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# How many values of $t$ does the particle change direction if a particle moves with acceleration $a\left( t \right) = 3{t^2} - 2t$ and its initial velocity is 0?

Last updated date: 26th Feb 2024
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Hint: Here, we will find the velocity by integrating the given acceleration. Then, we will find the general equation of the velocity using the given initial velocity. Then solving it further, we will be able to find the values of $t$. We will then substitute the points between the intervals of those points in the equation to find the required point where the particle changes its direction.

According to the question, the acceleration of the particle is $a\left( t \right) = 3{t^2} - 2t$.
Therefore, in order to find the velocity of the particle, we will find the integration.
Thus, we get, Velocity of the particle: $\int {a\left( t \right)} dt = \int {\left( {3{t^2} - 2t} \right)} dt = {t^3} - {t^2} + C$
This is because, $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
Now, according to the question, the initial velocity is 0, i.e. $t = 0$
Hence, substituting this in the velocity of the particle, we get,
${0^3} - {0^2} + C = 0$
$\Rightarrow C = 0$
Therefore, our equation becomes, $v\left( t \right) = {t^3} - {t^2}$
Now, we know that every point where the velocity is 0 is a potential turning point, thus, we get,
$v\left( t \right) = 0$
$\Rightarrow {t^3} - {t^2} = 0$
Taking ${t^2}$ common,
$\Rightarrow {t^2}\left( {t - 1} \right) = 0$
Hence,
${t^2} = 0$
$\Rightarrow t = 0$
And,
$t = 1$
Now, in order to verify whether the particle changes direction at each of these points or not, we will substitute any point between the intervals of these points to check the interval between them.
Also, we are not required to test before $t = 0$ because this is when the particle starts moving.
Thus, substituting $t = \dfrac{1}{2}$ in $v\left( t \right) = {t^3} - {t^2}$, we get,
$v\left( {\dfrac{1}{2}} \right) = {\left( {\dfrac{1}{2}} \right)^3} - {\left( {\dfrac{1}{2}} \right)^2}$
$\Rightarrow v\left( {\dfrac{1}{2}} \right) = \dfrac{1}{8} - \dfrac{1}{4} = \dfrac{{1 - 2}}{8} = \dfrac{{ - 1}}{8}$
Hence, in the interval from $t = 0$to $t = 1$, the particle moves in the negative direction.
Now, substituting $t = 2$ in $v\left( t \right) = {t^3} - {t^2}$, we get,
$v\left( 2 \right) = {\left( 2 \right)^3} - {\left( 2 \right)^2} = 8 - 4 = 4$

Therefore, in the interval beyond $t = 1$, the particle moves in a positive direction.
Hence, clearly, the particle changes its direction at $t = 1$.
Therefore, this is the required answer.

Note:
If a function gives the position of something as a function of time, the first derivative gives its velocity, and the second derivative gives its acceleration. Hence, we differentiate position to get velocity, and we differentiate velocity to get acceleration. Velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.