Question

A car of mass \(m\) accelerates from rest along a level straight track, not at constant acceleration but with constant engine power, \(P\). Assume that air resistance is negligible. a) Find the car's velocity as a function of time. b) A second car starts from rest alongside the first car on the same track, but maintains a constant acceleration. Which car takes the initial lead? Does the other car overtake it? If yes, write a formula for the distance from the starting point at which this happens. c) You are in a drag race, on a straight level track, with an opponent whose car maintains a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2} .\) Both cars have identical masses of \(1000 . \mathrm{kg} .\) The cars start together from rest. Air resistance is assumed to be negligible. Calculate the minimum power your engine needs for you to win the race, assuming the power output is constant and the distance to the finish line is \(0.250 \mathrm{mi}\)

Short answer

Answer: The minimum constant power needed by the first car's engine to win the race is approximately 285,966 W.

Step-by-step solution

a) Finding the car's velocity as a function of time

To derive an expression for the car's velocity as a function of time, we will use the equation for power, which is given as: \(P = \frac{dW}{dt}\) Where \(P\) is the power, \(W\) is the work done, and \(t\) is time. The work done on an object is equal to the force applied multiplied by the distance (or the displacement) it moves in the direction of the force. \(W = F \cdot d\) Now, we can substitute this equation for \(W\) into the power equation: \(P = \frac{d(F \cdot d)}{dt}\) In this particular problem, the force applied is equal to the mass of the car times its acceleration (\(F = ma\)), and the distance moved is related to the velocity (v) and acceleration (a). Thus, the displacement can be given as: \(d = \frac{1}{2} at^{2}\) Substitute \(F = ma\) and \(d = \frac{1}{2} at^2\) to the power equation: \(P = \frac{d(ma \cdot \frac{1}{2} at^2)}{dt}\) Now, we need to find an expression for the car's velocity using this equation. First, recall that: \(a = \frac{dv}{dt}\) We can replace the acceleration (a) in the power equation: \(P = \frac{d(\frac{1}{2} m (\frac{dv}{dt}) v t^{2})}{dt}\) Now, to solve this equation for the velocity (v), we will use a technique called separation of variables. First, take the derivative of both sides with respect to time: \(P = m (\frac{dv}{dt}) vt\) Now, divide both sides by \(mvt\): \(\frac{P}{mvt} = \frac{dv}{dt}\) Next, we integrate both sides with respect to their respective variables: \(\int{\frac{Pdt}{mvt}} = \int{dv}\) If we assume the initial velocity \(v_{0}=0\), on integrating the equation we get: \(\frac{P}{m} \int_{0}^{t}{\frac{dt}{v}} = \int_{0}^{v}{dv}\) Which gives us: \(v = \sqrt{2Pt/m}\) This is the expression for the car's velocity as a function of time.

b) Comparing with a car with constant acceleration

Now let's compare this car with another car that has a constant acceleration (a'). The other car's velocity as a function of time would be: \(v' = a't\) To find which car takes the initial lead, we can compare their velocities at the initial time, say, \(t = 0\). In this case, the velocity of both cars is 0. However, as time progresses, the second car (car with constant acceleration) will initially be accelerating faster than the first car, because it doesn't require time to "build up" its acceleration. Therefore, the second car will initially be in the lead. To determine if the first car overtakes the second, we need to examine when their velocities are equal: \(\sqrt{2Pt/m} = a't\) Solving for time t, we get: \(t = \sqrt{\frac{2mP}{a'^{3}}}\) Now, we can find the distance traveled by both cars when their velocities are equal. For the first car, the distance traveled is given by integrating the velocity function: \(d = \int{v dt} = \int_{0}^{t}{\sqrt{\frac{2P}{m}} \sqrt{t} dt}\) \(d = \frac{4}{3} \sqrt{\frac{2P}{m}} (\frac{2mP}{a'^{3}})^{3/2}\) And for the second car, the distance traveled can be computed as: \(d' = \int{v' dt} = \int_{0}^{t}{a't dt}\) \(d' = \frac{1}{2}a't^2 = \frac{1}{2}a'\frac{2mP}{a'^2} = \frac{2mP}{a'}\) Comparing the distances, we can see that \(d > d'\), meaning the first car overtakes the second car at the distance given by the formula: \(d = \frac{4}{3} \sqrt{\frac{2P}{m}} (\frac{2mP}{a'^{3}})^{3/2}\)

c) Calculating minimum power to win the race

Both cars have identical masses (m) of 1000 kg. The opponent's car accelerates at a constant rate of 12 m/s². The distance to the finish line is 0.250 mi, which is equal to 402 meters. Let's denote the time it takes for each car to reach the finish line as \(t_{1}\) and \(t_{2}\) for the first and second car, respectively. To win the race, the first car must cover the same distance (d) in the same or lesser time than the second car: \(d = \frac{1}{2}a't_{2}^{2}\) For the second car, we can find the time it takes to reach the finish line: \(t_{2} = \sqrt{\frac{2d}{a'}}\) As both cars have identical masses, we can substitute the distance formula for the first car: \(\frac{4}{3}(\frac{2P}{m})^{3/2}t_{1}^{1/2} = d\) Now, since the first car must reach the finish line at the same or lesser time, our inequality becomes: \(\frac{4}{3}(\frac{2P}{m})^{3/2}t_{1}^{1/2} \leq \frac{1}{2}a't_{2}^{2}\) Next, substitute the expression for \(t_{2}\) in terms of d and a': \(\frac{4}{3}(\frac{2P}{m})^{3/2}t_{1}^{1/2} \leq \frac{1}{2}a'\frac{2d}{a'}\) Now, solving for \(P\): \(P \geq \frac{9m^{2}a'^{3}}{16d} = \frac{9(1000\,\mathrm{kg})^{2}(12\,\mathrm{m/s}^2)^{3}}{16(402\,\mathrm{m})}\) \(P \geq 285966 \,\mathrm{W}\) Therefore, the minimum constant power required by the first car's engine to win the race is approximately \(285,966 \,\mathrm{W}\).

Key concepts

Kinematics

Kinematics is the branch of physics that describes the motion of objects without considering the forces causing the motion. It focuses on parameters like velocity, acceleration, time, and displacement.

In the given problem, we start with the concept of velocity—a measure of how fast an object changes its position. Here, the car’s velocity is derived using power equations. Power provides the energy per unit time to move the car, and we understand it through the relationship with work and force. Specifically, work is force times displacement, and power is the rate at which this work is done.

The car starts from rest; thus, its initial velocity is zero. To find how velocity depends on time, we integrate the power formula, revealing how the car's speed increases over time due to constant power output, separate from any impacts of forces like friction or resistance. This understanding underpins the concept of motion in mechanics.

Power and Motion

Power in physics is the rate of doing work or transferring energy with respect to time. In our exercise, the engine's constant power affects how the car accelerates over time.

1. **Understanding Power:** Power is defined as the work done per unit of time. Mathematically, it is expressed as:
\(P = \frac{dW}{dt}\)
2. **Work Formula:** Work relates to the force applied times the distance covered. It incorporates both the force of the engine and how effectively this force can move the car.
3. **Velocity Relationship:** When power is constant, the relationship between power, mass, and time shapes the velocity of the car, leading to a velocity-time equation:
\(v = \sqrt{\frac{2Pt}{m}}\)
This equation shows that as power is maintained, velocity increases as a function of time.
By understanding this concept, one grasps how continuous power translates to increasing velocity, ultimately influencing how a vehicle moves across time and space.

Drag Race Physics

Drag race physics revolves around analyzing the swift acceleration and high-speed motion of vehicles competing over short distances. This problem illustrates important aspects:

1. **Constant Power vs. Constant Acceleration:** One car uses constant power, and the other employs constant acceleration. While initially, constant acceleration offers quicker starts, over time, constant power provides growing velocity, helping the constant-power car catch up and surpass.

2. **Comparison of Performance:** To assess which car leads initially and how one overtakes the other, we examine their velocity functions:
- The constant power car: \(v = \sqrt{\frac{2Pt}{m}}\)
- Constant acceleration car: \(v' = a't\)

The constant acceleration car may initially lead, but over time, as velocity accumulates with less resistance from initial lag, the constant power car overtakes.

3. **Distance Calculation:** Convergence on timing pertains to calculating at what distance the overtaking happens. This involves setting the velocities equal and solving for the corresponding distance, revealing scenarios often seen in racing setups.

Velocity Equations

Velocity equations are crucial for describing motion dynamics in physics. They assess how speed changes over time under different conditions.

1. **Velocity with Power:** For the constant power scenario, velocity relates to power with the equation derived:
\(v = \sqrt{\frac{2Pt}{m}}\)

This shows the root dependence on both time and mass, indicating a non-linear change in velocity as time proceeds.

2. **Velocity with Constant Acceleration:** Conversely, with constant acceleration, we use:
\(v' = a't\)

Here, velocity increases linearly with time, a direct result of constant acceleration.

These equations form the backbone to comprehend how different forces influence speed over time. They determine strategic approaches in competitive racing situations and explain distinct advantages, helping to optimize performance based on vehicle mechanics and physics principles.