Exercise 1:

Father drives his car at a speed v_{1} of 100 miles/h. The son sets off on his motorbike half an hour later at a speed v_{2} of 150 miles/h. When and after what distance did the son catch up with his father?

Exercise 2:

Father starts with his car at a speed of 100 miles/h from place B. At the same time, his son starts with his motorcycle from place A to meet his father. His speed is 150 miles/h. At which point do they meet if the two places are 100 miles apart?

Exercise 1

When the motorcycle catches up with the car, both have covered the same distances at this point, i.e. the condition is: **∆s _{1} = ∆s_{2}**

Until the motorcycle starts after ∆t = 0.5 h, the car with v_{1} = 100 miles/h has a lead of: 100 miles/h × 0.5 h = 50 miles. The following must therefore apply to the meeting point:

∆s_{1} = ∆s_{2} => **v _{2} × ∆t = 50 miles +v_{1} × ∆t**

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

v_{2} × ∆t – v_{1} × ∆t = 50 miles => (v_{2} – v_{1})∆t = 50 miles => ∆t = 50 miles / (150 mph - 100 mph) = 1h

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

∆s = 150 miles/h × 1h = 150 miles

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.

Thus, the location functions for both vehicles are:

s_{2} = v_{2} × t and s_{1} = s_{0} - v_{1} × t

Cond. for meeting point: t_{1} = t_{2} = t

⇒ v_{2} × t = s_{0} - v_{1} × t ⇒ t(v_{1}+v_{2}) = s_{0} ⇒ t = s_{0} / (v_{1} + v_{2}) ⇒ t = 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 s_{2}:

s_{2} = v_{2} × t = 150 miles / h × 0.4 h = 60 miles