Question

1.A train starting from rest 
attains a velocity of 72km/h in 5 minutes.Assuming  that the  
acceleration is uniform,
Find (i)The acceleration and  (ii)The distance travelled by the train for
attaining this velocity.

Answer 1

$V = U + a t \\ \\ 72 * \frac{1000}{3600} m/sec = 0 + a * 5 * 60 sec \\ \\ a = 0.0667 m/sec^2 \\ \\ S = u t + \frac{1}{2}\ a\ t^2 = 0 t + \frac{1}{2} 0.0667 * 300^2 = 3,000 meter = 3 km$

Answer 2

u=0 v=72x1000/60x60=20m/s  t=5x60s=300s   a=? using the eqnv=u+at   a= v-u/t ie  a=20-0/300=20/300=1/15m/s2   s=ut+1/2at2  ies=0x300+1/2x1/15x300x300=3000m=3km