A man travels 360m along a straight road. He walks for the first 120m at 1.5ms-1, runs the next 180m at 4.5ms-1, and then walks the final 60m at 1.5ms-1. A women travels the same route, in the same time. At what time does the man overtake the women?

In order to answer this question we will break it down into several pieces. Firstly, using physical arguments, can we narrow down which interval the man must overtake the women? After some thought, it must be within a time interval in which the man is travelling faster than the women. So, let us calculate the velocity of the women. Seen as she is travelling at a constant speed, we can use the well known equation, v = d/t (speed = distance/time). We know that she travels a distance of 360m, but how long does it take her? From the question, we know it takes her the same length of time as the man, so we must calculate how long it took the man in order to calculate her speed. Given that the man travelled at different speeds at different times we need to add up all his subsequent times. So, his total journey took, T_{total} = (120 + 60)m/(1.5)ms^{-1} + (180)m/(4.5)ms^{-1} = (120 + 40)s = 160s. Therefore, her speed along the journey was: v = 360m/160s = 2.25ms^{-1}. So, from our previous discussion, the man must overtake the women in the time interval in which he is travelling at 4.5ms^{-1}. We now have all the information to write the equations of motion of both the man and the women. We first write the equation of motion of the women. Since she is travelling at constant speed we have: s_{Women} = 2.25t. Now, using our SUVAT equations, we are able to write the position of the man, as a function of time, where t is now between 80s and 120s. So, the position of the man is given by the equation : s_{Man} = 120 + 4.5(t-80). Finally, at the exact point in which the man overtakes the women, they must have travelled exactly the same distance. Therefore, s_{Women} = s_{Man} => 2.25t = 120 + (t-80)(4.5) => 240 = 9/4 t => t=(160) (2/3) s.