Video Transcript
A passenger in an airport baggage reclaim area notices her baggage moving toward her on a carousel. She walks toward the baggage at 0.30 meters per second, parallel to the carousel, and picks up the baggage. Right after that, she slows down while moving a distance of 0.15 meters against the direction of the carousel. After that, she continues walking at 0.10 meters per second parallel to the carousel. What is her acceleration in the direction of the carousel while picking up the baggage?
So in this question, we have a passenger in the airport baggage reclaim area who slows down while picking up the baggage from the carousel. We are asked to work out what her acceleration is in the direction of the carousel while she’s picking up the baggage. We are given some information about the speed that she starts out at before collecting the baggage, the speed that she ends up at, and the distance that she travels. So let’s use this information to draw a sketch showing the situation.
We are told that she walks toward the baggage at a speed of 0.30 meters per second and that she is moving parallel to the carousel. We’ll label this initial speed 𝑢. And this is the speed that she is moving at the instant that she picks up the baggage before she starts slowing down. The question also tells us that the baggage is moving toward her on the carousel, so we can label this carousel direction in our sketch.
We are then told that she slows down while moving a distance of 0.15 meters against the direction of the carousel. We’ll call this distance 𝑠. After traveling this 0.15 meters, we are told that she continues walking at 0.10 meters per second still parallel to the carousel. We’ll label this final velocity as 𝑣. So we know the passenger’s initial velocity and her final velocity, as well as how far she walks while this velocity changes.
We are asked to work out the passenger’s acceleration. We’ll label this acceleration as 𝑎. To find the value of 𝑎, we’re going to need to use one of the kinematic equations. Specifically, we’re going to want to use this one here, which says that the square of the final velocity is equal to the square of the initial velocity plus two times the acceleration times the distance traveled. Now, in our case, we can see from the sketch that we know the values of the final velocity 𝑣, the initial velocity 𝑢, and the distance traveled 𝑠. And the quantity that we’re trying to find is the value of the acceleration. So we want to take this equation and rearrange it to make the acceleration 𝑎 the subject. We begin by subtracting 𝑢 squared from both sides of the equation. Then, we divide both sides of the equation by two 𝑠.
On the right-hand side of the equation, the twos in the numerator and denominator and the 𝑠’s in the numerator and denominator cancel each other out. Then, if we swap the left- and right-hand sides of this equation over, we have that the acceleration 𝑎 is equal to the square of the final velocity 𝑣 minus the square of the initial velocity 𝑢 divided by two times the distance traveled 𝑠. We should point out that, in our situation, we have our initial velocity 𝑢 and our final velocity 𝑣 pointing to the right in the opposite direction to the motion of the carousel. So when we put these values for 𝑢 and 𝑣 into this equation for the acceleration 𝑎, then the acceleration that we calculate will be in the direction of the initial and final velocities, in other words, to the right.
Now, the question is actually asking us to find the acceleration in the direction of the carousel. So the way that we’ve drawn our sketch that would be to the left. What we are going to do is take these values of 𝑢 and 𝑣 as they are and substitute them into this equation to calculate the acceleration to the right opposite to the carousel direction. We’ll label this acceleration 𝑎 subscript 𝑝 as it’s the acceleration in the direction that the passenger is traveling. Then, since the carousel is traveling in the opposite direction to the passenger, then we have that the acceleration in the direction of the carousel, which we’ve labeled 𝑎 subscript 𝑐, is equal to negative 𝑎 subscript 𝑝, the acceleration in the direction of the passenger.
So now let’s take our values of 𝑢, 𝑣, and 𝑠 and substitute them into this equation to calculate the value of 𝑎 subscript 𝑝. When we do this, we get that 𝑎 subscript 𝑝 is equal to the square of our final velocity, 0.1 meters per second, minus the square of our initial velocity, 0.3 meters per second, divided by two times the distance traveled, 0.15 meters. In the numerator, 0.1 meters per second all squared gives us 0.01 meters squared per second squared. And similarly, the square of 0.3 meters per second gives us 0.09 meters squared per second squared. Meanwhile, in the denominator, two times 0.15 meters gives 0.30 meters. Doing the subtraction in the numerator gives us that 𝑎 subscript 𝑝 is equal to negative 0.08 meters squared per second squared divided by 0.30 meters.
Then, evaluating this expression gives a result of negative 0.26 recurring meters per second squared. This result gives the acceleration in the direction that the passenger is walking. So the fact that it’s negative tells us that she is slowing down. However, we should recall that we weren’t actually asked to find the acceleration in this direction, but rather in the direction of the carousel. Now, we’ve already said that the acceleration in the direction of the carousel, 𝑎 subscript 𝑐, is equal to the negative of the acceleration in the direction of the passenger. We therefore have that the acceleration in the direction of the carousel is equal to 0.26 recurring meters per second squared.
If we round this result to two decimal places, we can give our answer for the acceleration of the passenger in the direction of the carousel as 0.27 meters per second squared.
