it cover in the period assume that the the tracks are straigh

Answer 1

Answer:

832.8 m or, 0.832 km.

Explanation:

Initial Velocity (u) = 10 kph

= [10*5/18] mps

= 2.77 mps

Final Velocity (v) = 40 kph

= [40*5/18] mps

= 11.11 mps

Time (t) = 2 mins or, 120 seconds

Acceleration = (v-u)/t

= (11.1-2.77)/120

= 0.0695 m/s²

According to formula,

= v² = u²+2aS                                .... Here, 'S' is the distance.

= (11.11)² = (2.77)² + [2*0.0695*S]

= 123.4321 = 7.6729 + 0.139S

= [123.4321-7.6729] = 0.139S

= 115.7692 = 0.139S

= [115.7692 / 0.139] m = S

= 832.8 m = S

Hope this helps.