Question

A train acceleration from 20 km/hr to 80km/hr in 4 min. how does distance does it cover in this period? Assuming that the tracks are straight?

Answer 1

Initial velocity(u) = 20 km/hr

final velocity (v) = 80km/hr

time take (t) = 4/60 hr
= 1/15 hr

Distance(s )= ?
so firstly we have to find acceleration

Acceleration(a) = (v - u)/t
= ( 80-20)/1/15
= 60× 15
= 900 km/hr ^2

Now by using formula

v^2 = u^2 + 2as

(80)^2 = (20)^2 + 2× 900×s

6400 = 400 + 1800 × s

6400-400/1800 = s

6000/1800 = s

60/18 = s

10/3 = s

3.33 = s

s= 3.33 km is your answer