Problem 40
Question
The position in feet of a car along a straight racetrack after \(t\) seconds is approximated by \(s(t)\) Find the car's velocity in feet per second after 3 seconds. $$s(t)=4 t^{2}+5 t+1$$
Step-by-Step Solution
Verified Answer
The car's velocity after 3 seconds is 29 feet per second.
1Step 1: Understand the Formula,
The function given, \(s(t) = 4t^2 + 5t + 1\), describes the position of the car in feet at any time \(t\) seconds. The task is to find the velocity of the car, which is the rate of change of position with respect to time.
2Step 2: Find Derivative for Velocity Function,
Velocity of the car is the derivative of the position function \(s(t)\) with respect to time \(t\). Using differentiation rules, the derivative of \(s(t) = 4t^2 + 5t + 1\) is \(s'(t) = 8t + 5\).
3Step 3: Calculate Velocity at \(t=3\),
Now substitute \(t = 3\) seconds into the velocity function \(s'(t) = 8t + 5\). This gives us \(s'(3) = 8(3) + 5 = 24 + 5 = 29\).
4Step 4: Conclusion,
The velocity of the car after 3 seconds is 29 feet per second.
Key Concepts
DerivativeVelocityRate of Change
Derivative
In calculus, a derivative represents how a function changes as its input changes. It's like a tool that helps us understand the rate at which one quantity changes with another. When given a function like the position of a car along a racetrack, the derivative helps us find how fast the position is changing over time.
To compute a derivative, we use certain rules that describe how to differentiate various forms of functions. In the problem, this involves using the power rule, which is fundamental in calculus. For example, the power rule states that the derivative of a function like \(ax^n\) is \((n \cdot ax^{n-1})\).
Applying this to our position function \(s(t) = 4t^2 + 5t + 1\):
To compute a derivative, we use certain rules that describe how to differentiate various forms of functions. In the problem, this involves using the power rule, which is fundamental in calculus. For example, the power rule states that the derivative of a function like \(ax^n\) is \((n \cdot ax^{n-1})\).
Applying this to our position function \(s(t) = 4t^2 + 5t + 1\):
- For \(4t^2\), the derivative is \(8t\), because \(2 \times 4 = 8\), and we reduce the power by one to get \(t\).
- For \(5t\), it's simply \(5\), since the derivative of \(t\) is \(1\).
- The constant term \(1\) has a derivative of \0\, as constants don't change.
Velocity
Velocity is a measure of speed that includes direction. In the context of race cars and other moving objects, it's essential to know not just how fast something is moving but also in what direction. Mathematically, velocity is the derivative of the position function with respect to time, reflecting how position changes as time progresses.
For the car on the racetrack, the position function \(s(t) = 4t^2 + 5t + 1\) describes how the car's position changes over time. By differentiating this function, we acquire the velocity function \(s'(t) = 8t + 5\), which tells us the car's speed at any given moment.
To find the velocity at a specific instant, such as at \(t = 3\) seconds, we simply substitute \(t\) with \(3\) in the velocity function:
For the car on the racetrack, the position function \(s(t) = 4t^2 + 5t + 1\) describes how the car's position changes over time. By differentiating this function, we acquire the velocity function \(s'(t) = 8t + 5\), which tells us the car's speed at any given moment.
To find the velocity at a specific instant, such as at \(t = 3\) seconds, we simply substitute \(t\) with \(3\) in the velocity function:
- Calculate \(s'(3) = 8(3) + 5 = 24 + 5\).
- This results in a velocity of \29\ feet per second at 3 seconds.
Rate of Change
The rate of change is a concept that indicates how one quantity changes in relation to another. In simple terms, it's a comparison between changes in value over changes in time. In calculus, the rate of change is often expressed by a derivative, especially in the context of position and velocity.
In the given problem, the rate of change is represented by the velocity of the car, derived from the position function. The process of finding this involves calculating the derivative of the position function, giving us the car's velocity at any moment in time.
Understanding this idea:
In the given problem, the rate of change is represented by the velocity of the car, derived from the position function. The process of finding this involves calculating the derivative of the position function, giving us the car's velocity at any moment in time.
Understanding this idea:
- Position as a function of time: \(s(t)\), changes over time.
- Rate of change of position: is determined by \(s'(t)\), the derived velocity function.
- Evaluating rate of change (velocity) at \(t=3\) yields \29\ feet per second, showing a specific snapshot of how position changes at that time.
Other exercises in this chapter
Problem 39
Determine each limit, if it exists. $$\lim _{x \rightarrow 5}\left[\log _{3}(2 x-1)\right]$$
View solution Problem 40
Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-6 x+9}\)
View solution Problem 40
Determine each limit. $$\lim _{x \rightarrow-\infty} \frac{1-2 x+3 x^{2}}{2 x^{2}+5 x}$$
View solution Problem 40
Determine each limit, if it exists. $$\lim _{x \rightarrow 4}\left[\log _{2}(14+\sqrt{x})\right]$$
View solution