### Video Transcript

A particle moves along the π₯-axis. At time π‘ seconds, itβs displacement from the origin is given by negative π‘ cubed plus four metres, where π‘ is greater than or equal to zero. The letter π is used to denote the displacement. Determine the time at which the particleβs acceleration is nine metres per second squared.

Letβs have a look at the wording in the question in more detail. The particle moves along the π₯-axis. This means it moves either left or right. After π‘ seconds, where π‘ is greater than or equal to zero, its displacement from the origin β i.e. its change in position from the origin in metres β is given by the function π of π‘ equals negative π‘ cubed plus four.

Letβs substitute some values of π‘ into the particleβs displacement function π in order to get a better idea of the motion of the particle. Substituting π‘ equals zero into π of π‘, we obtained negative zero cubed plus four, which is just four. So, the displacement of the particle at time π‘ equals zero seconds is four metres from the origin. Similarly, the displacement of the particle from the origin at time π‘ equals one second, π‘ equals two seconds, and π‘ equals three seconds is three metres, negative four metres, and negative 23 metres, respectively.

Letβs mark these four values for the displacement of the particle from the origin on the π₯-axis. We can clearly see that the particle is moving in the negative π₯-direction since as π‘ gets larger, the displacement function π gets smaller in the negative π₯-direction. More rigorously, if π‘ one is bigger than π‘ two, then π‘ one cubed is bigger than π‘ two cubed. And therefore, negative π‘ one cubed is smaller than negative π‘ two cubed, as multiplying both sides up in an inequality by a negative number flips the inequality sign.

Adding four to both hand sides of the inequality, we obtain that π of π‘ one is smaller than π of π‘ two whenever π‘ one is greater than π‘ two. The question asks us to determine the time at which the particleβs acceleration is nine metres per second squared. Since the particle is moving in the negative π₯-direction, we can rephrase the question as follows for greater clarity. Determine the time at which the particle is accelerating at a rate of nine metres per second squared in the negative π₯-direction.

Now, recall how acceleration is related to displacement. If we differentiate the displacement function of an object with respect to time, we obtain the function which measures the velocity of the object. If we differentiate with respect to time the function which measures the velocity, then we obtain the function which measures the acceleration. This means that acceleration is obtained from displacement by differentiating displacement twice. Letβs work out the velocity of the particle in question.

The velocity of the particle is negative three π‘ squared metres per second. Here, we have just used the standard formula for differentiating terms of the form ππ‘ to the power of π with respect to π‘, where we multiply the constant π by the exponent π and decrease the exponent by one. Note that the velocity of this particle is always non-positive as the term negative three π‘ squared is zero for π‘ equals zero and always negative for any value of π‘ strictly greater than zero. This means that the particle is moving at a rate of three π‘ squared metres per second in the negative π₯-direction.

Now, letβs work out the function which measures the acceleration of the particle. The acceleration of the particle is negative six π‘ metres per second squared. Note that substituting any value of π‘ greater than or equal to zero into this function would result in a non-positive number. Since the particle is moving in the negative π₯-direction, this means that the particle is accelerating at a rate of six π‘ metres per second squared in the negative π₯-direction.

We want to find the time π‘ at which the particle is accelerating at a rate of nine metres per second squared in the negative π₯-direction. So, we need to solve the equation six π‘ equals nine. Doing so, we obtained that π‘ equals nine over six seconds, which we can rewrite as three over two seconds or one and a half seconds. So, we have determined that the particle is accelerating at a rate of nine metres per second squared in the negative π₯-direction at time π‘ equal to one and a half seconds.