Acceleration of car. One dimensional motion easy problem

That equation is not the most straight foreward for this situation but it can be used, here's how. If you divide the displacement, 110 m, by the time of 5.21 s you will get 21.1 m/s. But that is the average velocity over the entire displacement. The average veloctiy, when v is changing uniformly, is found by adding the initial and final velocities and dividing by two. Since the initial velocity is 0 m/s, you will need to double the 21.1 m/s for the final velocity such that the average will be the calculated 21.1 m/s. So the final velocity, and therefore the change in velocity over that displacement, is 42.2 m/s. When you divide that delta v by the time over which it occurs, you will have how much delta v per second, acceleration.

While it's true there is another equation that does this all in one step, it would be good for you to understand the process I've outlined above.

What is your $V_f$ ?

$V_f$ is not given in the question so you can't use this equation, $$V_f =V_i +at$$

But the distance is given which will allow you to use $$S=ut+1/2 at^2 $$

Yes, I think so. Below is the proper formula for the distance an object accelerating at a constant rate goes over time. I used it to get a formula where you enter distance and time traveled to get the acceleration. m=0.5at^2 2m/t^2=a (2*110)/5.21^2=8.10489203915 m/s^2

Hope this helps.

The displacement is equal to the area under a velocity $v$ against time $t$ graph as shown below.

If the body starts from rest and its final velocity is $v_{\rm f}$ then the average velocity is $\dfrac{v_{\rm f}}{2}$ and that is were your missing "$2$" comes from.