Approaching a station At t=0, a train approaching a station...

Approaching a station At $t=0,$ a train approaching a station begins decelerating from a speed of $80 \mathrm{mi} / \mathrm{hr}$ according to the acceleration function $a(t)=-1280(1+8 t)^{-3},$ where $t \geq 0$ is measured in hours. How far does the train travel between $t=0$ and $t=0.2 ?$ Between $t=0.2$ and $t=0.4 ?$ The units of acceleration are $\mathrm{mi} / \mathrm{hr}^{2}.$

Key Concepts

Initial Conditions in Motion Problems

When solving differential equations in kinematics, initial conditions are needed to determine the specific constants of integration. These conditions, such as the initial velocity or the starting position, ensure that the solution accurately reflects the physical situation. Applying initial conditions transforms the general solution into a particular solution applicable to the given problem.

Definite Integration for Displacement

Definite integration is used to calculate the net change of a quantity, such as displacement, over a specific interval of time. By evaluating the integral of the velocity function between two time limits, one can determine the exact distance traveled during that interval. This technique is critical for solving problems where motion is evaluated over a defined period.

Kinematic Relationships

In kinematics, acceleration, velocity, and displacement are interconnected. Acceleration is defined as the rate of change of velocity with respect to time, and velocity is the rate of change of displacement (or position). Understanding these relationships is essential because it allows one to use calculus methods to move from acceleration functions to velocity and subsequently to displacement.

Integration in Kinematics

Integration is the mathematical tool used to reverse the process of differentiation. When given an acceleration function, integrating it provides the velocity function, and integrating the velocity function gives the displacement function. This method of finding antiderivatives is key in solving many motion problems where the acceleration varies with time.

Transcript

00:02 Okay, here we have a problem.

00:05 So we have a train.

00:07 The train is decelerating from a speed of 80 miles per hour, which is why we have b of 0 is equal to 80.

00:15 And we know that the acceleration, or deceleration in this case, is negative is written as the function.

00:21 A of t equals negative 1 ,280 times 1 plus 8t to the power of negative 3.

00:29 So what we're going to do first is we're going to find the formula.

00:32 Or the function of the velocity.

00:34 To do that, i'm going to rewrite a of t as negative 1280 divided by 1 plus 8t to the power of 3, which if we take the anti -derivative of this, we're going to be getting negative 1280 divided by 1 plus 8t to the power of 2, 2 times 2 times 8.

01:16 The reason why we get 2 is because of this here, negative 2, which would be negative 2, and 8 comes from here.

01:24 So if we do that, we get 16, and if we divide 1 ,280 by 16, we get 80.

01:40 So our velocity, our function of velocity, is equal to negative or positive 80 divided by 1 plus 8t squared plus c.

02:07 Now we want to figure out c.

02:09 We know that when it's equal to 0, this is equal to 80.

02:16 So you get 80 equals 80 divided by 1 plus c.

02:28 So c is equal to 0.

02:33 There.

02:35 What we're going to do is we're going to figure out how far the train travels from t equals 0 to t equals 0 .2.

02:46 To do that, we're going to use an integral.

02:48 We're going to integrate this from 0 to 0 .2 of 80 divided by 1 plus 8t squared, which is equal to negative 10 divided by 1 plus 8t.

03:12 Oh, not 1 plus 8, sorry.

03:16 It's going to be 1 plus 8 times 0 .2...

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