So initial velocity = 0m/s$0m/s$.
Total time taken = 1s + 4s = 5s.$1s + 4s = 5s.$
accelerationAcceleration, assuming it is constant = 6m/s^2.$6m/s^2.$
How far will the car be from the starting line = displacement = ut + 1/2 at^2 = (0m/s)(5s) + 1/2 (6m/s^2)(5s)^2 = 75m$displacement = ut + 1/2 at^2 = (0m/s)(5s) + \frac12 (6m/s^2)(5s)^2 = 75m$
Your question I suppose is why must a square the time? For this, we need to understand the derivation of this formula of s = ut + 1/2at^2.$s = ut + 1/2at^2.$ where s=displacement, u=inital velocity, a=acceleration and t=time.$s=displacement,\ u=initial\ velocity,\ a=acceleration\ and\ t=time.$
It originates from the formula of s = 1/2(u+v)t.$s = \frac12(u+v)t.$ This is the formula of a trapezium. It represents the area under the graph of a velocity-time graph. If I were to find the are under the graph of a velocity-time graph, Im actually finding the displacement because velocity X time = displacement.$velocity\ *\ time = displacement.$
Now you also need to know the formula v = u + at.$v = u + at.$ The final velocity, v$v$, is equals to the initial velocity plus the acceleration and the time taketaken.
If I were to substitute v = u + at$v = u + at$ into s = 1/2(u+v)t$s = \frac12(u+v)t$, I would get s = ut + 1/2at^2.$s = ut + \frac12at^2.$
That is the explanation. I hope this helps. If you require a diagram to visualisevisualize the area under the graph, I will add it in if you ask for it.
