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?
Best Answer
$v=v_0+at$ 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