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 a time \(t=2\) s?

Short answer

Answer: The velocity vector as a function of time is v(t) = [9.8t + 2, 3], and the acceleration vector at t=2s is [9.8, 0].

Step-by-step solution

Find the velocity vector as a function of time

To find the velocity vector, we need to take the derivative of the position functions with respect to time, t. The x(t) function is given by: x(t) = 4.9t^2 + 2t + 1 Taking the derivative with respect to time, we get the x-component of the velocity vector: v_x(t) = \frac{dx}{dt} = 9.8t + 2 The y(t) function is given by: y(t) = 3t + 2 Taking the derivative with respect to time, we get the y-component of the velocity vector: v_y(t) = \frac{dy}{dt} = 3 Now, we can combine the x and y components to form the velocity vector: v(t) = [9.8t + 2, 3]

Find the acceleration vector at t=2s

To find the acceleration vector, we need to take the derivative of the velocity functions with respect to time, t, or take the second derivative of the position functions. Taking the derivative of v_x(t) (which is the second derivative of x(t)) with respect to time, we get the x-component of the acceleration vector: a_x(t) = \frac{d^2x}{dt^2} = 9.8 Taking the derivative of v_y(t) (which is the second derivative of y(t)) with respect to time, we get the y-component of the acceleration vector: a_y(t) = \frac{d^2y}{dt^2} = 0 Now, we can combine the x and y components to form the acceleration vector at any given time t: a(t) = [9.8, 0] To find the acceleration vector at t=2s, we simply plug t=2 into the acceleration vector equation: a(2) = [9.8, 0] The acceleration vector at time t=2s is [9.8, 0].

Key concepts

Understanding Kinematics Equations

Kinematics equations are the backbone of classical mechanics, especially when it's about motion in physics. They link the position, velocity, and acceleration of an object with time, which are essential to understanding its motion.

Kinematics is not concerned with the forces that cause the motion but rather describes the motion itself. A primary set of equations that describe this motion in one dimension are:

• The position equation: \( x(t) = x_0 + v_0t + \frac{1}{2}at^2 \)
• The velocity equation: \( v(t) = v_0 + at \)
• The acceleration equation: \( a(t) = a \), where \( a \) is constant.

But what if acceleration isn't constant? This leads us to our next key topic.

Time-Dependent Acceleration Concepts

While the kinematic equations assume constant acceleration, many real-world scenarios involve time-dependent acceleration, which means the acceleration changes as time progresses. To tackle this, calculus becomes an indispensable tool.

For time-dependent acceleration, you need to integrate acceleration over time to get the velocity, and similarly, integrate the velocity to find position. If you have an acceleration as a function of time, \( a(t) \), the velocity at time \( t \) is found by integrating \( a(t) \):

• \( v(t) = \int a(t) \, dt \)

And for position:

• \( x(t) = \int v(t) \, dt \)

The beauty of calculus lies in its ability to deal with such dynamic variables effectively. This concept is pivotal when acceleration is not uniform, which is the case in many complex motions.

Derivative of Position - Velocity Vector

In the realm of physics, particularly in the study of kinematics, the derivative of position with respect to time gives us an object's velocity vector. It is a vector because velocity has both magnitude and direction, which tells us not just how fast the object is traveling but also the direction of its travel.

Considering an object moving in the xy-plane, with its position described by functions \( x(t) \) and \( y(t) \), the velocity is found by differentiating these position functions with respect to time:

• \( v_x(t) = \frac{dx}{dt} \)
• \( v_y(t) = \frac{dy}{dt} \)

These components \( v_x(t) \) and \( v_y(t) \) form the velocity vector \( v(t) = [v_x(t), v_y(t)] \). Thus, if an object's position equations in terms of time are known, its velocity at any moment can be accurately determined by taking the time derivative of its position.