Question

a train starting from rest attains a velocity of 72 km h-1 in 5 min . assuming that the acceleration is uniform. find the distance traveled by the train for attaining this velocity 

Answer 1

INITIAL VELOCITY(U) = 0
FINAL VELOCITY (V)= 72 KM/H = 72×(5/18) M/S= 20 M/S
TIME(T) = 5 MIN = 300 SEC 
ACCELERATION(A) = (V-U)/T = 20/300 = 1/15 M/S²

DISTANCE TRAVELLED = (V²-U²)/(2A) = (20² - 0) / (2/15) = (400 × 15)/2 = 3000M = 3KM
                                          

Answer 2

S = average velocity * time traveled = (0 + 72 kmph) /2  * 5 /60 hours
     = 36*5/60 = 3 km

======================

a = acceleration = (v - u) / t = (72 - 0) / (5/60) = 12 * 72 km/hour²

Distance traveled = S = u t + 1/2 a t² = 0 + 1/2 * 12 * 72 * (5/60)² 
                                    = 3 km