Final answer:
The slowest constant speed at which the passenger can run to start catching up with the train immediately after 4.5 seconds is greater than the train's speed at that moment, calculated using the train's acceleration and the time. This minimum speed is 3.15 m/s.
Step-by-step explanation:
To find the slowest constant speed a passenger must run to catch a train that is accelerating away from a station, we first need to determine the distance that the train has covered by the time the passenger starts to run. The train has an acceleration of 0.7 m/s2. Using the equation of motion s = ut + 0.5at2, where s is the position (distance) from the starting point, u is the initial velocity (0 m/s, as the train starts from rest), a is the acceleration, and t is the time, we can find that distance. In this case, since the passenger starts running 4.5 s after the train has left, the distance the train will have covered in that time is s = 0*4.5 + 0.5*0.7*4.52, which equals 7.0875 meters.
To determine the necessary running speed v of the passenger, we assume a constant speed meaning no acceleration. Thus, the additional distance that the train will cover while the passenger attempts to catch it (let's call that distance d), will be equal to the distance covered by the running passenger. This gives the equation d = 0.5*a*t2, with t being the time it takes for the passenger to reach the train. Since the train will continue to accelerate, its ongoing distance covered after the 4.5 s will be d = 7.0875 + 0.5*0.7*t2 while the passenger's distance will be d = v*t. Equating these two we get 7.0875 + 0.5*0.7*t2 = v*t, which simplifies to v = (7.0875 + 0.5*0.7*t2)/t.
Since we desire to find the slowest speed at which the passenger can run to just catch up with the train, we want to find the minimum value of v. However, as time t gets larger, the term 0.5*0.7*t will dominate, and it will require a higher speed v to catch the train. Hence, the running speed v must be greater than the speed of the train at 4.5 s, which is simply the product of acceleration and time: 0.7 m/s2*4.5 s = 3.15 m/s. This is the minimum speed the passenger must run to start catching up with the train immediately after 4.5 s, without falling further behind.