Дано:

υ = 36 км/ч = 10 м/с

t1 = 5 c

t2 = 25 c

t3 = 10 c

L - ?

Решение:

L = L1 + L2 + L3

L1 = at1²/2

υ = υ0 + at1; υ0 = 0 => υ = at1 => a = υ/t1

L1 = ((υ/t1)*t1²)/2 = υt1/2

L2 = υt2

L3 = υt3 - at3²/2

υ' = υ - at3; υ' = 0 => υ = at3 => a = υ/t3

L3 = υt3 - ((υ/t3)t3²)/2 = υt3 - υt3/2 = υt3/2

L = υt1/2 + υt2 + υt3/2 = υ*(t1/2 + t2 + t3/2) = 10*(5/2 + 25 + 10/2) = 10*(2,5 + 25 + 5) = 10*32,5 = 325 м

Ответ: 325 м.