Question Video: Determining the Length of a Track with Given Velocities | Nagwa Question Video: Determining the Length of a Track with Given Velocities | Nagwa

Question Video: Determining the Length of a Track with Given Velocities Physics • First Year of Secondary School

Two cars are racing on a straight track. The winning car has an average velocity of 180 m/s, while the other car has an average velocity of 144 m/s and reaches the finish line 5 seconds after the winner. Find the length of the track.

08:31

Video Transcript

Two cars are racing on a straight track. The winning car has an average velocity of 180 meters per second, while the other car has an average velocity of 144 meters per second and reaches the finish line five seconds after the winner. Find the length of the track.

To begin, it’ll be a good idea to draw a diagram to help show what’s happening in this question. Here we have the two cars and their respective velocities. We’ll call the winning car car one, and the other slower car we’ll call car two. Now, to answer this question, it’ll be helpful to recall that the velocity, 𝑣, of an object is related to the displacement, Δ𝑠, of the object in a time interval, Δ𝑡, by the formula 𝑣 equals Δ𝑠 divided by Δ𝑡. Also recall that displacement is the straight-line distance from a point to another point.

Since we’ve been told that the cars are racing on a straight track and therefore do not change direction, we know that the cars’ displacement is simply the length of the track. Since this is the quantity we want to solve for, let’s make displacement, Δ𝑠, the subject of the equation. We can do this by multiplying both sides of the equation by Δ𝑡. So that term cancels out of the numerator and denominator of the right-hand side, leaving Δ𝑠 by itself. We can now write this equation as Δ𝑠 equals 𝑣 times Δ𝑡.

Let’s apply this equation to each car. To stay organized, we’ll use the subscripts one and two for cars one and two so that we have two separate equations that relate the velocity of each car to its displacement. Δ𝑠 sub one equals 𝑣 sub one times Δ𝑡 sub one and Δ𝑠 sub two equals 𝑣 sub two times Δ𝑡 sub two. We should also note that the length of the track 𝑑 is the same for both cars. So Δ𝑠 sub one and Δ𝑠 sub two are both equal to 𝑑.

Now let’s consider car one and its equation first. We can label the time taken for car one to reach the finish line as 𝑡. We were told that car one has an average velocity of 180 meters per second, so this is the value of 𝑣 sub one. Now, substituting these values into the equation, Δ𝑠 sub one equals 𝑣 sub one times Δ𝑡 sub one, we find that the length of the track 𝑑 is equal to 180 meters per second times 𝑡. We don’t know the value of 𝑡 yet, but that’s okay; we’ll be able to find it after we fill out the equation for car two.

So let’s now consider car two. We’ve been told that car two reaches the finish line five seconds after car one. Remember that we labeled the time taken for car one to reach the finish line as 𝑡. So the time taken for car two to reach the finish line, which is Δ𝑡 sub two, must be 𝑡 plus five seconds. We were also told that car two has an average velocity of 144 meters per second, so this is the value of 𝑣 sub two. Now, remembering that Δ𝑠 sub two equals 𝑑 and substituting these values into the equation Δ𝑠 sub two equals 𝑣 sub two times Δ𝑡 sub two, we find that the length of the track 𝑑 is equal to 144 meters per second times 𝑡 plus five seconds.

Okay, so now we have equations that incorporate all the information given to us in the question, but what’s next? We need to solve for 𝑑. But notice that we still have an unknown variable, that’s the time 𝑡, in each of the formulas. This means the next thing we should do is find a value for 𝑡. And once we do that, we’ll be able to substitute it back into one of the equations for displacement and reach the final answer to this question. To do so, we should notice that we have two equations each solved for the length of the track 𝑑. So we can solve these equations simultaneously.

Since we know that both of these expressions are equal to 𝑑, we can set them equal to each other. Doing this, we have that 180 meters per second times 𝑡 is equal to 144 meters per second times 𝑡 plus five seconds. Let’s now work to solve for 𝑡 by getting that term by itself on one side of the equation. First, on the right-hand side, we’ll multiply through to distribute the terms that are currently grouped together. Thus, we have 144 meters per second times 𝑡 plus 144 meters per second times five seconds. Then, to get 𝑡 on only one side of the equation, we’ll subtract this term, 144 meters per second times 𝑡, from both sides. Then, simplifying the left-hand side, we have 36 meters per second times 𝑡 equals 144 meters per second times five seconds.

Next, to get 𝑡 by itself, we divide both sides by 36 meters per second, so that term cancels out on the left-hand side. Thus, we have 𝑡 is equal to 144 meters per second times five seconds divided by 36 meters per second. Notice that units of meters per second cancel out of the numerator and denominator of the right-hand side. So now, the only units associated with this expression is seconds, which is a good sign, since we’re solving for a value of time right now. To simplify, we calculate 144 times five divided by 36, which comes out to just 20. Therefore, we’ve found that 𝑡 equals 20 seconds. This is the time taken for car one to reach the finish line.

Now that we’ve found 𝑡, we can substitute it into one of the two equations that we have for the displacement 𝑑. We could use either equation to reach our final answer. But for now, let’s choose the equation for car one: 𝑑 equals 180 meters per second times 𝑡. Substituting in 𝑡 equals 20 seconds, we find that the length of the track is equal to 180 meters per second times 20 seconds, which is equal to 3600 meters. This is our final answer. We’ve found that the length of the track is 3600 meters.

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