The KIT23 competes in the acceleration event with a constant power of 80 kW. The acceleration is build up rules complient. It weighs 200 kg including the driver. Its drag and rolling resistance can be neglected. What is its velocity when the foremost part oft the vehicle passes the finishing line?


To determine the velocity of the KIT23 vehicle when it crosses the finishing line, we'll use the given data and principles from physics. Given:

- Power, \( P = 80 \text{ kW} = 80,000 \text{ W} \)
- Mass of the vehicle, \( m = 200 \text{ kg} \)
- Drag and rolling resistance are negligible.

The velocity of a vehicle with constant power can be found using the work-energy theorem and the relationship between power and velocity:

### Step 1: Relating Power and Kinetic Energy
Power is the rate of doing work or transferring energy. With no resistive forces, the power is used to increase the kinetic energy of the vehicle. 

The kinetic energy (\( E_k \)) of the vehicle is given by:
\[
E_k = \frac{1}{2}mv^2
\]

### Step 2: Relating Power to Acceleration
The instantaneous power delivered to the vehicle can also be expressed as:
\[
P = \frac{dE_k}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right)
\]

Expanding this derivative:
\[
P = mv \frac{dv}{dt}
\]
where \( a = \frac{dv}{dt} \) is the acceleration. Rearranging gives:
\[
a = \frac{P}{mv}
\]

### Step 3: Finding the Final Velocity
To find the final velocity, we need to integrate the acceleration with respect to distance. Since \( a = \frac{P}{mv} \), we have:
\[
\frac{dv}{dt} = \frac{P}{mv}
\]

Integrating both sides:
\[
v \, dv = \frac{P}{m} \, ds
\]

The integral becomes:
\[
\int_{0}^{v_f} v \, dv = \frac{P}{m} \int_{0}^{s} \, ds
\]

\[
\frac{v_f^2}{2} = \frac{P s}{m}
\]

Solving for \( v_f \), we get:
\[
v_f = \sqrt{\frac{2Ps}{m}}
\]

### Step 4: Plug in the Values
Now, substitute \( P = 80,000 \text{ W} \), \( m = 200 \text{ kg} \), and \( s = 75 \text{ m} \):
\[
v_f = \sqrt{\frac{2 \times 80,000 \times 75}{200}}
\]

\[
v_f = \sqrt{60,000}
\]

\[
v_f \approx 245 \text{ m/s}
\]

Thus, the final velocity of the KIT23 as it crosses the finish line is approximately 245 m/s.

Would you like a more detailed breakdown, or have any questions?

### Related Questions:
1. How would including drag force affect the velocity calculation?
2. What is the role of the power-to-weight ratio in vehicle acceleration?
3. How does changing the mass of the vehicle impact the final velocity?
4. If the power were variable instead of constant, how would you approach this problem?
5. How can this problem be solved using work-energy principles instead of integration?

### Tip:
Remember that the relationship \(P = F \cdot v\) can help understand how power directly affects acceleration, especially when forces vary.

The KIT23 competes in the acceleration event with a constant power of 80 kW. The acceleration is build up rules compliant. It weighs 200 kg including the driver. Its drag and rolling resistance can be neglected. What is its velocity when the foremost part of the vehicle passes the finishing line?

Calculate the final velocity of the KIT23 vehicle as it crosses the finish line in a 75-meter race, using a constant power of 80 kW and a mass of 200 kg. This detailed solution uses kinetic energy, the work-energy theorem, and integration techniques. Suitable for undergraduate physics or engineering students.

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