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
is the derivative of a position function
and the acceleration
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
ft/sec
mph) when the brakes are applied. The car begins decelerating at a constant rate of
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
be the time (in seconds) after the brakes are first applied. Let
be the acceleration of the car (in feet per seconds squared) at time
Let
be the velocity of the car (in feet per second) at time
Let
be the car’s position (in feet) beyond the point where the brakes are applied at time
The car is traveling at a rate of
Therefore, the initial velocity is
ft/sec. Since the car is decelerating, the acceleration is
The acceleration is the derivative of the velocity,
Therefore, we have an initial-value problem to solve:
Integrating, we find that
Since
Thus, the velocity function is
To find how long it takes for the car to stop, we need to find the time
such that the velocity is zero. Solving
we obtain
sec.
To find how far the car travels during this time, we need to find the position of the car after
sec. We know the velocity
is the derivative of the position
Consider the initial position to be
Therefore, we need to solve the initial-value problem
Integrating, we have
Since
the constant is
Therefore, the position function is
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