### Video Transcript

A particle moves along the 𝑥-axis. It is initially at rest at the origin. At time 𝑡 seconds, the particle’s acceleration is given by 𝑎 is equal to 24 minus 19𝑡 meters per second squared, where 𝑡 is greater than or equal to zero. How long does it take for the particle to return to the origin?

We’re told that the particle is moving in a straight line along the 𝑥-axis. We’re told this particle is initially at rest at the origin. What this means is when 𝑡 is equal to zero, the velocity of our particle is equal to zero. And when 𝑡 is equal to zero, the displacement of our particle is also equal to zero. We’ll call the velocity 𝑣 of 𝑡 and the displacement 𝑠 of 𝑡. We’re then given the acceleration of our particle for values of 𝑡 greater than or equal to zero; the acceleration is 24 minus 19𝑡 meters per second squared.

We’re tasked with finding how long it takes for our particle to return to the origin. Since we chose our function 𝑠 of 𝑡 to be the displacement of our particle from the origin, our particle will be at the origin when this displacement is equal to zero. So we want to find an expression for our displacement function 𝑠 of 𝑡. Solving this equal to zero will give us all of the times where our particle is at the origin. And we’re interested in the first time it returns to the origin where 𝑡 is greater than zero, since we already know it starts at the origin.

To help us find this information, let’s start by recalling the acceleration is the rate of change of the velocity with respect to time. And if this is true, since integrating is the opposite process to differentiating, we know the integral of the acceleration with respect to time will be equal to the velocity of our particle up to a constant of integration. We’re given the acceleration function of our particle. So we can use this to find the velocity of our particle. By integrating our acceleration function with respect to time, we get the velocity of our particle at the time 𝑡 is equal to the integral or 24 minus 19𝑡 with respect to 𝑡. And this will be up to a constant of integration.

We can then integrate this by using the power rule for integration. We add one to our exponents of 𝑡 and then divide by this new exponent of 𝑡. This gives us 24𝑡 minus 19 over two 𝑡 squared plus our constant of integration 𝑐. Remember, though, we were told our particle was initially at rest. This means the velocity when 𝑡 is equal to zero should be equal to zero. So we can find our value of 𝑐 by substituting 𝑡 is equal to zero into our velocity function. Doing this, we get the particle’s initial velocity is equal to 24 times zero minus 19 over two times zero squared plus 𝑐.

We know that the initial velocity of our particle is equal to zero. And we know 24 times zero minus 19 over two times zero squared is equal to zero. So this simplifies to give us that 𝑐 is equal to zero. So substituting 𝑐 is equal to zero into our equation for the particle’s velocity, we’ve shown that the velocity of our particle at the time 𝑡 is equal to 24𝑡 minus 19 over two 𝑡 squared. And it’s also worth noting since the acceleration function of our particle was only valid when 𝑡 was greater than or equal to zero, our velocity function will also only be valid when 𝑡 is greater than or equal to zero.

But remember, we want to find the displacement of our particle from the origin. So we recall the velocity of our particle will be equal to the rate of change of displacement with respect to time. We can use the same logic we did before. We’ll integrate with respect to 𝑡. This gives us the integral of the velocity of our particle with respect to 𝑡 will be equal to the displacement of our particle at the time 𝑡. Remember, we’re using 𝑠 of 𝑡 to be the displacement of our particle from the origin. But we could’ve used any point as our reference point, since we’ll have a constant of integration. So let’s apply this. We have 𝑠 of 𝑡 is the integral of 𝑣 of 𝑡 with respect to 𝑡. And we’ve shown that 𝑣 of 𝑡 is 24𝑡 minus 19 over two 𝑡 squared.

Again, we can integrate this term by term by using the power rule for integration. We get 12𝑡 squared minus 19 over six 𝑡 cubed plus our constant of integration, we’ll call 𝑑. We want to find the value of 𝑑. Remember, our particle was initially at rest at the origin. So when 𝑡 is equal to zero, 𝑠 of 𝑡 is equal to zero. So we’ll substitute 𝑡 is equal to zero. We get zero is equal to 12 times zero squared minus 19 over six times zero cubed plus 𝑑. And this simplifies to give us that 𝑑 is equal to zero. So we’ll substitute 𝑑 is equal to zero into our expression for 𝑠 of 𝑡. This gives us 𝑠 of 𝑡 is equal to 12𝑡 squared minus 19 over six 𝑡 cubed.

And remember, 𝑠 of 𝑡 is a measure of the displacement of our particle from the origin. So when this is equal to zero, our particle is at the origin. So solving this equation equal to zero will tell us how long it takes our particle to return to the origin. Remember, our displacement function will only be valid when 𝑡 is greater than or equal to zero. And in our case, we’re interested in the solution where 𝑡 is positive, since we already know the particle was at rest at the origin when we started. To solve this equation equal to zero, we’ll start by taking out a shared factor of 𝑡 squared. This gives us 𝑡 squared times 12 minus 19 over six 𝑡 is equal to zero. We know if the product of factors is equal to zero, then one of our factors must be equal to zero.

When 𝑡 squared is equal to zero, 𝑡 is equal to zero. But we already know we’re looking for values of 𝑡 greater than zero. So we must have our other factor of 12 minus 19 over six 𝑡 is equal to zero. We can solve this for 𝑡. We get 𝑡 is equal to 72 divided by 19. And all of our units were given in terms of meters and seconds. So we can give this a unit. It’s 72 over 19 seconds. And this is our final answer. So we’ve shown if a particle is moving along the 𝑥-axis which is initially at rest at the origin. And at time 𝑡 seconds, the particle’s acceleration is given by 𝑎 is equal to 24 minus 19𝑡 meters per second squared, where 𝑡 is greater than or equal to zero. Then it will take 72 over 19 seconds for our particle to return to the origin.