### Video Transcript

A moving particle is defined by the
two equations π₯ equals π‘ cubed minus five π‘ minus five and π¦ equals seven π‘
squared minus three. Find the magnitude of the
acceleration of the particle at π‘ equals one.

In this question, weβve been given
the position of the particle as defined by a pair of parametric equations. So given a value of π‘, we obtain a
coordinate pair π₯π¦ for the position of our particle. We could choose to consider this in
vector terms and say that the position of the particle at time π‘, π of π‘, is
given by π‘ cubed minus five π‘ minus five π plus seven π‘ squared minus three
π. Now, in this question, weβre
looking to find the acceleration. Well, in fact, we want the
magnitude of the acceleration, but weβll deal with that in a moment. So we recall that the acceleration
is equal to the derivative of the velocity with respect to time. But we also know that the velocity
is equal to the first derivative of the displacement or the position vector.

We can in turn say that, to find a
function for acceleration, weβre going to need to differentiate our function for
position twice with respect to time. Weβll do it once to find the
function for velocity. We can differentiate each component
function in turn. When we differentiate π‘ cubed
minus five π‘ minus five, we get three π‘ squared minus five. And when we differentiate seven π‘
squared minus three, we get 14π‘. So our function for velocity is
three π‘ squared minus five π plus 14π‘ π. But what does this actually
mean? Well, it means that the velocity
can be defined in terms of its horizontal velocity and its vertical velocity. Horizontally, its velocity is three
π‘ squared minus five, but vertically, itβs given by the function 14π‘.

Okay, great. Letβs differentiate again to find
our function for acceleration. And when we differentiate three π‘
squared minus five, we find that the acceleration in the horizontal direction is six
π‘. We then differentiate 14π‘. And we see that the acceleration in
the vertical direction is 14. Weβre now able to find a vector
acceleration for our particle at π‘ equals one. We simply substitute π‘ equals one
into a vector function for acceleration. And we find that the acceleration
at π‘ equals one is given by the vector six π plus 14π. We, of course, want to find the
magnitude of the acceleration though.

So we recall that the magnitude of
a vector in two dimensions given by π₯ π plus π¦π is the square root of π₯ squared
plus π¦ squared. This means the magnitude of our
acceleration at π‘ equals one is the square root of six squared plus 14 squared,
which is two root 58. There are no units here. So we found the magnitude of the
acceleration of the particle at π‘ equals one to be two root 58.