# 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?

• Since you are interested in the difference between the positions at two different times, the value of $C$ doesn't matter. Just set $C=0$ and move on... – rogerl Sep 25 '17 at 14:46
• You know $v=3t+4$. At what time does the particle have velocity 4? At what time is the velocity 10? I hope this helps. – Gerhard S. Sep 25 '17 at 14:46

$v=v_0+at$ 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

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!

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.