A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h −1 in 10 minutes. Find its acceleration.

Given that:

Initial velocity,u=0

Final velocity,v=40km/h

Time taken t=10 min=10/60=1/6 hours

Using v=u+at

a=(v-u)/t=(40-0)/(1/6) = 240 Km/h2

  • 28

final velocity=11.11 ms-1 , time taken=600 sec. acceleration=(11.11/600) ms-2 =0.0185 ms-2

  • 11

u=0

v=40

  • -18

u=0

v=40 km/hr

t=10 min =1/6hr

a=v-u/t

a=40-0/1/6

a=40*6

a=240km/hr2

  • 2

 intial speed of train,u=0

final speed,v=40km/hr-1

                    =40*5/18 m/s-1

                    =100/9 m/s-1

time,t =10min

          =10*60 s=600 s  

acceleration

                   a= v-u/t

                   a=(100-0/9) / 600

                   a=100/9 /  600/1

                   =100/9 * 1/600

                   =1/54 m/s-2

  • -1

initial speed of train, u=0

final speed, v=40kmh-1=40x1000m/3600s= 40x5/18=11.11m/s

time,t=10 minutes=300 seconds

acceleration = change in velocity

  time taken

a=v-u

  t

a=(11.11m s-1-0m s-1/300)

  =11.11m s-2

  therefore, the acceleration of the train is 11.11m s-2  

  • -13
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