El padre conduce su coche a una velocidad v1 de 100 km/h. El hijo sale en moto media hora más tarde a una velocidad de 150 km/h. ¿Cuándo y a qué distancia alcanza el hijo a su padre?
Exercise - Motion with constant speed
Ejercicio 2:
El padre conduce su automóvil a una velocidad de 100 km/h desde el lugar B. Al mismo tiempo, su hijo conduce su motocicleta desde el lugar A hasta el padre a una velocidad de 150 km/h. ¿En qué punto se encuentran si los dos lugares están a 100 km el uno del otro?
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.
Ejercicio movimiento uniforme solución
Thus, the location functions for both vehicles are:
Δs2 = v2 × t and Δs1 = s0 - v1 × t
Cond. for meeting point: t2 = t1 = t
⇒ v2 × t = s0 - v1 × t ⇒ t (v2+v1) = s0 ⇒ t = s0 / (v2 + v1)
⇒ t = 100 km / (100 km/h + 150 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:
