A Particle Travelling in a Straight Line I was studying for some quizzes when a wild question apperas. It goes like this:

A particle travels in a straight line with a constant acceleration of 3 meters per second per second. If the velocity of the particle is 
  10 meters per second at the time 2 seconds, how far does the particle travel during time interval when its velocity increases from 4 meters per 
  second to 10 meters per second?

My work
The given was acceleration $a$ of 3 meters per second per second and  the velocity of the particle is 10 meters per second at the time 2 seconds.
With that in mind, I get the integral of $a$ to get the velocity $v$.
$$v = \int a = \int\frac{dv}{dt} = \int 3 = 3t + C$$
Since the velocity of the particle is 10 meters per second at the time 2 seconds, we can now get the value of $C$:
$$ v = 3t + C$$
$$(10) = 3(2) + C$$
$$C = 4$$
So...the equation of velocity is $v = 3t + 4$
The distance $s$ travelled is the integral of speed $v$, so....
$$s = \int \frac{ds}{dt} = \int v = \int 3t + 4 = \frac{3t^2}{2} + 4t + C $$
Now this is my problem.....there is another given that when its velocity increses from 4 meters per second to 10 meters per second, and asking what might be the distance travelled while this acceleration happened. 
With the equation I got now $\left( \frac{3t^2}{2} + 4t + C\right) $, I don't think I could proceed because of this additional information.
How do you answer the above question?
 A: $v=v_0+at$ since when $t=2$s we have $v=10$m/s then $v_0=4$m/s
So it took $2$s to increase from $4$m/s to $10$m/s
According to the formula 
$s=\dfrac{1}{2}at^2+v_0t$ 
we plug the data and get
$s=14$m
Hope this helps
A: As you correctly computed, the velocity is given by 
$$v(t)=3t+4.$$
You already know that at $t=2$, the velocity is $10$ meters per second. Analogously, at what time is the velocity 4 meters per second?
Once you have both times, you have to compute the distance. You are saying that the travelled distance is given by
$$s(t)=\frac{3t^2}{2} + 4t+C.$$
But being more specific, $s$ gives you the position with respect to some unkown origin (since you do not know the value of $C$). The travelled distance is $s(t_{end})-s(t_{begin})$. Thus:
$$\begin{align}
s(t_{end}) - s(t_{begin}) &=\frac{3t_{end}^2}{2} + 4t_{end}+C - \left( \frac{3t_{begin}^2}{2} + 4t_{begin}+C\right) \\ &= \frac{3}{2}\left(t_{end}^2-t_{begin}^2\right) + 4(t_{end} - 4t_{begin}).
\end{align}$$
And you do not have $C$ anymore!
A: This is more of a physics question, so here is a physics answer. For movement along a straight line with constant accelration there are four formulas that are worth remembering:
$$
v^2 - v_0^2 = 2as\\
v-v_0 = at\\
s = v_0t + \frac12at^2\\
(v+v_0)t = 2s
$$
where $v_0$ is initial velocity, $v$ is final velocity, $a$ is the accelration, $t$ is the difference in time and $s$ is the difference in position from the beginning of the movement until the end.
In this case, we know $v_0, v$ and $a$, and we're asked about $s$. This means that the first formula will be most convenient. Insert and calculate, and you get your answer.
A: Given the constant acceleration, the average speed of the particle from the time it's going $4$ m/s until the time it's going $10$ m/s must just be the average of those two speeds: $7$ m/s.  (Note that you don't need to know what the acceleration is to determine this... only that it's constant.)  Now, the time over which this happens is $(10 - 4)/3=2$ seconds.  So the distance traveled is ($7$ m/s) $\times$ ($2$ seconds) $=14$ meters.  The other fact given, that the velocity is $10$ m/s at $t=2$ seconds, is a red herring.
