A particle is moving in a straight line such that its displacement 𝑥 at time 𝑡 seconds is given by 𝑥 equals 9.6𝑡 minus 𝑡 squared meters, for 𝑡 is greater than or equal to zero. What distance does the particle travel in the first 9.6 seconds?
We need to be really careful here. We’ve been given an expression for the displacement of the particle. And we’re looking to find the distance traveled. So what’s the difference between distance and displacement?
Displacement has a direction. It’s the distance we are from some starting point, such as the origin, whereas distance doesn’t have a direction. It’s just the total distance traveled. If we think about the link between displacement and velocity 𝑣, displacement is the integral between 𝑎 and 𝑏 of the velocity with respect to time, whereas the distance traveled between 𝑎 seconds and 𝑏 seconds is the definite integral between 𝑎 and 𝑏 of the absolute value of 𝑣 with the respect to 𝑡.
So let’s find an expression for 𝑣, given our expression for 𝑥. Velocity is change in displacement with respect to time, so it’s d𝑥 by d𝑡. In this case, we need to differentiate 9.6𝑡 minus 𝑡 squared. We’ll do this term by term. The derivative of 9.6𝑡 is just 9.6. Then to differentiate a power term, we multiply the entire term by the exponent and reduce that exponent by one. So the derivative of negative 𝑡 squared is negative two 𝑡. And that’s great. We have an expression for velocity at time 𝑡.
Now, of course, we’re interested in the absolute value of this. Velocity, remember, is a vector quantity. It changes sign depending on direction. So we need to find the value of 𝑡 at which the particle changes direction. When the particle changes direction, there will be a moment where its velocity is equal to zero. It will change from being positive to negative, or vice versa.
So we substitute 𝑣 equals zero into our equation. And we get zero equals 9.6 minus two 𝑡. To solve for 𝑡, we’ll add two 𝑡 to both sides. Then we divide through by two. So we find 𝑡 is equal to 4.8 seconds. So between zero and 4.8 seconds, the particle is traveling in one direction. Then between 4.8 and 9.6 seconds, it’s traveling in the opposite direction.
We’re going to integrate our expression for velocity between zero and 4.8 and 4.8 and 9.6. So the distance traveled in the first 4.8 seconds is the definite integral between zero and 4.8 of 9.6 minus two 𝑡.
Now, in fact, we know that differentiation and integration are reverse processes of one another. So we go back to our original expression for 𝑥. The distance traveled will be 9.6 minus 𝑡 squared evaluated between zero and 4.8. Substituting zero and 4.8 into this expression and finding the difference, we get 9.6 times 4.8 minus 4.8 squared. This means the particle travels 23.04 meters in the first 4.8 seconds.
We repeat this process, this time integrating between 9.6 and 4.8. When we substitute 9.6 and 4.8 into our expression and find their difference, we get negative 23.04. Now that’s the displacement, and it’s in meters. A displacement of negative 23.04 means the particle has just traveled 23.04 meters but in the opposite direction to its first part of its journey.
So the total distance traveled will be the distance traveled in the first part of the journey plus the absolute value of the distance traveled in the second part. That’s 23.04 plus 23.04, which is 46.08. And so we see that the particle travels a total distance of 46.08 meters in the first 9.6 seconds.