Question

a train starting from rest ,attains velocity of 72Km/h in 5 min .assuming that the acceleration is uniform,find the acceleration and the distance travelled by the train while it attained the velocity answer should be 1/15 ms-2,3km i want the answer with procedure​

Answer 1

Answer:

Given :

Initial Velocity=u=0m/s

Final velocity= V= 72km/h

Time =5 min =5/60 h =1/12 h

acceleration=a=?

As we know that , a=v-u/t

⇒a=72-0/1/12

=72*12

=864km/h²

Distance travelled  =s=ut+1/2at²

=0x1/12 +1/2(864)(1/12)²

=(1/2)x864x(1/144)

=3km

∴Distance travelled by the train for attaining this velocity is 3km.

Answer 2

Answer:

u=0

v=72 kmph=20 m/s

t=5min=300sec

acceleration=a=v-u/t

                    =20-0/300

                    =1/15 m/s²

s=ut+1/2×a×t²

 =1/2×1/15×90000

 =3000 m

 =3 km

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