### Video Transcript

A body started moving along the
π₯-axis from the origin at an initial speed of 10 meters per second. When it was π meters away from the
origin and moving at π£ meters per second, its acceleration was 45 times π to the
power of negative π meter per square second in the direction of increasing π₯. Determine π when π£ equals 11
meters per second.

Letβs begin by defining the
acceleration to be π such that π is equal to 45π to the power of negative π
meters per square second. Weβre given information about the
speed of the body at various points during its motion. And so we begin by recalling how we
can link acceleration with velocity. In particular, the acceleration is
the rate of change of velocity. In other words, acceleration is the
derivative of velocity with respect to time.

Now, of course, weβre interested in
the initial speed, and this is the velocity. But since speed is the magnitude of
velocity and the object is moving in the direction of increasing π₯, then these are
going to be the same thing. So we can begin to answer this
question by replacing π with dπ£ by dπ‘ in our earlier equation. So dπ£ by dπ‘ is equal to 45 times
π to the power of negative π . And then what we might look to do
is find an expression for π£ here by integrating.

However, we notice that we have dπ£
by dπ‘, and then our expression for dπ£ by dπ‘ is actually in terms of π . And so we recall that we can
alternatively represent dπ£ by dπ‘ as dπ£ by dπ times dπ by dπ‘. This, of course, is essentially a
special version of the chain rule. And this allows us to rewrite our
equation even further as dπ£ by dπ times dπ by dπ‘ equals 45π to the power of
negative π . But we also know that we can link
velocity with displacement. Velocity is said to be the rate of
change of displacement. In other words, velocity is the
derivative of displacement with respect to time. And so our equation is now dπ£ by
dπ times π£ equals 45π to the power of negative π .

We now notice we have dπ£ and π£
and dπ and π . And so what weβre going to do is
treat dπ£ and dπ as differentials. This allows us to separate our
equation and then integrate both sides. In particular, weβre allowed to
treat dπ£ by dπ here a little bit like a fraction, allowing us to multiply through
by dπ . Then, we have an equation that we
can begin to solve by integrating both sides with respect to their respective
variables. In other words, weβre going to
integrate π£ with respect to π£ and 45π to the power of negative π with respect to
π .

Now, of course, the indefinite
integral of π£ with respect to π£ is a half π£ squared plus some constant of
integration π. But actually weβre going to create
a constant of integration on the right-hand side as well. And what weβll actually do is just
combine them into one final constant. We also know that if we integrate
π to the power of negative π , we get negative π to the power of negative π . And so the right-hand side of our
equation becomes negative 45π to the power of negative π plus π.

And this is really useful because
we now know that we have an equation for π£ in terms of π . And if we rearrange this, we can
determine the value of π for some π£ if we know the value of π. But of course we can calculate the
value of π by using the fact that the initial speed of the particle is 10 meters
per second. In particular, when π , the
displacement, is equal to zero, we can say that π£ is equal to 10. Substituting this into our
equation, we get a half times 10 squared equals negative 45π to the power of
negative zero plus π. This simplifies to 50 equals
negative 45 plus π, and we can then solve for π by adding 45. So π is 95. And so we have the equation that
links π£ with π .

Weβre now going to substitute π£
equals 11 into this equation and rearrange to make π the subject. This gives us a half times 11
squared equals negative 45π to the power of negative π plus 95. A half times 11 squared then we
subtract 95 gives us negative 69 over two. This is equal to negative 45π to
the power of negative π . So weβre next going to divide
through by negative 45. And we get 23 over 30 equals π to
the power of negative π .

To solve this equation for π ,
letβs take the natural logarithm of both sides. Weβll clear some space when we do
so. The left-hand side becomes the
natural logarithm of 23 over 30. And then the natural logarithm of
π to the power of negative π is simply negative π . We multiply through by negative
one. So π is negative the natural
logarithm of 23 over 30. And we can simplify this a little
bit further if we think about this as negative one times the natural logarithm of 23
over 30. Using the laws of logarithms, it
allows us to write that left-hand side as the natural logarithm of 23 over 30 to the
power of negative one. And then 23 over 30 to the power of
negative one is 30 over 23. So the left-hand side simplifies
even further to the natural logarithm of 30 over 23.

And so we found the displacement of
the object when π£ equals 11 meters per second. In meters, itβs the natural
logarithm of 30 over 23.