Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function
$v\left(t\right)$ is the derivative of a position function
$s\left(t\right),$ and the acceleration
$a\left(t\right)$ is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.
Decelerating car
A car is traveling at the rate of
$88$ ft/sec
$(60$ mph) when the brakes are applied. The car begins decelerating at a constant rate of
$15$ ft/sec
^{2} .
How many seconds elapse before the car stops?
How far does the car travel during that time?
First we introduce variables for this problem. Let
$t$ be the time (in seconds) after the brakes are first applied. Let
$a\left(t\right)$ be the acceleration of the car (in feet per seconds squared) at time
$t.$ Let
$v\left(t\right)$ be the velocity of the car (in feet per second) at time
$t.$ Let
$s\left(t\right)$ be the car’s position (in feet) beyond the point where the brakes are applied at time
$t.$ The car is traveling at a rate of
$88\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ Therefore, the initial velocity is
$v\left(0\right)=88$ ft/sec. Since the car is decelerating, the acceleration is
Since
$v\left(0\right)=88,C=88.$ Thus, the velocity function is
$v\left(t\right)=\mathrm{-15}t+88.$
To find how long it takes for the car to stop, we need to find the time
$t$ such that the velocity is zero. Solving
$\mathrm{-15}t+88=0,$ we obtain
$t=\frac{88}{15}$ sec.
To find how far the car travels during this time, we need to find the position of the car after
$\frac{88}{15}$ sec. We know the velocity
$v\left(t\right)$ is the derivative of the position
$s\left(t\right).$ Consider the initial position to be
$s\left(0\right)=0.$ Therefore, we need to solve the initial-value problem
requires us first to find the set of antiderivatives of
$f$ and then to look for the particular antiderivative that also satisfies the initial condition.
For the following exercises, show that
$F\left(x\right)$ are antiderivatives of
$f\left(x\right).$
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject.
Well, this is what I guess so.
A conical container of radius 10 ft and height 30 ft is filled with water to a depth of 15 ft. How much work is required to pump all the water out through a hole in the top of the container if the unit weight of the water is 62.4 lb/ft^3?