1a) A formula describing the cars displacement in the first t seconds is simply a constant acceleration equation:


b) If the van starts when the gap between the two vehicles is 4m, it must start when the car has moved 3m, since there is a 1m starting gap. So, the time interval is, using the above equation:



c) Both the car and the van will travel the same distance in accelerating to 15 m/s and take the same time to do so because they both have the same acceleration. Since the van starts 2 s later than the car, once the car reaches 15 m/s it will have travelled for a further 2 s at 15 m/s before the van reaches the same speed. Thus the gap between the cars at this point will be:


2) This is a classic question. By classic I mean that it involves simultaneous equations, which crop up nearly all the time in mechanics! To find the distance at which the dog catches up with the bone, we need to know the time at which this happens. To calculate this time, all we need to realise is that the distances travelled by the dog and the bone must be equal.
So firstly, the distance travelled by the stone as a function of time:


And the same for the distance travelled by the dog:


So, to calculate how long it takes the dog to catch the bone:




Thus, to answer the question, the actual distance travelled out of the bank by both the dog and the bone is, using (2):



