Pociąg zwiekszył swą prędkość od =5m/s do v=10m/s na odcinku drogi s=500m. Oblicz przyspieszenie ruchu pociągu
Δv=v-v₀
Δv=10m/s-5m/s=5m/s
a=Δv/t
s=1/2 at²
s=1/2 Δv/t*t²
s=1/2Δvt
t=2s/Δv
t=1000m/5m/s
t=200s
a=5m/s/200s
a=0,025m/s²
dane:
vo = 5 m/s
v = 10 m/s
s = 500 m
szukane:
a = ?
a = dv/t = (v-vo)/t
s = vo x t + at^2 /2
s = vo x t + (v-vo)/t x t^2 /2
s = vo x t + (v-vo) x t /2 I*2
2vo x t + (v-vo) x t = 2s
t (2vo+v-vo) = 2s
t(vo+v) = 2s /:(vo+v)
t = 2s/(vo+v) = 2 x 500m/(5m/s+10m/s) = 200/3 s
a = (15m/s-10m/s) / 200/3s = 10m/s / 200/3 s
a = 0,15 m/s2
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Odp.Szukane przyspieszenie wynosi 0,15 m/s2.
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Δv=v-v₀
Δv=10m/s-5m/s=5m/s
a=Δv/t
s=1/2 at²
s=1/2 Δv/t*t²
s=1/2Δvt
t=2s/Δv
t=1000m/5m/s
t=200s
a=Δv/t
a=5m/s/200s
a=0,025m/s²
dane:
vo = 5 m/s
v = 10 m/s
s = 500 m
szukane:
a = ?
a = dv/t = (v-vo)/t
s = vo x t + at^2 /2
s = vo x t + (v-vo)/t x t^2 /2
s = vo x t + (v-vo) x t /2 I*2
2vo x t + (v-vo) x t = 2s
t (2vo+v-vo) = 2s
t(vo+v) = 2s /:(vo+v)
t = 2s/(vo+v) = 2 x 500m/(5m/s+10m/s) = 200/3 s
a = (15m/s-10m/s) / 200/3s = 10m/s / 200/3 s
a = 0,15 m/s2
============
Odp.Szukane przyspieszenie wynosi 0,15 m/s2.