### Video Transcript

A particle is moving in a straight
line such that its velocity π£ at time π‘ seconds is given by π£ equals two π‘
squared minus 68 meters per second for π‘ 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 π£ in
terms of time π‘ seconds. And weβre interested in calculating
the acceleration of the particle. So letβs 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. Letβs differentiate our expression
for velocity term by term. We know that to differentiate a
power term such as ππ₯ to the πth power for real constants π and π, we multiply
the entire term by the exponent and then reduce that exponent by one.

The same holds when we
differentiate two π‘ squared. We multiply it by the power, thatβs
two, and then reduce the power by one. So we get two times two π‘ to the
power of one or two times two π‘, which is simply four π‘. 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 π‘ or four π‘ 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 π‘. So what weβre going to do is to
find the value of π‘ such that the velocity is equal to 94.

We replace π£ in our original
equation with 94, and that gives us 94 equals two π‘ squared minus 68. To solve for π‘, we add 68 to both
sides, and then we get 162 equals two π‘ squared. Dividing through by two gives us 81
equals π‘ squared, and then we take the square root of both sides. Now, normally, weβll look to find
both the positive and negative square root. But we were told that π‘ has to be
greater than or equal to zero. So in fact, we just find the
positive square root, and we get π‘ 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 π‘ equals nine into our equation for π. We get four times nine, which is
36. And so the acceleration of the
particle is 36 meters per square second. Now, we neednβt worry too much
about this word magnitude here. When we consider the magnitude of
the acceleration, weβre looking for its size. In other words, weβre looking for
its absolute value. So had we achieved a negative
result, we wouldβve simply made it positive.