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After four seconds of motion the speed, which is proportional to the time, is equal to \(1 \mathrm{~cm} / \mathrm{s}\). What is the distance travelled in the first ten seconds ?

Short Answer

Expert verified
Answer: The distance traveled by the object in the first ten seconds is 100/8 cm.

Step by step solution

01

Find the proportionality constant k.

Since the speed is proportional to the time, we can write the speed as \(v(t) = kt\), where k is the proportionality constant. We know that after four seconds, the speed is \(1 \mathrm{~cm} / \mathrm{s}\). So, to find the proportionality constant k, we can use the equation: $$v(4) = k \cdot 4.$$ Plug in the given speed value: $$1 = k \cdot 4.$$ Now, we can find the value of k: $$k = \frac{1}{4} \mathrm{~cm} / \mathrm{s^2}.$$
02

Find the speed function

Now that we found the proportionality constant k, we can write the speed function as: $$v(t) = \frac{1}{4}t \mathrm{~cm} / \mathrm{s}.$$
03

Calculate the distance traveled

To calculate the distance traveled in the first ten seconds, we can find the integral of the speed function with respect to time from \(0\) to \(10\) seconds: $$d = \int_{0}^{10} \frac{1}{4}t \, dt.$$ Integrate the function: $$d = \frac{1}{8}t^2 \Big|_{0}^{10}.$$ Evaluate the integral at the bounds: $$d = \frac{1}{8}(100) - \frac{1}{8}(0) = \frac{100}{8} \mathrm{~cm}.$$ Finally, the distance traveled in the first ten seconds is \(\frac{100}{8} \mathrm{~cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportionality Constant
In integral calculus, the concept of a proportionality constant is essential when dealing with relationships where one quantity varies linearly with another. In our exercise, the velocity is directly proportional to time. This means as time increases, the velocity increases linearly, and the factor by which time is multiplied to obtain the velocity is the proportionality constant.

This constant is represented by the variable 'k' in the equation v(t) = kt. Finding 'k' is crucial, as it defines the specific velocity-time relationship for a given situation. Once we know that at 4 seconds the speed is 1 cm/s, we use this information to calculate 'k'. Simply put, if you double the time, the speed doubles, and so on, maintaining a consistent ratio defined by 'k'. Understanding the role of the proportionality constant helps us construct the velocity function needed to find the distance traveled, which is just another step along the way.
Velocity-Time Relation
The velocity-time relation in physics expresses how velocity changes over time. In our problem, we found the proportionality constant and established that the velocity changes with time according to the rule v(t) = kt. Here, v(t) reflects the velocity at any instant 't', and it is crucial to note that this relationship is the foundation for finding displacement when acceleration is not constant.

To truly grasp the concept, consider a simple analogy: if the speed of a car is directly proportional to the time since it started (just like in our exercise), then the car goes faster the longer it's been driving. Similarly, we can describe the car’s velocity at any point in time using the velocity-time relation. This linear relationship forms the basis for predicting the car's future speed or retracing its past speed, making it an indispensable tool in kinematics and integral calculus.
Definite Integral
The definite integral is a fundamental concept in calculus, representing the net area under a curve within specified limits. It is used to calculate various physical quantities such as distance, area, or even volume. In the context of our problem, the integral of the velocity function over a given time interval gives us the total distance traveled during that interval.

Applying the definite integral to the velocity function from 0 to 10 seconds, \(d = \int_{0}^{10} \frac{1}{4}t \, dt\), allows us to calculate the area under the velocity-time graph, which corresponds to the distance. The process involves finding the antiderivative of the integrand, applying the limits, and evaluating the difference, which gives us the final answer for the distance traveled. Through a definite integral, we can interpret and solve practical problems in physics, engineering, and beyond, making it a cornerstone of applied mathematics.

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Most popular questions from this chapter

Supose that \(\mathrm{f}\) and \(\mathrm{g}\) are integrable on \([\mathrm{a}, \mathrm{b}]\), but neither \(f(x) \geq g(x)\) nor \(g(x) \geq f(x)\) holds for all \(x\) in \([a, b]\) [i.e., the curves \(y=f(x)\) and \(y=g(x)\) are intertwined]. (a) What is the geometric significance of the integral \(\int_{a}^{b}[f(x)-g(x)] d x\) ? (b) What is the geometric significance of the integral \(\int_{a}^{b}|f(x)-g(x)| d x\) ?

Find the area included between the curves \(y=\sin ^{-1} x, y=\cos ^{-1} x\) and the \(x\)-axis.

Let \(\mathrm{A}\) and \(\mathrm{B}\) be the points of intersection of the parabola \(y=x^{2}\) and the line \(y=x+2\), and let \(C\) be the point on the parabola where the tangent line is parallel to the graph of \(\mathrm{y}=\mathrm{x}+2 .\) Show that the area of the parabolic segment cut from the parabola by the line four-thirds the area of the triangle \(\mathrm{ABC}\).

(a) If \(f(y)=-y^{2}+y+2\), sketch the region bounded by the curve \(x=f(y)\), the \(y\)-axis, and the lines \(\mathrm{y}=0\) and \(\mathrm{y}=1\). Find its area. (b) Find the area bounded by the curve \(x=-y^{2}+\) \(\mathrm{y}+2\) and the \(\mathrm{y}\)-axis. (c) The equation \(\mathrm{x}+\mathrm{y}^{2}=4\) can be solved for \(\mathrm{x}\) as a function of \(\mathrm{y}\), or for \(\mathrm{y}\) as plus or minus a function of \(x\). Sketch the region in the first quadrant bounded by the curve \(x+y^{2}=4\) and find its area first by integrating a function of \(\mathrm{y}\) and then by integrating a function of \(\mathrm{x}\).

Which of the following statements are true? (i) If \(\mathrm{C}>0\) is a constant, the region under the line \(\mathrm{y}=\mathrm{C}\) on the interval \([\mathrm{a}, \mathrm{b}]\) has area \(\mathrm{A}=\mathrm{C}(\mathrm{b}-\mathrm{a})\) (ii) IfC> 0 is a constant and \(\mathrm{b}>\mathrm{a} \geq 0\), the region under the line \(\mathrm{y}=\mathrm{Cx}\) on the interval \([\mathrm{a}, \mathrm{b}]\) has area \(\mathrm{A}=\frac{1}{2} \mathrm{C}(\mathrm{b}-\mathrm{a})\) (iii) Theregion under the parabolay \(=\mathrm{x}^{2}\) on the interval \([\mathrm{a}, \mathrm{b}]\) has area less than \(\frac{1}{2}\left(\mathrm{~b}^{2}+\mathrm{a}^{2}\right)(\mathrm{b}-\mathrm{a})\). (iv) The region under the curve \(\mathrm{y}=\sqrt{1-\mathrm{x}^{2}}\) on the interval \([-1,1]\) has area \(\mathrm{A}=\frac{\pi}{2}\). (v) Let f be a function that satisfies \(\mathrm{f}(\mathrm{x}) \geq 0\) for \(\mathrm{x}\) in the interval \([a, b]\). Then the area under the curve \(\mathrm{y}=\mathrm{f}^{2}(\mathrm{x})\) on the interval \([\mathrm{a}, \mathrm{b}]\) mustalways be greater than the area under \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) on the same interval. (vi) If f is even and \(\mathrm{f}(\mathrm{x}) \geq 0\) throughout the interval \([-a, a]\), then the area under the curve \(y=f(x)\) on this interval is twice the area under \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) on \([0, a]\).

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