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Chapter 2: Problem 37

A car is traveling at a constant speed of 33 \(\mathrm{m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.5 \(\mathrm{km}\) away?

Short Answer

Expert verified
The second car must maintain an acceleration of approximately 0.871 m/s虏.

Step by step solution

01

Convert Distance to Meters

To work in consistent units, convert the distance from kilometers to meters. Since there are 1000 meters in a kilometer, \(2.5 \text{ km} = 2500 \text{ m}\).
02

Determine Time for the First Car

The first car travels at a speed of 33 m/s. To find the time it takes to travel 2500 meters, use the formula \(t = \frac{d}{v}\), where \(d = 2500 \text{ m}\) and \(v = 33 \text{ m/s}\). Thus, \(t = \frac{2500}{33}\approx 75.76 \text{ seconds}\).
03

Set Up the Equation for the Second Car

The second car starts from rest, so its initial velocity is 0 m/s. It travels 2500 meters in the same time as the first car, with constant acceleration. Use the equation for distance traveled with constant acceleration: \(d = ut + \frac{1}{2}at^2\). Since \(u = 0\), the formula simplifies to \(2500 = \frac{1}{2}a(75.76)^2\).
04

Solve for Acceleration

Rearrange the equation \(2500 = \frac{1}{2}a(75.76)^2\) to solve for \(a\):\[ a = \frac{2 \times 2500}{(75.76)^2} \approx \frac{5000}{5740.58} \approx 0.871\text{ m/s}^2\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Speed
When we talk about constant speed, it means that an object is moving at the same rate without speeding up or slowing down. For example, in our problem, the first car travels at a constant speed of 33 meters per second (m/s).
This means it covers 33 meters every second, like a steady march.
  • No change in velocity: The speed doesn鈥檛 increase or decrease.
  • Time and Distance Relation: With constant speed, distance is always the product of speed and time.
  • Straightforward Calculations: Makes it easy to predict how long a given distance will take.
Understanding constant speed is crucial because it tells us how we can calculate time and distance easily. Using the formula \(t=\frac{d}{v}\), where \(d\) is distance and \(v\) is velocity, we can find out how long it takes for the car to travel the given distance.
Acceleration
Acceleration is the rate of change in velocity. In simpler terms, it indicates how quickly an object speeds up or slows down.
In the exercise, the second car accelerates to catch up with the first car traveling at a constant speed.
  • Starts from Rest: Initial speed is 0 m/s.
  • Constant Acceleration: The second car's acceleration doesn鈥檛 change over time.
  • Equation in Use: Distance can be expressed as \(d = ut + \frac{1}{2}at^2\).
Here, \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is time. Since the second car starts from rest, \(u = 0\). The equation simplifies to \(d = \frac{1}{2}at^2\) for our problem. Solving for \(a\) lets us determine the necessary acceleration for the second car to meet the first car at the exit.
Distance Conversion
Distance conversion ensures consistency across units and simplifies computations.
In physics problems like this one, converting kilometers to meters aligns with the standard metric system and makes calculations more straightforward.
  • 1000 meters in a kilometer: This is the basic conversion factor.
  • Converted Distance: 2.5 kilometers becomes 2500 meters.
  • Consistency: Ensures all units match, crucial for solving equations accurately.
Becoming familiar with converting units is important in physics, as it helps avoid errors due to inconsistent units and makes real-world problem-solving more intuitive.
Equations of Motion
Equations of motion describe the relationship between an object's motion parameters, such as displacement, velocity, and acceleration.
These equations are fundamental tools in solving kinematics problems like the one presented.
  • Three Key Equations: Utilize different variables depending on the information given.
  • Used for Precise Calculations: Help calculate unknowns when certain parameters are known.
  • Relatable to Real-Life Scenarios: Useful in determining trajectories, speed-time relations, and more.
In the exercise, the essential equation is \(d = ut + \frac{1}{2}at^2\) since it incorporates initial velocity, acceleration, and time to find distance. This specific motion equation helps us figure out the required acceleration for the second car to meet the first car at a fixed distance.

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Most popular questions from this chapter

A Boeing 747 Jumbo Jet has a length of 59.7 \(\mathrm{m}\) . The runway on which the plane lands intersects another runway. The width of the inter- section is 25.0 \(\mathrm{m}\) . The plane decelerates through the intersection at a rate of 5.70 \(\mathrm{m} / \mathrm{s}^{2}\) and clears it with a final speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . How much time is needed for the plane to clear the intersection?

A hot-air balloon is rising upward with a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . When the balloon is 3.00 \(\mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

While standing on a bridge 15.0 \(\mathrm{m}\) above the ground, you drop a stone from rest. When the stone has fallen \(3.20 \mathrm{m},\) you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.

A ball is thrown straight upward. At 4.00 \(\mathrm{m}\) above its launch point, the ball's speed is one-half its launch speed. What maximum height above its launch point does the ball attain?

A golfer rides in a golf cart an average speed of 3.10 \(\mathrm{m} / \mathrm{s}\) for 28.0 s. She then gets out of the cart and starts walking at an average speed of 1.30 \(\mathrm{m} / \mathrm{s}\) . For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 1.80 \(\mathrm{m} / \mathrm{s} ?\)
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