# Video: APCALC03AB-P1B-Q43-342154060531

A particle moving in a straight line has a velocity of 𝑣(𝑡) = 3^(−𝑡) sin 3𝑡 feet per minute. What is the total distance in feet, travelled by the particle when 0 ≤ 𝑡 ≤ 2 minutes?

02:56

### Video Transcript

A particle moving in a straight line has a velocity of 𝑣 of 𝑡 equals three to the power of negative 𝑡 times sin three 𝑡 feet per minute. What is the total distance in feet travelled by the particle when 𝑡 is greater than or equal to zero and less than or equal to two minutes?

We have a particle moving in a straight line. We sometimes call that rectilinear motion. And we know that we can use calculus to evaluate motion in a straight line. We also see that we’ve been given a function for velocity in feet per minute and asked to find the total distance travelled by the particle.

What we should recall that we can find an expression for the displacement of a particle at 𝑡 seconds by integrating the function for velocity with respect to time. However, we’re looking for the total distance travelled. And distance is equal to the magnitude of the displacement. So we can’t just go ahead and integrate our function between the limits of zero and two. Instead, we’ll need to consider the shape of the curve before we do any integration at all.

We begin then by sketching the graph of 𝑣 equals three to the power of negative 𝑡 times sin of three 𝑡 out using our graphical calculator. Notice how between 𝑡 equals zero and two minutes we have two distinct regions on the graph. The first — let’s call that 𝑅 one — sits above the 𝑥-axis. The second — let’s call that 𝑅 two — sits below the 𝑥-axis.

We know that when we integrate a function that sits below the 𝑥-axis, we end up with a negative value. But we’re looking for distance, not displacement. So what we’re going to do is integrate our function twice. The first time, we’ll evaluate it between the limits of 𝑡 equals zero and the 𝑡 value at the point of which the curve crosses the 𝑡-axis here. We’ll then integrate again between whatever this 𝑡-value is and two. And then we’ll change the sign for this value.

Now if we look at the point at which our graph crosses the 𝑡-axis, we see that this occurs at the point where 𝑡 equals 1.047198, correct to six decimal places. In fact, this is 𝜋 by three radians. And so this means the total distance travelled is equal to the integral evaluated between zero and 1.047198 of three to the power of negative 𝑡 times sin three 𝑡 minus the integral evaluated between 1.047198 and two of three to the power of negative 𝑡 times sin of three 𝑡. And we can use our calculators to evaluate each of these integrals.

The first is equal to 0.38693 and so on. And the second is equal to negative 0.12103. The difference between these is 0.50796. And we’ve obtained the total distance travelled by the particle in our time interval. It’s 0.508 feet correct to three decimal places.