Overtaking Distance & Time: Car Chase Calculation

by Kenji Nakamura 50 views

Hey guys! Let's dive into a fun physics problem involving two cars, distance, speed, and time. We've got a scenario where one car is chasing another, and we need to figure out where and when the chaser catches up. Buckle up, it's gonna be a mathematical ride!

Problem Setup

So, here's the deal: A car leaves town A heading towards town B at a speed of 80 kilometers per hour (km/h). Ninety minutes later, another car starts from the same spot, chasing the first one at a speed of 27.78 meters per second (m/s). Our mission, should we choose to accept it, is to calculate:

  • At what distance from town A will the second car overtake the first?
  • At what time will the second car overtake the first?

This is a classic relative motion problem, and we're going to break it down step by step so it's super clear. Get your thinking caps on!

Step 1: Converting Units for Consistency

Alright, the first thing we need to do is make sure all our units are playing nice together. We've got speeds in both km/h and m/s, and time in minutes and potentially hours. To avoid confusion, let's convert everything to meters and seconds. This is crucial for accurate calculations in physics problems.

Converting the First Car's Speed

The first car is cruising at 80 km/h. To convert this to meters per second, we use the following conversion factors:

  • 1 kilometer (km) = 1000 meters (m)
  • 1 hour (h) = 3600 seconds (s)

So, we multiply 80 km/h by (1000 m / 1 km) and (1 h / 3600 s):

80 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 22.22 m/s (approximately)

Therefore, the first car's speed is approximately 22.22 m/s. Now both speeds are in the same unit, which is excellent for comparing and calculating relative motion. Understanding unit conversions is fundamental in physics, and getting this step right ensures our subsequent calculations will be accurate. This conversion process not only aligns the units but also simplifies the problem, making it easier to visualize and solve.

Converting the Time Delay

The second car starts 90 minutes after the first. Let's convert those minutes into seconds:

90 minutes * 60 seconds/minute = 5400 seconds

So, the second car has a 5400-second head start. This time difference is a key factor in determining when and where the second car will catch up. Accurate time conversion is just as vital as speed conversion because it directly affects the distance the first car travels before the second car even starts. Having this value in seconds allows us to directly use it in our equations later on, providing a cohesive and clear approach to solving the problem.

Step 2: Setting Up the Equations of Motion

Now for the fun part: creating the equations that describe the cars' movements. We'll use the basic equation for distance traveled at a constant speed:

Distance = Speed × Time

Let's define our variables:

  • d = the distance from town A where the second car overtakes the first (what we want to find).
  • t = the time (in seconds) the second car travels until it overtakes the first car.

Equation for the First Car

The first car travels for t + 5400 seconds (the second car's time plus the initial 90-minute delay). Its distance from town A is:

d = 22.22 * (t + 5400)

This equation represents the total distance covered by the first car, considering both its speed and the head start it had. Understanding this equation is crucial because it lays the foundation for comparing the distances traveled by both cars. It effectively captures the cumulative distance the first car covers over time, which we can then equate with the distance covered by the second car to find the overtaking point.

Equation for the Second Car

The second car travels for t seconds at a speed of 27.78 m/s. Its distance from town A is:

d = 27.78 * t

This equation shows the distance the second car covers from the starting point up to the point it overtakes the first car. This simple yet powerful equation highlights the direct relationship between the second car's speed and time traveled, allowing us to calculate its displacement. By equating this distance with the distance covered by the first car, we can determine the precise moment and location of the overtaking, making it a critical component of our problem-solving strategy.

Step 3: Solving for Time (t)

We now have two equations with two unknowns (d and t). To find when the second car overtakes the first, we can set the two distance equations equal to each other:

22.22 * (t + 5400) = 27.78 * t

Let's solve for t:

  1. Expand the left side: 22.22t + 119988 = 27.78t
  2. Subtract 22.22t from both sides: 119988 = 5.56t
  3. Divide both sides by 5.56: t = 21580.58 seconds (approximately)

So, the second car will overtake the first car approximately 21580.58 seconds after the second car starts its journey. Accurately solving for t is pivotal because it gives us the time frame within which the overtaking occurs. This value serves as the cornerstone for our subsequent calculations, allowing us to pinpoint the distance traveled by both cars up to the moment of overtaking. This meticulous algebraic manipulation ensures that we arrive at a precise temporal solution, which is crucial for answering the original question.

Step 4: Calculating the Overtaking Distance (d)

Now that we know the time t, we can plug it back into either of our distance equations to find the overtaking distance d. Let's use the second car's equation:

d = 27.78 * 21580.58

d = 600528.54 meters (approximately)

Therefore, the second car will overtake the first car approximately 600528.54 meters from town A. This distance, calculated with precision, provides the spatial answer to our problem. It represents the exact location where the second car catches up to the first, giving us a clear understanding of the scale of the chase. By substituting the calculated time t into the equation, we seamlessly bridge the temporal and spatial aspects of the problem, culminating in a definitive answer to the overtaking distance.

To get a better sense of this distance, let's convert it to kilometers:

600528.54 meters / 1000 meters/kilometer = 600.53 kilometers (approximately)

So, the overtaking occurs about 600.53 kilometers from town A.

Step 5: Converting Time Back to a More Understandable Unit

21580.58 seconds is a bit hard to grasp, isn't it? Let's convert it back to hours, minutes, and seconds:

  1. Hours: 21580.58 seconds / 3600 seconds/hour = 5.99 hours (approximately)
  2. Minutes: The decimal part of the hours (0.99) multiplied by 60 minutes/hour: 0.99 * 60 = 59.4 minutes (approximately)
  3. Seconds: The decimal part of the minutes (0.4) multiplied by 60 seconds/minute: 0.4 * 60 = 24 seconds (approximately)

So, the second car overtakes the first approximately 5 hours, 59 minutes, and 24 seconds after the second car starts its journey. Converting the time back into hours, minutes, and seconds transforms the abstract numerical result into a relatable and comprehensible timeframe. This step is essential for practical understanding, as it allows us to contextualize the duration of the chase in everyday terms. By breaking down the total time into more familiar units, we gain a better appreciation for the scale of the event and the effort involved in the second car's pursuit.

Final Answer

Alright, guys, we did it! Let's summarize our findings:

  • Distance from town A: Approximately 600.53 kilometers
  • Time after the second car starts: Approximately 5 hours, 59 minutes, and 24 seconds

We've successfully calculated the distance and time at which the second car overtakes the first. This problem demonstrates the power of using equations of motion and unit conversions to solve real-world scenarios. Great job everyone!