Calculating Acceleration Hi I have just started my GCSE Physics and Maths course and I am hoping to confirm my understanding of the formula to calculate acceleration. Now I understand $a= \frac{(v_f - v_i)}{t}$. However I am not sure how I use this formula or change it to accomodate the following question.

A toy car is acceleratingat $2\frac{m}{s^2}$. It starts at a velocity of $5\frac{m}{s}$. What will its velocity be after $3s$?

I didn't know how to rearrange the formula to answer the question, so I just thought it through (so im sure its probably wrong). Could someone say yes/no and explain or link me to an explination to this scenario?
So my thinking was, if its traveing at $5\frac{m}{s}$ for $3s$ that would be $6\frac{m}{s}$ plus the initial velocity of $5\frac{m}{s}$. So the answer is $11\frac{m}{s^2}$?
 A: Yes, you did it right.
One thing that helps give confidence is to check your units.  Here, you multiplied $2 \frac{\mbox{m}}{\mbox{s}^2} \cdot 3 \mbox{s} = 6 \frac{\mbox{m}}{\mbox{s}} $
so you know your units are fine.
A: Let's work on $a = \frac{v_f - v_i}{t}:$
$$a = \frac{v_f - v_i}{t} \Rightarrow \\
\Rightarrow a \cdot t = v_f - v_i \Rightarrow \\
\Rightarrow v_f = a \cdot t + v_i.$$
Since $a = 2$, $v_i = 5$ and $t = 3$, then:
$$v_f = 2 \cdot 3 + 5 = 11 \frac{m}{s}.$$

Long version
Your objective is to find out $v_f$ (this is what the problem is asking to you). Then do something to obtain $v_f = $ something... Let's start!
$$a = \frac{v_f - v_i}{t}$$
First step: multiply both side by $t$:
$$a \cdot t = \frac{v_f - v_i}{t} \cdot t.$$
Second step: simplify $t$ on the right hand side:
$$a \cdot t = v_f - v_i.$$
Third step: add $v_i$ to both side:
$$a \cdot t + v_i= v_f - v_i + v_i.$$
Fourth step: simplify $v_i$ on the right hand side
$$a \cdot t + v_i= v_f.$$
Fifth step: invert the order of the equation:
$$v_f = a \cdot t + v_i$$.
Notice that these are very elementary algebraic steps...
