Acceleration of car. One dimensional motion easy problem A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
My attempt at solving the problem:
$$a(x) = \frac{v - u}{t}$$
where
$v =$ final velocity
$u =$ initial velocity
$$$$
I get the answer as $4.05 \space ms^{-2}$
But the correct answer given to the problem is $8.10 \space ms^{-2}$.
They used a different equation to reach that answer.
Did I use the wrong equation? I have the average velocity and not the instantaneous veolcity? 
 A: That equation is not the most straight foreward for this situation but it can be used, here's how. If you divide the displacement, 110 m, by the time of 5.21 s you will get 21.1 m/s. But that is the average velocity over the entire displacement. The average veloctiy, when v is changing uniformly, is found by adding the initial and final velocities and dividing by two. Since the initial velocity is 0 m/s, you will need to double the 21.1 m/s for the final velocity such that the average will be the calculated 21.1 m/s. So the final velocity, and therefore the change in velocity over that displacement, is 42.2 m/s. When you divide that delta v by the time over which it occurs, you will have how much delta v per second, acceleration.
While it's true there is another equation that does this all in one step, it would be good for you to understand the process I've outlined above.
A: What is your $V_f$ ?
$V_f$ is not given in the question so you can't use this equation,
$$V_f =V_i +at$$
But the distance is given which will allow you to use
 $$S=ut+1/2 at^2 $$
A: 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.
A: 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.
