1957 Corvette Physics: Speed, Acceleration & Forces
Hey guys! Let's dive into a cool physics problem featuring a classic 1957 Chevrolet Corvette. This beauty has a mass of 1240 kg and starts from rest. It's tearing around a circular test track with a constant tangential acceleration of 2.00 m/s². The track has a radius of 60.0 m. We're going to treat this awesome car as a particle (for simplicity, of course!). So, buckle up and let's solve some physics!
a) Speed After 2.00 Seconds
Okay, first up, we need to figure out the Corvette's speed after just 2.00 seconds. This is a classic kinematics problem, and we've got all the info we need. We know the initial velocity (0 m/s since it starts from rest), the acceleration (2.00 m/s²), and the time (2.00 s). We're hunting for the final velocity.
The formula that jumps to mind is: v = vâ‚€ + at
Where:
- v is the final velocity (what we're trying to find)
- vâ‚€ is the initial velocity (0 m/s)
- a is the acceleration (2.00 m/s²)
- t is the time (2.00 s)
Let's plug in those values:
v = 0 m/s + (2.00 m/s²)(2.00 s) v = 4.00 m/s
So, after 2.00 seconds, our classic Corvette is cruising at 4.00 m/s. Not bad, huh? This calculation showcases the direct application of basic kinematic principles, which are fundamental to understanding motion in physics. The simplicity of this formula, v = vâ‚€ + at, belies its power in describing uniformly accelerated motion. Understanding this relationship is crucial for more complex physics problems down the road.
But what does this speed mean in the context of the circular track? Well, it tells us how quickly the car is moving along the circumference of the circle at that particular moment. It's important to remember that this is the tangential speed, meaning it's the speed along the direction tangent to the circle at the car's position. As the car accelerates, this tangential speed will increase, making the Corvette zip around the track faster and faster.
This initial calculation sets the stage for the rest of the problem. Now that we know the speed at a specific time, we can start to explore other aspects of the car's motion, like its radial acceleration and the forces acting upon it. This step-by-step approach is key to tackling more complex physics problems – break them down into smaller, manageable chunks, solve each part, and then put it all together for the bigger picture.
b) Magnitude of Radial Acceleration
Next up, we're tasked with finding the magnitude of the Corvette's radial (centripetal) acceleration at this same moment (2.00 seconds). Now, this is where things get a little more circular – literally! Remember, anything moving in a circle experiences an acceleration directed towards the center of the circle. That's the radial acceleration, and it's what keeps the car from flying off the track.
The formula for radial acceleration (aáµ£) is:
aᵣ = v² / r
Where:
- aáµ£ is the radial acceleration (what we're looking for)
- v is the tangential speed (we just found it! 4.00 m/s)
- r is the radius of the circular track (60.0 m)
Let's plug in those numbers:
aᵣ = (4.00 m/s)² / 60.0 m aᵣ = 16.0 m²/s² / 60.0 m aᵣ = 0.267 m/s²
So, the magnitude of the Corvette's radial acceleration after 2.00 seconds is 0.267 m/s². This value might seem small compared to the tangential acceleration, but it's crucial for maintaining the car's circular path. Without this inward acceleration, the car would simply move in a straight line, ignoring the curve of the track. The radial acceleration constantly changes the direction of the car's velocity, keeping it on the circular path.
This calculation highlights a key difference between tangential and radial acceleration. Tangential acceleration changes the speed of the object, while radial acceleration changes the direction of the object's motion. In this case, the tangential acceleration is making the car go faster and faster around the track, while the radial acceleration is ensuring it stays on the circular path. Understanding this distinction is vital for grasping the nuances of circular motion.
Furthermore, the magnitude of the radial acceleration depends on both the speed of the object and the radius of the circular path. A higher speed or a smaller radius will result in a greater radial acceleration. Think about it – if the car were going much faster, or if the track had a much tighter turn, the radial acceleration needed to keep it on the track would have to be much larger. This relationship between speed, radius, and radial acceleration is fundamental to understanding many real-world phenomena, from the motion of planets around the sun to the design of amusement park rides.
c) Tangential and Radial Forces
Alright, now we're getting to the forces! We need to determine the tangential and radial forces acting on the Corvette at 2.00 seconds. This is where Newton's Second Law of Motion comes into play: F = ma (Force equals mass times acceleration).
First, let's tackle the tangential force (Fₜ). We know the car's mass (1240 kg) and its tangential acceleration (2.00 m/s²). So, we can simply plug these values into Newton's Second Law:
Fₜ = (1240 kg)(2.00 m/s²) Fₜ = 2480 N
Therefore, the tangential force acting on the Corvette is 2480 N. This force is what's causing the car to speed up along the circular path. It's likely provided by the friction between the tires and the track, pushing the car forward.
Now, let's calculate the radial force (Fᵣ). This force is also known as the centripetal force, and it's what's causing the car to change direction and move in a circle. We know the car's mass (1240 kg) and its radial acceleration (0.267 m/s²), which we calculated earlier. Again, we use Newton's Second Law:
Fᵣ = (1240 kg)(0.267 m/s²) Fᵣ = 331 N
So, the radial (centripetal) force acting on the Corvette is approximately 331 N. This force is directed towards the center of the circle and is also provided by the friction between the tires and the track. However, in this case, the friction is acting perpendicular to the car's motion, rather than in the direction of motion like the tangential force.
These force calculations provide a deeper understanding of the dynamics of the Corvette's motion. The tangential force is responsible for the change in speed, while the radial force is responsible for the change in direction. Together, these forces dictate the car's trajectory on the circular track. It's crucial to remember that forces are interactions between objects. In this case, the forces we calculated are the forces exerted on the car by the track (through friction). The car, in turn, exerts equal and opposite forces on the track (Newton's Third Law!).
d) Angle of Total Acceleration
Okay, we're almost there! Now we need to determine the angle of the Corvette's total acceleration relative to the tangential direction. This is a bit trickier because the total acceleration is the vector sum of the tangential and radial accelerations. Think of it like this: the car is accelerating both forward (tangentially) and inward (radially), so its total acceleration is somewhere in between those two directions.
To find the angle, we can use trigonometry. The tangential and radial accelerations are perpendicular to each other, so they form the legs of a right triangle. The total acceleration is the hypotenuse of this triangle, and the angle we're looking for is the angle between the tangential acceleration leg and the hypotenuse.
We can use the tangent function to find this angle (θ):
tan(θ) = (opposite side) / (adjacent side) tan(θ) = aᵣ / aₜ
Where:
- θ is the angle we're trying to find
- aᵣ is the radial acceleration (0.267 m/s²)
- aₜ is the tangential acceleration (2.00 m/s²)
Let's plug in the values:
tan(θ) = 0.267 m/s² / 2.00 m/s² tan(θ) = 0.1335
To find the angle, we need to take the inverse tangent (arctan) of 0.1335:
θ = arctan(0.1335) θ ≈ 7.61 degrees
So, the angle of the Corvette's total acceleration relative to the tangential direction is approximately 7.61 degrees. This relatively small angle indicates that the tangential acceleration is much larger than the radial acceleration at this particular moment. In other words, the car is primarily accelerating forward, with only a small component of acceleration directed inward to keep it on the circular path.
Visualizing this with a vector diagram can be helpful. Draw a horizontal arrow representing the tangential acceleration and a vertical arrow representing the radial acceleration. The total acceleration is the vector pointing from the tail of the tangential acceleration arrow to the head of the radial acceleration arrow. The angle we just calculated is the angle between the horizontal arrow and this resultant vector. As the car speeds up, the radial acceleration will increase (since it depends on the square of the speed), and this angle will also increase. This problem beautifully illustrates how vector addition and trigonometry are essential tools for analyzing motion in two dimensions.
Conclusion
And there you have it! We've successfully tackled a classic physics problem involving a gorgeous 1957 Chevy Corvette. We calculated its speed, radial acceleration, tangential and radial forces, and the angle of its total acceleration. This problem demonstrates the interplay of several key physics concepts, including kinematics, circular motion, Newton's Laws of Motion, and trigonometry. By breaking the problem down into smaller, manageable steps, we were able to gain a comprehensive understanding of the car's motion. Hope you enjoyed the ride, guys! Remember, physics is all around us, even in classic cars tearing up the track!
