Question

A car of mass 942.4 kg accelerates from rest with a constant power output of 140.5 hp. Neglecting air resistance, what is the speed of the car after 4.55 s?

Short answer

Answer: The final speed of the car after 4.55 seconds is approximately 30.49 m/s.

Step-by-step solution

Convert the power to Watts

Firstly, we need to convert the power from horsepower (hp) to Watts (W), since SI-units are preferred. We know that 1 hp = 746 Watts. So, the power output in Watts: Power (in Watts) = 140.5 hp * 746 W/hp = 104863.7 W

Find the work done

To find the work done, we use the equation: Work = Power * Time. In this case, the power is 104863.7 W, and the time is 4.55 seconds. Work Done = 104863.7 W * 4.55 s = 477125.835 J (Joules)

Use the work-energy theorem

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. Since the car starts from rest, its initial kinetic energy is 0. Therefore, the final kinetic energy is equal to the work done. Final Kinetic Energy = 477125.835 J

Find the final velocity

To find the final velocity, we use the kinetic energy formula, which is: KE = (1/2) * m * v^2, where m is the mass, v is the velocity, and KE is the kinetic energy. We have the mass (942.4 kg), and the final kinetic energy (477125.835 J). Now, we will solve for v: (1/2) * m * v^2 = Final Kinetic Energy v^2 = (2 * Final Kinetic Energy) / m v = sqrt((2 * 477125.835 J) / 942.4 kg) After calculating, we get: v = 30.49 m/s The speed of the car after 4.55 seconds is approximately 30.49 m/s.

Key concepts

Work-Energy Theorem

The Work-Energy Theorem is a crucial concept in physics that helps us understand the relationship between work done on an object and its kinetic energy.
When work is done on an object, it can cause a change in the object’s kinetic energy. This is precisely what the Work-Energy Theorem states: the work done on an object is equal to its change in kinetic energy.

In the exercise above, as the car accelerates, the power generated performs work over the given time period. Since the car starts from rest, its initial kinetic energy is zero. Hence, all the work done contributes to its final kinetic energy. This final kinetic energy is calculated using the work done, derived from constant power and time.

• Theorem emphasizes the transformation of energy.
• Useful in solving problems where kinetic energy changes occur.

The overall concept helps us predict how an object’s speed will change over time due to the work applied.

Power Conversion

Power Conversion is the process of transforming power from one unit to another. In physical problems, it’s essential to convert units to SI Units, which are internationally standardized.
Horsepower (hp) is a common unit of power, but it needs to be converted to Watts (W) in scientific equations because Watt is the SI unit of power. Knowing the conversion factor, where 1 hp equals 746 Watts, is vital in such exercises.

In our exercise, 140.5 horsepower was converted to 104,863.7 Watts to be used in further calculations. This conversion is necessary as equations involving work, energy, and power typically need inputs in SI units.

• Maintains consistency in calculations.
• Essential for communicating scientific results globally.

By converting power units, it ensures accuracy and compatibility in physics calculations.

Kinematics

Kinematics examines the motion of objects without considering the forces causing that motion. In the context of this exercise, kinematics helps in determining the speed of the car from the work done and resulting kinetic energy.
When applying kinematic equations, one often uses the relationships between acceleration, velocity, displacement, and time. However, in this scenario, kinetic energy is used instead.

The final velocity of the car is derived from the car’s kinetic energy using the formula \[ KE = \frac{1}{2} m v^2 \]where

• KE is kinetic energy,
• m is the mass of the object,
• v is the velocity.

Since the work done equates to the change in kinetic energy, we adjust the formula to solve for velocity:\[ v = \sqrt{\frac{2 \times \text{Final Kinetic Energy}}{m}} \]This form links kinematic variables through energy concepts.

SI Units

The International System of Units, known as SI Units, serves as the standard for measurements in science.
SI Units ensure that calculations are performed using universally accepted numbers without conversions between different measurement systems complicating matters.

In this exercise, units like Watts (power) and Joules (energy) are used to maintain this standard.

• Ensures precision and accuracy in scientific work.
• Makes data comparable and shareable across the world.

By using SI Units, we prevent mistakes that might occur due to unit conversion errors and promote uniformity in scientific communication.