Exercício – Movimento uniforme

Exercise 1

When the motorcycle catches up with the car, both have covered the same distances at this point, i.e. the condition is: ∆s₁ = ∆s₂

Until the motorcycle starts after ∆t = 0.5 h, the car with v₁ = 100 km/h has a lead of: 100 km/h × 0.5 h = 50 km. The following must therefore apply to the meeting point:

∆s₁ = ∆s₂  =>  v₂ × ∆t = 50 km +v₁ × ∆t

∆t is the time that passes from the start of the motorcycle. Solving the equation for ∆t:

v₂ × ∆t – v₁ × ∆t = 50 km  =>  (v₂ – v₁)∆t = 50 km  =>  ∆t = 50 km / (150 km/h - 100 km/h ) = 1h

The distance is easy to calculate. We consider the motorcycle:

∆s = 150 km/h × 1h = 150 km

Exercise 2

We choose location A as the coordinate origin (reference point) and designate the direction of travel from A to B as positive. The speed of the car is evaluated negatively, because it runs in the opposite direction.

Física - Movimento uniforme - exercício 2

Thus, the location functions for both vehicles are:

s₂ = v₂ × t and s₁ = s₀ - v₁ × t

Cond. for meeting point:  t₁ = t₂ = t

⇒ v₂ × t = s₀ - v₁ × t ⇒ t (v₂+v₁) = s₀ ⇒ t = s₀ / (v₁ + v₂)

⇒ t = 100 km / (150 km/h + 100 km/h) = 0.4 h

Thus, both vehicles meet after 0,4 h. To dermine the distances, that both vehicles have at this time from the coordinate origin (location A), we take the simpler equation for s₂:

s₂ = v₂ × t = 150 km/h × 0,4 h = 60 km