Respuesta :
Answer:
A) t = 900 s, the meeting time is 2 h 15 min p.m, B) x = 15000 m
Explanation:
A) This is an exercise in kinematics, we must assume that car A is behind so that it can catch up with the slower car B.
car A
x = x₀ₐ + v₀ₐ t
car B
x = x_{ob} + v_{oB} t
in this case x₀ₐ = 5 km and the initial position. B is zero
at the bare point the position of the two vile cars is the same.
x₀ₐ + v₀ₐ t = v_{ob) t
x₀ₐ = (v_{ob} - v₀ₐ) t
t = [tex]\frac{x_{oa}}{v_{ob} - v_{oa} }[/tex]
let's reduce the magnitudes to the SI system
x₀ₐ = - 5 km (1000m / 1 km) = -5000 m
v₀ₐ = 80 km / h (1000 m / 1km) (1 h / 3600 s) = 22.22 m / s
v_{ob} = 60 km / h (1000 m / 1 km) (1 h / 3600s) = 16,667 m / s
let's calculate
t = [tex]- \frac{5000}{16.667 - 22.222}[/tex]
t = 900 s
let's reduce to minutes
t = 900 s (1 min / 60 s) = 15 min
Therefore, the meeting time is 2 h 15 min p.m.
B) at what point are they
x = v_{ob} t
x = 16,667 900
x = 15000 m
