Question

A 2000 kg experimental car can accelerate from 0 to $10\frac{\mathrm{m}}{\mathrm{s}}$ in $6\mathrm{s}$ What is the average power of the engine needed to achieve this acceleration? (A) 150 W (B) 150 kW (C) 900 W (D) 900 kW

Short answer

Calculated result is 16.725 kW, thus none of the given options.

Step-by-step solution

- Calculate the acceleration

To find the acceleration, use the formula: $$a=\frac{v_f-v_i}{t}$$ where $v_f=10\frac{\mathrm{m}}{\mathrm{s}}$, $v_i=0\frac{\mathrm{m}}{\mathrm{s}}$, and $t=6\mathrm{s}$. Substituting the values, $$a=\frac{10-0}{6}=\frac{10}{6}\approx 1.67\frac{\mathrm{m}}{{\mathrm{s}}^2}$$

- Calculate the force required

Use Newton's second law to calculate the force: $$F=m\cdot a$$ where $m=2000\mathrm{kg}$ and $a=1.67\frac{\mathrm{m}}{{\mathrm{s}}^2}$. Substituting the values, $$F=2000\times 1.67\approx 3340\mathrm{N}$$

- Calculate the work done

Use the work formula: $$W=F\cdot d$$ First, find the distance $d$ using the formula: $$d=v_{i}t+\frac{1}{2}a{t}^2$$ Substituting the values, $$d=0\cdot 6+\frac{1}{2}\cdot 1.67\cdot 6^2=\frac{1}{2}\cdot 1.67\cdot 36\approx 30.06\mathrm{m}$$ Then calculate the work done: $$W=3340\cdot 30.06\approx 100351.6\mathrm{J}$$

- Calculate the average power

Use the power formula: $$P=\frac{W}{t}$$ where $W=100351.6\mathrm{J}$ and $t=6\mathrm{s}$. Substituting the values, $$P=\frac{100351.6}{6}\approx 16725.27\mathrm{W}$$ Converting to kilowatts, $$P\approx 16.725\mathrm{kW}$$ Hence, the correct option is not listed, but calculated power is approximately 16.725 kW.

Key concepts

Kinematics

Kinematics is a branch of mechanics that describes the motion of points, objects, and systems of bodies without considering the forces that caused the motion. In essence, kinematics focuses on the 'how' of motion. In the given exercise, we need to calculate the acceleration as part of the kinematic analysis. We use the formula:
a = $\frac{v_f-v_i}{t}$
Here:

• $v_f$ represents the final velocity ( $10\frac{\mathrm{m}}{\mathrm{s}}$),
• $v_i$ denotes the initial velocity ($0\frac{\mathrm{m}}{\mathrm{s}}$),
• and $t$ is the time taken ($6\mathrm{s}$).

Substituting the known values, we find the acceleration to be approximately $1.67\frac{\mathrm{m}}{{\mathrm{s}}^2}$. Understanding this calculation is essential for the subsequent steps, as acceleration influences every other variable in our problem.

Newton's Second Law

Newton's second law of motion states that the force acting on an object is equal to the mass of that object times its acceleration. Mathematically represented as:
F = m \cdot a
In our exercise, we have:

• $m$ is the mass of the car (2000 kg)
• $a$ is the acceleration ( $1.67\frac{\mathrm{m}}{{\mathrm{s}}^2}$)

By substituting these values, we calculate the force required to accelerate the car:
$F\approx 3340\mathrm{N}$
Understanding Newton's second law helps explain how the force calculated will be utilized in the consequent calculations of work and power.

Work and Energy

The concept of work in physics refers to the amount of energy transferred by a force over a distance. The formula to calculate work is:
W = F \cdot d
Firstly, we need to find the displacement $d$ using the kinematic equation:
d = v_i t + \frac{1}{2}at^2
By substituting the given values, we get:

• $v_i=0\frac{\mathrm{m}}{\mathrm{s}}$
• $t=6\mathrm{s}$
• $a=1.67\frac{\mathrm{m}}{{\mathrm{s}}^2}$

We find the distance to be approximately 30.06 meters. Using this distance in our work formula, and the force calculated earlier ( $3340\mathrm{N}$), we calculate work done:
$W\approx 100351.6\mathrm{J}$
Understanding work and energy is crucial, as it explains how energy is transferred within a system due to applied forces.

Average Power

Power is the rate at which work is performed or energy is transferred. Average power can be calculated using the formula:
P = \frac{W}{t}
Here:

• $W=100351.6$ J (work done)
• $t=6\mathrm{s}$ (time taken)

Substituting these values, we find:
$P\approx 16725.27\mathrm{W}$
Converting this to kilowatts (kW), we get:
$P\approx 16.725\mathrm{kW}$
Understanding power is important, as it dictates how swiftly work is done or energy is utilized in a process. In this exercise, the calculated power provides insight into the engine's performance capability.