Motion Analysis: Displacement And Velocity Calculation

Hey guys! Let's dive into a classic problem from calculus and physics – analyzing the motion of an object along a straight line. We're going to break down a scenario where a body's position is described by a mathematical function, and we'll calculate some key aspects of its movement, such as displacement and average velocity. This kind of problem is super important for understanding how things move and change over time, so buckle up and let's get started!

Understanding Displacement and Average Velocity

Before we jump into the calculations, let's make sure we're all on the same page with the basic concepts. When we talk about displacement, we're referring to the change in an object's position. It's the difference between the final position and the initial position. Think of it like this: if you start at your house and walk to the store, your displacement is the distance between your house and the store, regardless of the path you took to get there. Displacement is a vector quantity, meaning it has both magnitude (how far) and direction (which way). In our one-dimensional case, direction will be positive or negative depending on the direction of motion along the coordinate line.

Average velocity, on the other hand, is the displacement divided by the time interval over which the displacement occurred. It tells us how quickly the object's position changed, on average, during that time. It's also a vector quantity. The key word here is "average" – the object might have been moving faster or slower at different points during the interval, but the average velocity gives us an overall sense of its motion. To really understand the motion, we'll also need to look at instantaneous velocity, which we'll get to later. But for now, let's focus on getting a handle on displacement and average velocity.

Why are these concepts important? Well, they form the foundation for understanding more complex motion. Whether it's a car moving down the street, a ball being thrown, or even a planet orbiting a star, the principles of displacement and velocity are crucial for describing and predicting their movement. By mastering these basics, we can tackle more challenging problems in physics and engineering. So, let's get into the nitty-gritty of the problem at hand!

Problem Setup: Position as a Function of Time

Okay, so here's the scenario. We have a body moving along a coordinate line (think of it as a straight track). The body's position, denoted by s, is given by a function of time, t:

$s = f(t) = t^2 - 9t + 8$

This equation tells us exactly where the body is located at any given time t. The units for position, s, are in meters, and the units for time, t, are in seconds. We're interested in the body's motion over the time interval $0 extless t extless 6$. This means we'll be looking at what happens from the very beginning (t = 0 seconds) up to 6 seconds. Got it?

The function $f(t) = t^2 - 9t + 8$ is a quadratic function, which means its graph is a parabola. This tells us that the body's motion isn't constant – it's going to be changing its speed and possibly its direction. The t² term indicates that the body's velocity will change linearly with time (constant acceleration), making this a uniformly accelerated motion problem. The coefficient -9 in front of the t suggests that the body will initially move in the negative direction (decreasing s) before eventually turning around and moving in the positive direction.

Our task is twofold. First, we need to find the body's displacement and average velocity over the given time interval (0 to 6 seconds). This will give us an overall picture of how the body's position changed during this time. Second, we'll need to delve deeper into the specifics of the motion at particular points in time. This will involve finding the body's velocity and acceleration as functions of time, and analyzing when the body is speeding up or slowing down. By tackling both parts of this problem, we'll gain a solid understanding of the body's motion.

Part (a): Displacement and Average Velocity Calculation

Let's start with the first part of the problem: finding the displacement and average velocity of the body over the time interval $0 extless t extless 6$.

Calculating Displacement

As we discussed earlier, displacement is the change in position. To find it, we need to determine the body's position at the beginning (t = 0) and at the end (t = 6) of the time interval. We can do this by plugging these values of t into our position function, $f(t) = t^2 - 9t + 8$.

At t = 0 seconds:

$s(0) = f(0) = (0)^2 - 9(0) + 8 = 8 ext{ meters}$

So, at the beginning, the body is located 8 meters from the origin.

At t = 6 seconds:

$s(6) = f(6) = (6)^2 - 9(6) + 8 = 36 - 54 + 8 = -10 ext{ meters}$

At the end of the interval, the body is located -10 meters from the origin. This means it has moved to the other side of the origin (in the negative direction).

Now we can calculate the displacement, which is the final position minus the initial position:

Displacement = $s(6) - s(0) = -10 ext{ meters} - 8 ext{ meters} = -18 ext{ meters}$

The displacement is -18 meters. The negative sign indicates that the body's overall movement was in the negative direction along the coordinate line. It moved a total of 18 meters in the negative direction from its starting point.

Calculating Average Velocity

Next, we need to find the average velocity. Remember, average velocity is the displacement divided by the time interval.

We already know the displacement is -18 meters. The time interval is the difference between the final time (t = 6 seconds) and the initial time (t = 0 seconds), which is 6 seconds.

Average velocity = $\frac{\text{Displacement}}{\text{Time interval}} = \frac{-18 ext{ meters}}{6 ext{ seconds}} = -3 ext{ meters/second}$

The average velocity is -3 meters per second. This tells us that, on average, the body was moving in the negative direction at a rate of 3 meters every second during the time interval. Keep in mind that this is just the average – the body's instantaneous velocity might have been higher or lower at different points in time.

So, to recap, we've found that the body's displacement over the time interval $0 extless t extless 6$ is -18 meters, and its average velocity is -3 meters per second. We've got a good handle on the overall motion of the body during this time. Now, let's dive deeper into what's happening at specific points in time!

Part (b): Finding the ... (To be continued)

We've tackled the first part of our motion analysis problem, calculating the displacement and average velocity of the body. But to get a complete picture of the body's movement, we need to dig a little deeper. We need to understand how the body's velocity changes over time, and identify when it's speeding up or slowing down. This involves finding the body's velocity and acceleration as functions of time.

Instantaneous Velocity

The average velocity gives us an overview of the motion, but it doesn't tell us the velocity at a specific instant. To find the instantaneous velocity, we need to use calculus. Remember that velocity is the rate of change of position with respect to time. Mathematically, this means we need to find the derivative of the position function, $s(t)$, with respect to time, t. So, the instantaneous velocity, v(t), is given by:

$v(t) = \frac{ds}{dt} = f'(t)$

Our position function is $s(t) = t^2 - 9t + 8$. Let's find its derivative:

$v(t) = f'(t) = 2t - 9$

This equation, $v(t) = 2t - 9$, gives us the velocity of the body at any time t in the interval $0 extless t extless 6$. Notice that the velocity is a linear function of time, which means it changes at a constant rate. This is a consequence of the fact that our original position function was quadratic.

Acceleration

Now, let's consider acceleration. Acceleration is the rate of change of velocity with respect to time. In other words, it tells us how quickly the velocity is changing. Just like we found velocity by taking the derivative of position, we can find acceleration by taking the derivative of velocity. So, the acceleration, a(t), is given by:

$a(t) = \frac{dv}{dt} = v'(t) = f''(t)$

We already found the velocity function, $v(t) = 2t - 9$. Let's take its derivative to find the acceleration:

$a(t) = v'(t) = 2$

Wow! The acceleration is constant and equal to 2 meters per second squared. This means the body's velocity is increasing at a steady rate of 2 meters per second every second. This makes sense given that the coefficient of the t^2 term is positive.

When is the Body at Rest?

A crucial question we can answer now is: when is the body at rest? The body is at rest when its velocity is zero. So, we need to solve the equation $v(t) = 0$ for t:

$2t - 9 = 0$

$2t = 9$

$t = 4.5 ext{ seconds}$

The body is at rest at t = 4.5 seconds. This is an important moment in the body's motion because it's the point where the body changes direction. Before 4.5 seconds, the velocity is negative (body moving in the negative direction), and after 4.5 seconds, the velocity is positive (body moving in the positive direction).

Speeding Up and Slowing Down

Finally, let's determine when the body is speeding up and when it's slowing down. This depends on the signs of the velocity and acceleration. Here's the rule:

• If the velocity and acceleration have the same sign (both positive or both negative), the body is speeding up.
• If the velocity and acceleration have opposite signs (one positive and one negative), the body is slowing down.

We know that the acceleration, a(t), is always positive (2 m/s²). So, we only need to consider the sign of the velocity, v(t).

• Slowing down: The body is slowing down when $v(t) extless 0$. We already found that $v(t) = 0$ at $t = 4.5$ seconds. So, for $0 extless t extless 4.5$ seconds, $v(t)$ is negative, and the body is slowing down.
• Speeding up: The body is speeding up when $v(t) extgreater 0$. For $4.5 extless t extless 6$ seconds, $v(t)$ is positive, and the body is speeding up.

We've now completely analyzed the motion of the body over the given time interval. We've found its displacement, average velocity, instantaneous velocity, and acceleration. We've also determined when the body is at rest, speeding up, and slowing down. This is a pretty comprehensive understanding of the body's movement!

Conclusion

This problem demonstrates the power of calculus in analyzing motion. By using derivatives, we can move from position to velocity to acceleration, gaining a deeper understanding of how an object moves. This kind of analysis is crucial in many areas of science and engineering, from designing vehicles to predicting the trajectories of spacecraft. So, keep practicing these concepts, and you'll be well on your way to mastering the physics of motion!