A drag racer accelerates at a(t)=88 ft / s^2 . Assume v(0)=0, s(0)=0, and t is measured in secon

A drag racer accelerates at a(t)=88 ft / s^2 . Assume...

A drag racer accelerates at $a(t)=88 \mathrm{ft} / \mathrm{s}^{2} .$ Assume $v(0)=0, s(0)=0,$ and $t$ is measured in seconds. a. Determine the position function, for $t \geq 0$ b. How far does the racer travel in the first 4 seconds? c. At this rate, how long will it take the racer to travel $\frac{1}{4}$ mi? d. How long does it take the racer to travel 300 ft? e. How far has the racer traveled when it reaches a speed of $178 \mathrm{ft} / \mathrm{s} ?$

A drag racer accelerates at $a(t)=88 \mathrm{ft} / \mathrm{s}^{2} .$ Assume $v(0)=0, s(0)=0,$ and $t$ is measured in seconds.
a. Determine the position function, for $t \geq 0$
b. How far does the racer travel in the first 4 seconds?
c. At this rate, how long will it take the racer to travel $\frac{1}{4}$ mi?
d. How long does it take the racer to travel 300 ft?
e. How far has the racer traveled when it reaches a speed of
$178 \mathrm{ft} / \mathrm{s} ?$

Key Concepts

Kinematics Relationships

This concept involves understanding the interdependence of acceleration, velocity, and position. It is based on the fundamental kinematics equations in one-dimensional motion that state acceleration is the derivative of velocity and velocity is the derivative of position with respect to time. Conversely, obtaining velocity requires integration of the acceleration, and finding position requires integration of the velocity.

Integration in Calculus

In the context of this problem, integration plays a crucial role since it is used to determine the antiderivatives of acceleration and velocity to obtain velocity and position functions, respectively. Understanding the basic integration rules and how to apply them to constant and variable acceleration is essential in deriving these kinematic equations.

Initial Conditions and Definite Integration

Initial conditions such as the starting velocity and starting position are critical in solving constant integration problems because they help in finding the constants of integration. In kinematics, knowing that both initial velocity and position are zero provides a clear starting point and simplifies the process of integrating the acceleration function properly.

Solving Equations for Specific Variables

After obtaining expressions for velocity and position as functions of time, the next step involves solving these equations for unknown quantities, such as time or distance traveled. This requires algebraic manipulation and sometimes the application of inverse operations, ensuring that the outcomes address the specific requirements of the problem.

Unit Conversion and Consistency

When dealing with real-world problems, it is important to ensure that units are consistent throughout the calculations. Whether converting feet to miles or vice versa, maintaining consistent units in acceleration, velocity, and distance is crucial for obtaining correct and meaningful results. This concept underlines the importance of paying attention to unit conversions when solving problems involving physical quantities.

Transcript

00:01 The information that we start with is a known acceleration at any time t of 88 feet per square second, a velocity at zero of zero, so it starts at rest, and a position of zero is zero, so it's starting at some initial zero point.

00:18 The velocity function can be found just by integrating the acceleration.

00:24 And if i integrate the constant 88, i would have 88t plus a constant.

00:31 However, i know that the position is, or excuse me, the velocity is zero at the time of zero.

00:37 So if i fill in a zero for t, i end up with zero plus c.

00:42 And if the velocity is supposed to be zero at that point, that also means that c is zero.

00:50 For the position function, i would integrate the velocity.

00:55 The integral of 88t, oops, get rid of that t, get rid of that c there.

01:00 The integral of 88t is 44 t squared plus the constant.

01:09 But the position is zero when the time is zero.

01:12 And if i fill in zero for the time, that would give me the position of c.

01:17 So c is also zero.

01:21 Okay, now let's answer our questions.

01:24 A wanted us to find the position function, which we just did.

01:28 It's 44 t squared.

01:34 The second part asks us to figure out how far the racer traveled in four seconds.

01:42 Well, the distance he traveled, since there's no change in direction here, is just s of 4 minus s of 0.

01:52 S of 4, if i fill 4 in for the time, will give me a distance of 704 feet, and the starting position s of 0 is 0.

02:03 I'm subtracting zero...

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