A train starting from rest moves with a uniform acceleration of 0.2 m/s square for 5 minutes .Calculate the speed acquired and the distance travelled in this time
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Answered by
20
a = 0.2 m/s^2
t = 5*60 = 300 sec.
u = 0
then v= u + at
v = 0 + 0.2*300
v = 60 m/s
t = 5*60 = 300 sec.
u = 0
then v= u + at
v = 0 + 0.2*300
v = 60 m/s
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Answered by
21
u=0m/s
v= to be found
a=0.2m/a²
t=5 min = 300 sec
using first equation of motion
v=u+at
v=0+0.2×300= 0+ 60 = 60 m/s
now v = 60m/s
using second equation of motion
s = ut+1/2at²
s=0×0.2 + 1/2 × 0.2 × 300 × 300
s= 0+1/2 × 150 × 0.2 × 300
s= 9000m = 9km
Hope this helps
v= to be found
a=0.2m/a²
t=5 min = 300 sec
using first equation of motion
v=u+at
v=0+0.2×300= 0+ 60 = 60 m/s
now v = 60m/s
using second equation of motion
s = ut+1/2at²
s=0×0.2 + 1/2 × 0.2 × 300 × 300
s= 0+1/2 × 150 × 0.2 × 300
s= 9000m = 9km
Hope this helps
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