Chapter 4: Problem 39
A car travels s meters in the \(t\) seconds after the brakes are applied, where\(s=-9.9 t^{2}+79.2 t\) a. Find an expression for the instantaneous velocity as a function of time. b. How many seconds after the brakes are applied will the car stop? c. After the brakes are applied, how far will the car travel before it stops?
Short Answer
Expert verified
a. \(v(t) = -19.8t + 79.2\) ewline b. 4 seconds ewline c. 158.4 meters
Step by step solution
01
Find the expression for instantaneous velocity
To find the instantaneous velocity as a function of time, take the derivative of the position function with respect to time. The position function is given by ewline \(s(t) = -9.9 t^{2} + 79.2 t\) ewline Differentiating with respect to time, ewline \(v(t) = \frac{ds}{dt} = -19.8t + 79.2\)
02
Find when the car stops
The car stops when its instantaneous velocity is zero. Set the velocity function equal to zero and solve for t. ewline \(0 = -19.8t + 79.2\) ewline Solving for t, ewline \(t = \frac{79.2}{19.8} = 4\)
03
Find the distance traveled before stopping
To find the distance traveled before the car stops, substitute the time found in Step 2 into the original position function. Substituting \(t = 4\) into \(s(t) = -9.9 t^{2} + 79.2 t\), ewline \(s(4) = -9.9 (4)^{2} + 79.2 (4) = 158.4\)
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
differentiation
Differentiation is a fundamental concept in calculus used to find the rate at which something changes. In this exercise, we need to find the rate of change of the car's position, which gives us the car's instantaneous velocity. The position function of the car is given by \(s(t) = -9.9 t^{2} + 79.2 t\). To find the velocity as a function of time, we take the derivative of \(s(t)\) with respect to \(t\). This derivative gives us the formula for the instantaneous velocity \(v(t) \).
The differentiation process follows these steps:
- Identify the function you want to differentiate, which is \(s(t) = -9.9 t^{2} + 79.2 t\).
- Apply the power rule of differentiation. If you have a term like \(a t^n\), its derivative is \(n a t^{n-1}\).
Following these rules, the derivative of \(s(t)\) becomes
\[ v(t) = \frac{ds}{dt} = -19.8 t + 79.2 \].
This equation tells us how the velocity of the car changes with time after the brakes are applied.
kinematic equations
Kinematic equations describe the motion of objects without considering the causes of motion. They are essential tools in physics for analyzing the motion of vehicles, among other things. In this exercise, the kinematic equation provided is the position function of the car \(s(t) = -9.9 t^{2} + 79.2 t\).
This equation has useful information:
- The term \(-9.9 t^{2}\) relates to the decelerative effect after the brakes are applied.
- The term \(79.2 t \) gives the initial velocity component of the car before deceleration significantly affects its motion.
This function helps you track the exact position of the car at any time \(t\).
By differentiating this position function, we convert it into a velocity function that tells us how fast and in which direction the car is moving at any time.
calculus in physics
Calculus is widely used in physics to solve problems about motion, force, and energy. This exercise is a perfect example of how calculus helps in understanding physical phenomena. Here, we've used differentiation to move from a position function to a velocity function.
This shift helps solve questions like:
- When will the car stop? To answer this, we set the velocity function to zero and solve for time, giving \(t = 4 \) seconds.
- How far will it travel before stopping? By substituting \( t = 4 \) back into the position function \(s(4)\), we find that the car travels 158.4 meters before coming to a stop.
Integrating these calculations into real-life scenarios provides a keen insight into how initial speeds and deceleration rates can influence travel distances and stopping times. Calculus methods like differentiation and integration play crucial roles in predicting and analyzing these outcomes.
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