Răspuns :
V= d/ t
in conditii normale, v = d/t si t = S/v = 15 minute
in conditii de drum in reparatii, V1 = 40/100v = s/t1 t1 = 100v x S / 40 = 10S/4v
t1 - t = 10S/ 4v - S/v = S/v( 10/4 - 1) = S/v x 1,5
dar s/v = 15 min rezulta ca tx - t = 15 x 1,5 = 22,5 minute intarziere
in conditii normale, v = d/t si t = S/v = 15 minute
in conditii de drum in reparatii, V1 = 40/100v = s/t1 t1 = 100v x S / 40 = 10S/4v
t1 - t = 10S/ 4v - S/v = S/v( 10/4 - 1) = S/v x 1,5
dar s/v = 15 min rezulta ca tx - t = 15 x 1,5 = 22,5 minute intarziere
∆t'=15min
Distanta este aceeasi deci:
v•∆t'=40v/100•∆t"|÷v
∆t'=40/100•∆t"
∆t"=(15•100)÷40=37,5 min
37,5min-15min=22,5 min