### Video Transcript

A particle is moving in a straight line such that its displacement π in meters is given as a function of time π‘ in seconds by π is equal to five π‘ cubed minus 84π‘ squared plus 33π‘, where π‘ is greater than or equal to zero. Find the magnitude of the acceleration of the particle when the velocity is zero.

In this question, we are given the displacement of a particle as a function of time, π of π‘. This is equal to five π‘ cubed minus 84π‘ squared plus 33π‘. We are asked to find the magnitude of the acceleration of the particle, and we need to do this when the velocity of the particle is zero. We begin by recalling that the derivative of π of π‘ with respect to π‘ is π£ of π‘. And we can therefore find an expression for the velocity in terms of time by differentiating our function π of π‘ term by term. Using the power rule of differentiation, we have π£ of π‘ is equal to 15π‘ squared minus 168π‘ plus 33.

Next, we recall that the acceleration is the derivative of the velocity. This means that we can find an expression for the acceleration in terms of time by differentiating once again. Recalling that differentiating a constant gives us zero and using the power rule of differentiation once again, we have π of π‘ is equal to 30π‘ minus 168.

Our next step is to find the value of π‘ when the velocity is zero. We do this by setting the quadratic expression 15π‘ squared minus 168π‘ plus 33 equal to zero. Whilst this quadratic can be factored, it is quite complicated to do so. So we might choose to solve the equation using the quadratic formula. This states that if ππ₯ squared plus ππ₯ plus π equals zero, then π₯ is equal to negative π plus or minus the square root of π squared minus four ππ all divided by two π. In our equation, we have values of π, π, and π of 15, negative 168, and 33.

Substituting in these values and since we have a quadratic in terms of π‘, we have π‘ is equal to negative negative 168 plus or minus the square root of negative 168 squared minus four multiplied by 15 multiplied by 33 all divided by two multiplied by 15. Typing this expression for π‘ into our calculator, we get two values π‘ is equal to 0.2 and π‘ is equal to 11.

Had we decided to try and factor our expression, we could begin by dividing through by three. This gives us five π‘ squared minus 56π‘ plus 11 is equal to zero. The left-hand side factors into two sets of parentheses, five π‘ minus one and π‘ minus 11. Setting five π‘ minus one equal to zero gives us π‘ is equal to one-fifth, or 0.2. And if π‘ minus 11 equals zero, we have π‘ is equal to 11.

We now have two times where the velocity of the particle is zero. Substituting π‘ is equal to 0.2 into our expression for the acceleration, we have 30 multiplied by 0.2 minus 168. This is equal to negative 162. So the acceleration of the particle when π‘ is equal to 0.2 seconds is negative 162 meters per second squared, which can also be thought of as 162 meters per second squared in the negative direction.

When π‘ is equal to 11, the acceleration is equal to 30 multiplied by 11 minus 168. This is equal to positive 162. When the time is equal to 11 seconds, the acceleration of the particle is 162 meters per second squared. We were asked to find the magnitude of the acceleration, and we can therefore conclude that when the velocity of the particle is zero, the magnitude of the acceleration is 162 meters per second squared.