Question

Some rental cars have a GPS unit installed, which allows the rental car company to check where you are at all times and thus also know your speed at any time. One of these rental cars is driven by an employee in the company's lot and, during the time interval from 0 to \(10.0 \mathrm{~s}\), is found to have a position vector as a function of time given by $$ \begin{aligned} \vec{r}(t)=&\left((24.4 \mathrm{~m})-t(12.3 \mathrm{~m} / \mathrm{s})+t^{2}\left(2.43 \mathrm{~m} / \mathrm{s}^{2}\right)\right.\\\ &\left.(74.4 \mathrm{~m})+t^{2}\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right)-t^{3}\left(0.130 \mathrm{~m} / \mathrm{s}^{3}\right)\right) \end{aligned} $$ a) What is the distance of this car from the origin of the coordinate system at \(t=5.00 \mathrm{~s} ?\) b) What is the velocity vector as a function of time? c) What is the speed at \(t=5.00 \mathrm{~s} ?\)

Short answer

b) What is the velocity vector function of the car? c) What is the speed of the car at t=5s? a) To find the distance of the car from the origin at t=5s, calculate the magnitude of the position vector at t=5s (using the formula \(d=\sqrt{x^2+y^2}\)). b) The velocity vector function of the car is given by: $$\vec{v}(t) = \left(-12.3 + 2(2.43)t, 2(1.80)t - 3(0.130)t^2\right)$$ c) To find the speed of the car at t=5s, calculate the magnitude of the velocity vector at t=5s (using the formula \(s=|\vec{v}(t)|\)).

Step-by-step solution

Calculate the distance at t=5s

Given the position vector function \(\vec{r}(t)\). First, find the position of the car at t=5s by substituting t=5s in the given function: $$ \begin{aligned} \vec{r}(5)=&\left((24.4 \mathrm{~m})-5(12.3 \mathrm{~m} / \mathrm{s})+5^{2}\left(2.43 \mathrm{~m} / \mathrm{s}^{2}\right)\right.\\\ &\left.(74.4 \mathrm{~m})+5^{2}\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right)-5^{3}\left(0.130 \mathrm{~m} / \mathrm{s}^{3}\right)\right) \end{aligned} $$ Calculate the x and y components of the position vector at t=5s: $$\begin{aligned} x(5) =& 24.4 - 5(12.3) + 5^2(2.43) \\ y(5) =& 74.4 + 5^2(1.80) - 5^3(0.130) \end{aligned}$$ Now, use the distance formula \(d=\sqrt{x^2+y^2}\) to find the distance of the car from the origin at t=5s: $$d(5) = \sqrt{x(5)^2+y(5)^2}$$

Find the velocity vector function

To find the velocity vector function, we need to find the derivative of the position vector function with respect to time: $$\begin{aligned} \vec{v}(t) = \frac{d\vec{r}(t)}{dt} = \frac{d}{dt}\left( (24.4-t(12.3)+t^2(2.43)), (74.4+t^2(1.80)-t^3(0.130)) \right) \end{aligned}$$ Take the derivative of each component with respect to time: $$\begin{aligned} v_x(t) =& - 12.3 + 2(2.43)t \\ v_y(t) =& 2(1.80)t - 3(0.130)t^2 \end{aligned}$$ So the velocity vector function as a function of time is: $$\begin{aligned} \vec{v}(t) = \left(-12.3 + 2(2.43)t, 2(1.80)t - 3(0.130)t^2\right) \end{aligned}$$

Find the speed at t=5s

To find the speed of the car at t=5s, first find the velocity vector at t=5s by substituting t=5s into the velocity vector function: $$\begin{aligned} \vec{v}(5) = \left(-12.3 + 2(2.43)(5), 2(1.80)(5) - 3(0.130)(5^2)\right) \end{aligned}$$ Now, find the magnitude of the velocity vector at t=5s, which represents the speed: $$s(5) = |\vec{v}(5)| = \sqrt{ (v_x(5))^2 + (v_y(5))^2 }$$

Key concepts

Position Vector

Understanding the position vector is key when studying kinematics, which is essentially the branch of physics that describes the motion of points, bodies, and systems. Think of a position vector as an arrow drawn from a fixed point of reference, usually called the origin, to an object's location. It tells you exactly where the object is relative to that origin, in terms of the x, y, and sometimes z coordinates in space. In our exercise, the position vector \(\vec{r}(t)\) was given as a function that changes over time, reflecting the car's movement over time.

When calculating the position of the car at a specific time, like at 5 seconds, we plug in the value of t into the given equation. By simplifying the equation, we find the x and y components of the position vector. The magnitude of this vector then gives us the distance from the origin. In simpler terms, we're just finding out how far away the car is from where it started and in which direction it's moved.

Velocity Vector

If the position vector tells us where the car is, the velocity vector can tell us how fast the car's position is changing and in which direction. Velocity is a vector, meaning it has both magnitude and direction. It's obtained by taking the derivative of the position vector with respect to time.

Derivatives here act like a snapshot of the car's speedometer at any moment; they give us an instant reading of the speed and direction of the car. Mathematically, we denote the velocity vector as \(\vec{v}(t)\) and calculate it by differentiating each component of the position vector \(\vec{r}(t)\) with respect to time. This gives us the car's speed in the x-direction and the y-direction, which together form the velocity vector. The faster the car goes or the more it curves, the larger the magnitude of our velocity vector will be.

Time-Dependent Motion

In our exercise, the car's motion is time-dependent, meaning that its position and speed vary as time passes. Time-dependent motion is an essential concept in physics because most objects change their position and speed over time. To analyze this motion, like we did for the rental car, we need mathematical functions that explain how these properties change with time.

Our position and velocity equations are precisely these kinds of functions—they can be seen as recipes that tell us the car's position and velocity at any given moment. With these equations, we can graph the car's path over time, predict where it will be at a future moment, or determine how fast it was going at some point in the past. This sort of analysis is crucial for everything from designing car airbags to programming video games.

Derivatives in Physics

In physics, derivatives are everywhere. They are mathematical tools that help us understand rates of change. Whenever you see the word 'rate,' you're likely dealing with a derivative. It's like asking, 'how quickly or slowly is something happening?'

In our example, we found the derivative of the position vector function to get the velocity vector. The act of finding the derivative is known as differentiation. If we take the derivative of the velocity vector function next, we'd get the acceleration of the car—the rate at which the car's velocity is changing. By understanding derivatives, you unlock the ability to analyze motion at any level—whether it’s a car on a track, a planet orbiting the sun, or particles zooming inside an accelerator.