A train is travelling at a speed of 90 km per hour .Brakes are applied so as to produced a uniform acceleration of -0•5 m s-2. Find how far the train will go before it is brought to rest?
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Answered by
3
u= 90 km/h = 25 m/s
v= 0 km/h
a=-0.5 m/s2
so, t= (v-u)/a
= (0-25)/-0.5
=50 s
therefore, s= ut+1/2 at^2
= 25×50 -1/2 ×1/2× 2500
=1250- 625
= 625 m ans
v= 0 km/h
a=-0.5 m/s2
so, t= (v-u)/a
= (0-25)/-0.5
=50 s
therefore, s= ut+1/2 at^2
= 25×50 -1/2 ×1/2× 2500
=1250- 625
= 625 m ans
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Answered by
3
u=90km/h = 25m/s
v=0m/s
a= -0.5 m/s^2
s=?
by the third equation of motion:
v^2 -u^2 =2as
0^2-25^2= 2× (-0.5)×s
-625= -1 ×s
- s = - 625
minus minus gets cancelled so s=625m
hence it travelled 625 m before comming to rest.
v=0m/s
a= -0.5 m/s^2
s=?
by the third equation of motion:
v^2 -u^2 =2as
0^2-25^2= 2× (-0.5)×s
-625= -1 ×s
- s = - 625
minus minus gets cancelled so s=625m
hence it travelled 625 m before comming to rest.
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