Determining Velocity Of An Object With Variable Acceleration In Linear Motion
In the fascinating realm of physics, understanding the motion of objects is a fundamental concept. One common scenario involves an object moving in a straight line, experiencing a change in velocity over time. This change in velocity is known as acceleration, and when it's not constant, the motion becomes more intricate and requires a deeper analysis. Let's delve into a specific case: an object moving in a straight line with an initial velocity of 5 m/s, subjected to a time-dependent acceleration given by a(t) = -12t - 2 m/s². Our goal is to determine the object's velocity, v(t), at any time t seconds after the experiment begins.
Unveiling the Velocity Function: v(t)
To determine the velocity function, v(t), we must use the fundamental relationship between acceleration and velocity. Acceleration is the rate of change of velocity with respect to time. Mathematically, this is expressed as:
a(t) = dv(t) / dt
Where:
- a(t) is the acceleration at time t
- v(t) is the velocity at time t
- dv(t)/dt represents the derivative of the velocity function with respect to time
To find v(t), we need to perform the reverse operation of differentiation, which is integration. We integrate the acceleration function, a(t), with respect to time:
v(t) = ∫ a(t) dt
In our case, a(t) = -12t - 2 m/s². Substituting this into the integral, we get:
v(t) = ∫ (-12t - 2) dt
Now, we can integrate term by term:
v(t) = -12 ∫ t dt - 2 ∫ dt
Using the power rule of integration (∫ x^n dx = (x^(n+1))/(n+1) + C), we get:
v(t) = -12 (t²/2) - 2t + C
Simplifying, we have:
v(t) = -6t² - 2t + C
Where C is the constant of integration. To find the value of C, we use the initial condition: the object has an initial velocity of 5 m/s at t = 0. This means v(0) = 5.
Substituting t = 0 and v(0) = 5 into the equation, we get:
5 = -6(0)² - 2(0) + C
Therefore, C = 5. Now we have the complete velocity function:
v(t) = -6t² - 2t + 5
This equation describes the velocity of the object at any time t after the experiment begins.
Interpreting the Velocity Function
The velocity function, v(t) = -6t² - 2t + 5, provides valuable insights into the object's motion. Let's break down its components:
- -6t² term: This term indicates that the velocity is decreasing quadratically with time. The negative sign signifies that the acceleration is acting in the opposite direction to the initial velocity, causing the object to slow down. As time progresses, this term becomes more significant, leading to a greater decrease in velocity.
- -2t term: This term represents a linear decrease in velocity with time. It further contributes to the deceleration of the object.
- +5 term: This is the constant term, representing the initial velocity of the object at t = 0. It serves as the starting point for the velocity function.
By analyzing the velocity function, we can understand how the object's velocity changes over time. Initially, the object is moving at 5 m/s, but the negative acceleration causes it to slow down. The quadratic term (-6t²) dominates as time increases, indicating a rapid decrease in velocity. Eventually, the object's velocity will become zero, and it will start moving in the opposite direction (assuming the motion is along a straight line). This turning point can be found by setting v(t) = 0 and solving for t.
Connecting to Real-World Scenarios
This type of motion, with a time-dependent acceleration, is common in various real-world scenarios. Consider the following examples:
- A car braking: When a car brakes, it experiences a deceleration that can be approximated by a time-dependent function. The initial braking force might be stronger, leading to a higher deceleration, which gradually decreases as the car slows down.
- A projectile in motion: The vertical motion of a projectile (like a ball thrown upwards) is affected by gravity, which provides a constant downward acceleration. However, air resistance can also play a role, and its effect can be approximated by a time-dependent acceleration term.
- A rocket launch: During the initial phase of a rocket launch, the engine provides a thrust that generates acceleration. This acceleration can vary with time as the rocket burns fuel and the mass of the rocket decreases.
Understanding the concepts of linear motion with variable acceleration is crucial for analyzing and predicting the motion of objects in these and many other real-world situations.
Determining the Time When Velocity is Zero
A critical aspect of analyzing motion is identifying when the object's velocity becomes zero. This point often signifies a change in direction or a temporary pause in motion. To find the time (t) when the velocity is zero, we set the velocity function, v(t), equal to zero and solve for t:
v(t) = -6t² - 2t + 5 = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
t = [-b ± √(b² - 4ac)] / 2a
Where, in our equation:
- a = -6
- b = -2
- c = 5
Substituting these values into the quadratic formula, we get:
t = [2 ± √((-2)² - 4(-6)(5))] / 2(-6)
t = [2 ± √(4 + 120)] / -12
t = [2 ± √124] / -12
t = [2 ± 2√31] / -12
t = [-1 ± √31] / 6
This gives us two possible solutions for t:
- t₁ = (-1 + √31) / 6
- t₂ = (-1 - √31) / 6
Since time cannot be negative in this context (we're considering time after the experiment begins), we discard the negative solution. Therefore, the time when the velocity is zero is:
t = (-1 + √31) / 6 ≈ 0.76 seconds
This means that approximately 0.76 seconds after the experiment begins, the object's velocity will be zero. At this point, the object momentarily stops before potentially changing direction due to the continuous acceleration.
Calculating Displacement
Another important aspect of motion is displacement, which refers to the change in position of an object. To calculate the displacement of the object over a certain time interval, we need to integrate the velocity function, v(t), with respect to time over that interval. The displacement, denoted as Δx, between time t₁ and t₂ is given by:
Δx = ∫[t₁ to t₂] v(t) dt
In our case, v(t) = -6t² - 2t + 5. Let's calculate the displacement of the object from t = 0 to t = 1 second:
Δx = ∫[0 to 1] (-6t² - 2t + 5) dt
Integrating term by term, we get:
Δx = [-6 ∫[0 to 1] t² dt - 2 ∫[0 to 1] t dt + 5 ∫[0 to 1] dt]
Using the power rule of integration, we have:
Δx = [-6 (t³/3)|[0 to 1] - 2 (t²/2)|[0 to 1] + 5 (t)|[0 to 1]]
Now, we evaluate the definite integrals:
Δx = [-6 (1³/3 - 0³/3) - 2 (1²/2 - 0²/2) + 5 (1 - 0)]
Δx = [-6 (1/3) - 2 (1/2) + 5]
Δx = [-2 - 1 + 5]
Δx = 2 meters
This means that the object's displacement from t = 0 to t = 1 second is 2 meters. In other words, the object moves 2 meters in the positive direction during this time interval.
Understanding how to calculate displacement is crucial for determining the overall change in position of an object undergoing variable motion. It provides a comprehensive picture of the object's trajectory and its location at different points in time.
Conclusion: Mastering the Dynamics of Linear Motion
In conclusion, analyzing the motion of an object moving in a straight line with variable acceleration requires a thorough understanding of the relationships between acceleration, velocity, and displacement. By integrating the acceleration function, we can determine the velocity function, which provides insights into how the object's speed and direction change over time. Setting the velocity function to zero allows us to find the time when the object momentarily stops or changes direction. Furthermore, integrating the velocity function gives us the displacement, which quantifies the overall change in position.
In our specific example, the object started with an initial velocity of 5 m/s and experienced a time-dependent acceleration of a(t) = -12t - 2 m/s². We successfully derived the velocity function, v(t) = -6t² - 2t + 5, and used it to calculate the time when the velocity is zero and the displacement over a specific time interval. These calculations demonstrate the power of calculus in analyzing motion and predicting the behavior of objects in dynamic systems.
The concepts and techniques discussed in this article are fundamental to various fields of physics and engineering. They provide the foundation for understanding more complex motions, such as projectile motion, rotational motion, and oscillations. By mastering the dynamics of linear motion with variable acceleration, we gain a valuable toolset for analyzing and solving a wide range of real-world problems.
