A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceeration is uniform. Find the acceleration and the distance traveed by the train for attaining this velocity.

u=0

v=72 km/h=20 m/s

t=5 mins=300s

a=v-u/t

    =20-0/300

     =20/300

      =1/15

s=ut+1/2at2

    =0*300+1/2*1/15*300*300

     =0+20*150

      =3000m

       =3 km

  • 83

 what do you mean by v=72km h-1= 20 m s-1

  • -8

TO convert 72 h/m into m/s ,u have to multiply it by 5/18

  • 6
Please explain me how you answered the distance part
  • -8
Hii guys, 
i will tell u how he conveted.
see, v=72km/h
when we convert from km to h we multiply into 1000
from hr to sec divide by 60 to get min and then by 60 again to get sec
atlast multiply by 1000 and divide by 3600
so, 72*1000/3600=72000/3600=20m/s
 
  • -14
but i have a doubt,
how did u get the answer for 1/2*1/15
i got the answer has 1 .
 
  • -7

if u want to convert 72 km/h into m/s..then :

72*1000/60*60
=20 m/s..

 

  • -8
u= 0 ms-1
v=72 km/hr 

  =72×5/18
  =360/18 =20 ms-1
t=5×60 
=300s
a= v-u/t
 =20-0/300
=1/15
s=ut+1/2at​2
=0×300+1/2 × 1/15 × 300×300
=1/30×300×300
=10×300
=3000m
=3km
 
  • 18
HOPE HELPS CHEERS!!

  • 8
HOPE HELPS

  • 4

dont know
  • -1
3  km or 3000m
  • 0
3 km
  • 0
u=0m/s v=72km/h =20m/s t=5min =300sec A=v-u/t =20-0/300 =1/15m/s sq.
  • 0
U0mgjkmf
  • 0
ai chup
  • 0
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