# Video: Finding the Interval of Acceleration for a Particle Moving along the 𝑥-Axis

A particle moves along the 𝑥-axis. Its velocity, 𝑣, in meters per second is given by 𝑣 = 9(𝑡 − 4) − 12(𝑡 − 4)³, where 𝑡 is the time in seconds. Find the interval over which the particle is accelerating in the positive 𝑥-direction.

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### Video Transcript

A particle moves along the 𝑥-axis. Its velocity 𝑣 in meters per second is given by 𝑣 is equal to nine times 𝑡 minus four minus 12 times 𝑡 minus four all cubed, where 𝑡 is the time in seconds. Find the interval over which the particle is accelerating in the positive 𝑥-direction.

We’re told that a particle is moving along the 𝑥-axis. This means it’s moving in a straight line. We’re told the velocity of this particle in meters per second is given by 𝑣 is equal to nine times 𝑡 minus four minus 12 multiplied by 𝑡 minus four all cubed. And again, 𝑡 here is measured in seconds. We need to find the interval over which this particle is accelerating in the positive 𝑥-direction. First, remember, to find the acceleration of a particle moving in a straight line, we want to find the rate of change of velocity with respect to time. In our case, that’s d𝑣 by d𝑡. This will give us a function for the acceleration of our particle at time 𝑡.

But remember, we’re looking for where our particle is accelerating in the positive 𝑥-direction. And of course, for our particle to be accelerating in the positive 𝑥-direction, our rate of change of velocity with respect to time must be positive. So we’re looking for 𝑎 of 𝑡 to be greater than zero. So let’s find an expression for the acceleration of our particle. Remember, that’s the derivative of the velocity function with respect to time. This gives us the derivative of nine times 𝑡 minus four minus 12 multiplied by 𝑡 minus four all cubed with respect to 𝑡.

We’ll evaluate this derivative term by term. First, by expanding our first set of parentheses, we get 9𝑡 minus 36. And if we differentiate this, we see the derivative of 9𝑡 with respect to 𝑡 is nine and the derivative of the constant negative 36 is equal to zero. So the derivative of our first term is just equal to nine. Now, to evaluate our second term, we could use the chain rule. However, we’re going to use the general power rule. We recall the general power rule tells us, for any differentiable function 𝑓 of 𝑥 and any constant 𝑛, the derivative of 𝑓 of 𝑥 raised to the 𝑛th power with respect to 𝑥 is equal to 𝑛 times 𝑓 prime of 𝑥 multiplied by f of 𝑥 raised to the power of 𝑛 minus one.

In our case, we’re differentiating with respect to 𝑡. Our inner function 𝑓 of 𝑡 will be 𝑡 minus four. And we’re raising this to the power of three. So we have 𝑛 is equal to three and we see the derivative 𝑓 prime of 𝑡 is just equal to one. So by applying the general power rule, the derivative of our second term is equal to negative 12 times three multiplied by one times 𝑡 minus four all squared. And we can simplify the leading coefficient to be negative 36. So we’ve shown the acceleration of our particle at times 𝑡 is equal to nine minus 36 times 𝑡 minus four squared.

Remember, we want to find the values of 𝑡 where 𝑎 of 𝑡 is greater than zero. So we need to solve this is greater than zero. There’s a few different ways of doing this. We’ll start by adding 36 times 𝑡 minus four squared to both sides of the equation. This means we now need to solve the inequality nine is greater than 36 times 𝑡 minus four squared. Next, we’ll divide both sides of the inequality through by 36. And we can simplify nine divided by 36 to give us one-quarter. So we need to find the values of 𝑡 where one-quarter is greater than 𝑡 minus four all squared.

We could do this algebraically. However, we’re going to do this with a graphical solution. To find our values of 𝑡, let’s start by sketching a graph of the curve 𝑦 is equal to 𝑥 minus four all squared. To sketch this graph, remember that subtracting four from our values of 𝑥 is the same as translating four units to the right. So our sketch would just be a parabola translated four units to the right.

Now, let’s also sketch the line 𝑦 is equal to one-quarter. Remember, we’re looking for the values where one-quarter is greater than 𝑡 minus four all squared. And we can find these values from our sketch. We can see this will be when our line 𝑦 is equal to one-quarter is above the curve 𝑥 minus four all squared. And this is between the two points of intersection between our lines. So let’s find the points of intersection. We want to solve 𝑥 minus four squared is equal to one-quarter. We want to take the square roots of both sides of this equation. The square root of one-quarter is equal to a half. But remember, we’ll get a positive and a negative square root.

So we get 𝑥 minus four is equal to positive or negative one-half. And to solve for 𝑥, we’ll add four to both sides of the equation. So the 𝑥-coordinates of our points of intersection are four plus a half and four minus a half. We’ll write these as seven over two and nine over two. So from our sketch, one-quarter is greater than 𝑥 minus four all squared when 𝑥 is between nine over two and seven over two. This tells our particle will be accelerating in the positive 𝑥-direction when 𝑡 is between nine over two and seven over two.

And remember, the question wants us to give this as an interval. So we’ll write this as the open interval from seven over two to nine over two. Therefore, we were able to show if a particle is moving along the 𝑥-axis with a velocity function 𝑣 is equal to nine times 𝑡 minus four minus 12 multiplied by 𝑡 minus four all cubed. Then the particle will be accelerating in the positive 𝑥-direction when 𝑡 is in the open interval from seven over two to nine over two.