# Video: Finding the Go-and-Return Time of a Particle given the Relation between Its Velocity and Time

A particle starts from rest and moves in a straight line. At time 𝑡 seconds, its velocity is given by 𝑣 = (36𝑡 − 45𝑡²) cm/s, 𝑡 ≥ 0. How long does it take for the particle to return to its starting point?

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### Video Transcript

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

We’re given the velocity function of a particle which is moving in a straight line. And this velocity function is only valid for values of 𝑡 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 particle’s starting place?

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

So, we’ll need to find an expression for 𝑠 of 𝑡. We’ll 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 𝑡. We’ll get the integral of our velocity function with respect to 𝑡 is equal to our displacement function. And we’re given the velocity function of our particle. So, we’ll use this to find an expression for our displacement function 𝑠 of 𝑡.

It’s equal to the integral of 36𝑡 minus 45𝑡 squared with respect to 𝑡. We can then integrate this term by term by using the power rule for integration. We want to add one to our exponent of 𝑡 and then divide by this new exponent of 𝑡. This gives us 𝑠 of 𝑡 is equal to 18𝑡 squared minus 15𝑡 cubed plus our constant of integration 𝐶.

Remember, we’re trying to find out how long it takes our particle to return to its starting place. So, let’s find its starting place. We’ll substitute 𝑡 is equal to zero into our expression for 𝑠 of 𝑡. Substituting 𝑡 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 𝐶. 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 𝐶.

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

So, we just need to solve the equation 𝑠 of zero is equal to 𝑠 of 𝑡 where 𝑡 is greater than or equal to zero. First, we showed that 𝑠 of zero is equal to our constant of integration 𝐶. Next, we found an expression for 𝑠 of 𝑡. It’s equal to 18𝑡 squared minus 15𝑡 cubed plus 𝐶. And we want to solve this equation. We’ll start by subtracting 𝐶 from both sides of the equation. This gives us zero is equal to 18𝑡 squared minus 15𝑡 cubed.

And to solve this, we’ll want to fully factor the right-hand side of this equation. We’ll take out a factor of three and a factor of 𝑡 squared. Taking out our factor of three 𝑡 squared, we get zero is equal to three 𝑡 squared times six minus five 𝑡. So, now, we’re 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 𝑡 squared is equal to zero, then 𝑡 must be equal to zero.

But remember, we’re not interested in the solution where 𝑡 is equal to zero. So, we must have our other factor of six minus five 𝑡 is equal to zero. And we can solve this equation for 𝑡, we get that 𝑡 is equal to six divided by five. And we’ll 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, we’ve shown if a particle starts from rest and moves in a straight line. And at a time 𝑡 second, its velocity is given by 𝑣 is equal to 36𝑡 minus 45𝑡 squared centimeters per second, where 𝑡 is greater than or equal to zero. Then after 1.2 seconds, our particle will return to its starting place.