### Video Transcript

A particle starts from rest and moves in a straight line. At a time π‘ seconds, its velocity is given by π£ is equal to 36π‘ minus 45π‘ squared centimeters per second, where π‘ is greater than or equal to zero. How long does it take the particle to return to its starting place?

Weβre given the velocity function of a particle which is moving in a straight line. And this velocity function is only valid for values of π‘ greater than or equal to zero. Our particle starts from rest. And we need to work out how long it takes our particle to return to its starting place. The first question we should ask is, what is our particleβs starting place?

Weβll call the displacement function of our particle after π‘ seconds π of π‘. Then, its starting position is just π evaluated at π‘ is equal to zero. Now, we notice if π of π‘ is equal to π of zero, then our particle is at our starting place. So, we just need to solve π of zero is equal to π of π‘, where π‘ is not equal to zero.

So, weβll need to find an expression for π of π‘. Weβll start by recalling that the velocity of our particle is the rate of change of displacement with respect to time. If this is true, we can then integrate both sides of this equation with respect to π‘. Weβll get the integral of our velocity function with respect to π‘ is equal to our displacement function. And weβre given the velocity function of our particle. So, weβll use this to find an expression for our displacement function π of π‘.

Itβs equal to the integral of 36π‘ minus 45π‘ squared with respect to π‘. We can then integrate this term by term by using the power rule for integration. We want to add one to our exponent of π‘ and then divide by this new exponent of π‘. This gives us π of π‘ is equal to 18π‘ squared minus 15π‘ cubed plus our constant of integration πΆ.

Remember, weβre trying to find out how long it takes our particle to return to its starting place. So, letβs find its starting place. Weβll substitute π‘ is equal to zero into our expression for π of π‘. Substituting π‘ is equal to zero, we get the initial displacement of our particle is equal to 18 times zero squared minus 15 times zero cubed plus πΆ. And 18 times zero cubed minus 15 times zero squared is equal to zero. So, in actual fact, the initial displacement of our particle is equal to our constant of integration πΆ.

At this point, we might be worried. We donβt know how to find the value of πΆ in this case. But, in actual fact, we donβt need to find the value of πΆ. We just need to solve the equation π of zero is equal to π of π‘ where π‘ is not equal to zero. And, of course, since our velocity function was only valid when π‘ is greater than or equal to zero, our displacement function will also only be valid when π‘ is greater than or equal to zero.

So, we just need to solve the equation π of zero is equal to π of π‘ where π‘ is greater than or equal to zero. First, we showed that π of zero is equal to our constant of integration πΆ. Next, we found an expression for π of π‘. Itβs equal to 18π‘ squared minus 15π‘ cubed plus πΆ. And we want to solve this equation. Weβll start by subtracting πΆ from both sides of the equation. This gives us zero is equal to 18π‘ squared minus 15π‘ cubed.

And to solve this, weβll want to fully factor the right-hand side of this equation. Weβll take out a factor of three and a factor of π‘ squared. Taking out our factor of three π‘ squared, we get zero is equal to three π‘ squared times six minus five π‘. So, now, weβre solving an equation where the product of factors is equal to zero. This means one of our factors must be equal to zero. If three π‘ squared is equal to zero, then π‘ must be equal to zero.

But remember, weβre not interested in the solution where π‘ is equal to zero. So, we must have our other factor of six minus five π‘ is equal to zero. And we can solve this equation for π‘, we get that π‘ is equal to six divided by five. And weβll write this in its decimal form of 1.2. And since all of our units were given in terms of centimeters and seconds, we can give this the units of 1.2 seconds.

Therefore, weβve shown if a particle starts from rest and moves in a straight line. And at a time π‘ second, its velocity is given by π£ is equal to 36π‘ minus 45π‘ squared centimeters per second, where π‘ is greater than or equal to zero. Then after 1.2 seconds, our particle will return to its starting place.