Question

a train start from rest and attains velocity 72 km/hr in 5min find acceleration and distance travelled​

Answer 1

Answer: a=1/15 m/s^2 and S=3 m

Explanation

U=0,V=72 km/h

72×5/18=20 m/s

t=5 min

=5×60

=300 s

a=v-u/t

=20-0/300

=1/15 m/s^2

Now we have to find distance

By using S=ut+1/2at^2

S=0×300+1/2×1/15×300×300

=0+1/30×300×300

=3000 m

3000 m=3 km