Question

David parcurge cu bicicleta 2/5 din distanta dintre 2 localitati cu v1=10 m/s, 1/6 din distanta ramasa cu v2=0,9 km/h, iar restul drumului v3=300m/min.​

Answer 1

V1 = 10 m/s
V2 = 0,9km/h = 900m/3600s = 0,25 m/s
V3
= 100m/m = 100m/60s = 1(6) m/s
VI = d/t,=(2d/5)/t => t = 2d/5V,
d2 = 1/6 3d/5 = 3d/30 = d/10
V2 = de/t2 = (d/10)/t2 => t2 = d/10V2
da = 3d/5 - d/10 = 5d/10 = d/2
Va= delta = (d/2)/t => ta = d/2V3
um= d/t
t=t1+t2+t3
vm = d/(2d/5V, + 1d/10V2 + 1d/2V3)
= d/d(2/5V, + 1/10V2 + 1/2V3)
= 1/(4/10V, + 1/10V2 + 5/10V3)
=1/(4V2V3 + VIVa + 5V,V21/10VIV2V3
=10VIV2Va/(4V2⅓3 + VIV3 +5V1V2)
= 41.5/(1,66 + 16,6 + 12,5)
= 41.5/30.76
= 2.35 m/s
3d/5