### Video Transcript

A particle is moving in a straight
line such that its velocity ๐ฃ at time ๐ก seconds is given by ๐ฃ equals two ๐ก
squared minus 68 meters per second for ๐ก is greater than or equal to zero. Find the magnitude of the
acceleration of the particle when its velocity is 94 meters per second.

Here, we have an equation for ๐ฃ in
terms of time ๐ก seconds. And weโre interested in calculating
the acceleration of the particle. So letโs begin by recalling how we
link velocity and acceleration. Acceleration is the rate of change
of velocity. When we think about rate of change,
we think about finding the derivative. So acceleration is the first
derivative of velocity with respect to time. Letโs differentiate our expression
for velocity term by term. We know that to differentiate a
power term such as ๐๐ฅ to the ๐th power for real constants ๐ and ๐, we multiply
the entire term by the exponent and then reduce that exponent by one.

The same holds when we
differentiate two ๐ก squared. We multiply it by the power, thatโs
two, and then reduce the power by one. So we get two times two ๐ก to the
power of one or two times two ๐ก, which is simply four ๐ก. We also know that when we
differentiate a constant, we get zero. So the derivative of negative 68 is
zero. And therefore, acceleration is
equal to four ๐ก or four ๐ก meters per square second. Now, the question asks us to find
the acceleration when the velocity is equal to 94. But of course, our acceleration is
in terms of time ๐ก. So what weโre going to do is to
find the value of ๐ก such that the velocity is equal to 94.

We replace ๐ฃ in our original
equation with 94, and that gives us 94 equals two ๐ก squared minus 68. To solve for ๐ก, we add 68 to both
sides, and then we get 162 equals two ๐ก squared. Dividing through by two gives us 81
equals ๐ก squared, and then we take the square root of both sides. Now, normally, weโll look to find
both the positive and negative square root. But we were told that ๐ก has to be
greater than or equal to zero. So in fact, we just find the
positive square root, and we get ๐ก is equal to nine. In other words, the velocity of the
particle is 94 meters per second when the time is nine seconds.

The acceleration at this time then
is found by substituting ๐ก equals nine into our equation for ๐. We get four times nine, which is
36. And so the acceleration of the
particle is 36 meters per square second. Now, we neednโt worry too much
about this word magnitude here. When we consider the magnitude of
the acceleration, weโre looking for its size. In other words, weโre looking for
its absolute value. So had we achieved a negative
result, we wouldโve simply made it positive.