Question

A person attempting to maintain social distancing is walking at 1.20 m/s and is being overtaken by a person walking at 1.58 m/s. If the faster person starts 13.3 m behind the slower person, how long does it take for the faster person to come within 2.00 m of the slower person?

Answer 1

Speed of the person attempting to maintain social distancing = V₁ = 1.2 m/s

Speed of the faster person = V₂ = 1.58 m/s

Initial distance of the faster person from the slower person = D₁ = 13.3 m

Final distance of the faster person from the slower person = D₂ = 2 m

Distance the faster person covers more than the slower person = D

D = D₁ - D₂

D = 13.3 - 2

D = 11.3 m

Time taken by the faster person to come within 2 m of the slower person = T

Distance traveled by the slower person in this time = L₁

L₁ = V₁T

Distance traveled by the faster person in this time = L₂

L₂ = V₂T

D = L₂ - L₁

D = V₂T - V₁T

D = (V₂ - V₁)T

11.3 = (1.58 - 1.2)T

T = 29.7 sec

Time taken by the faster person to come within 2 m of the slower person = 29.7 sec