from a station x a train starts from rest and attains a speed of 54km/h in 10 s .Then by applying brakes negative acceleration of 2.5 ms power -2 is produced and it stops at station y in 6s Find the distance between station x and y
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Firstly here, u₁=initial velocity=0
v₁=final velocity=54km/h=54000/3600=15m/s
time=t₁=10s
∴, using v₁=u₁+f₁t₁ we get,
15=0+f₁×10
or, f₁=15/10
or, f₁=1.5m/s²
Again using s₁=u₁t₁+1/2f₁t₁² we get,
s₁=0×t₁+1/2×1.5×10²
or, s₁=1/2×1.5×100
or, s₁=1.5×50
or, s₁=75m
Finally, f₂=2.5ms², t₂=6s, u₂=15m/s, v₂=0
Using, s₂=u₂t₂-1/2f₂t₂² [(-) sign due to negative acceleration] we get,
s₂=15×6-1/2×2.5×6²
or, s₂=90-1/2×2.5×36
or, s₂=90-2.5×18
or, s₂=90-45
or, s₂=45m
∴, s=s₁+s₂=75+45=120m
∴, the distance between the stations x and y is 120m.
v₁=final velocity=54km/h=54000/3600=15m/s
time=t₁=10s
∴, using v₁=u₁+f₁t₁ we get,
15=0+f₁×10
or, f₁=15/10
or, f₁=1.5m/s²
Again using s₁=u₁t₁+1/2f₁t₁² we get,
s₁=0×t₁+1/2×1.5×10²
or, s₁=1/2×1.5×100
or, s₁=1.5×50
or, s₁=75m
Finally, f₂=2.5ms², t₂=6s, u₂=15m/s, v₂=0
Using, s₂=u₂t₂-1/2f₂t₂² [(-) sign due to negative acceleration] we get,
s₂=15×6-1/2×2.5×6²
or, s₂=90-1/2×2.5×36
or, s₂=90-2.5×18
or, s₂=90-45
or, s₂=45m
∴, s=s₁+s₂=75+45=120m
∴, the distance between the stations x and y is 120m.
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