Question

An object moves in the \(x y\) -plane. The \(x\) - and \(y\) -coordinates of the object as a function of time are given by the following equations: \(x(t)=4.9 t^{2}+2 t+1\) and \(y(t)=3 t+2\). What is the velocity vector of the object as a function of time? What is its acceleration vector at the time \(t=2 \mathrm{~s}\) ?

Short answer

The velocity vector as a function of time is \(\mathbf{V}(t) = (9.8t + 2)\mathbf{i} + 3\mathbf{j}\) and the acceleration vector at t = 2s is \(\mathbf{A}(2) = 9.8\mathbf{i}\).

Step-by-step solution

Differentiate x(t) and y(t)

To find the components of the velocity vector, we need to find the derivatives of x(t) and y(t) with respect to time t. Differentiate x(t) and y(t) with respect to time t: \(x'(t) = \frac{d(4.9t^2 + 2t + 1)}{dt} = 9.8t + 2\) \(y'(t) = \frac{d(3t + 2)}{dt} = 3\)

Find the velocity vector

Now that we have the derivatives, we can combine them to find the velocity vector as a function of time: \(\mathbf{V}(t) = (9.8t + 2)\mathbf{i} + 3\mathbf{j}\)

Differentiate the components of the velocity vector

Next, we need to differentiate the components of the velocity vector to find the components of the acceleration vector: \(x''(t) = \frac{d(9.8t + 2)}{dt} = 9.8\) \(y''(t) = \frac{d(3)}{dt} = 0\)

Calculate the acceleration vector at t = 2s

Now, we have the acceleration components as functions of time. To find the acceleration vector at t = 2s, we substitute t with 2 in the x and y components of the acceleration vector: \(\mathbf{A}(2) = (9.8)\mathbf{i} + 0\mathbf{j}\) So the velocity vector as a function of time is \(\mathbf{V}(t) = (9.8t + 2)\mathbf{i} + 3\mathbf{j}\) and the acceleration vector at t = 2s is \(\mathbf{A}(2) = 9.8\mathbf{i}\).

Key concepts

Kinematics in Physics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It's the study of how objects move through space over time. Key concepts in kinematics include displacement, velocity, acceleration, and time. The velocity of an object is a vector that describes both how fast the object is moving and in what direction it's traveling. When questions arise about the position of an object in a plane as it changes in time, like in the exercise, kinematics provides the tools to describe this motion algebraically and graphically using motion equations.

In our example, the path of the object is given by equations for both the x and y coordinates as functions of time, known as parametric equations. Calculating the velocity vector involves differentiating both of these parametric equations with respect to time. Through understanding kinematics, students can visualize and calculate an object's motion path, as well as its velocity and acceleration at any given point in time.

Vector Differentiation

Vector differentiation is fundamental when dealing with motion in multiple dimensions. It involves computing the derivative of vector-valued functions with respect to a variable, usually time in kinematic problems. For the velocity and acceleration vectors in our exercise, we differentiate the position vectors, which are dependent on time, to find these quantities.

The process involves taking the derivative of each component of the position vector separately. In our textbook solution, the position vector's x-component is differentiated to get the velocity in the x-direction, and similarly for the y-component. Once you have both velocity components, they can be combined into a velocity vector. Whenever a physical quantity varies with time and needs to be described as a vector, such as velocity or force, vector differentiation becomes a crucial tool in physics.

Motion Equations

Motion equations, also known as the equations of motion, describe the mathematical relationship between displacement, velocity, acceleration, and time. They are a core part of solving kinematic problems. In our exercise, the object's equations of motion are described with two separate equations for the x and y coordinates.

For one-dimensional motion, the equations are relatively simple, but for two-dimensional motion, like in the exercise, you need to use separate motion equations for each dimension. In the textbook solution, we differentiate the motion equations to determine the velocity and acceleration. The velocity as a function of time is easily extracted, and then acceleration is found by differentiating velocity, helping us understand how the object's motion changes with time. These equations are powerful because they can predict an object's future position and velocity based on its current state.

Time-Dependent Functions

Many quantities in physics depend on time, which is where time-dependent functions come into play. They are functions whose values change as time progresses. The position, velocity, and acceleration of a moving object can all be described by time-dependent functions.

In the given exercise, both the object's x and y coordinates are time-dependent, and they change as time increases. Calculating the rate of change of these coordinates with respect to time (using differentiation) gives us the object's velocity and acceleration, which are themselves time-dependent. It's a key concept in physics that allows us to model and predict how dynamic systems evolve over time. By understanding time-dependent functions, students can analyze and characterize the motion of objects and how it changes at any instant.