Chapter 4 | Applications of Derivatives
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We are interested in how long it takes for the car to stop. Recall that the velocity function v ( t ) is the derivative of a position function s ( t ), and the acceleration a ( t ) 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. Example 4.54 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 .
a. How many seconds elapse before the car stops? b. How far does the car travel during that time?
Solution a. First we introduce variables for this problem. Let t be the time (in seconds) after the brakes are first applied. Let a ( t ) be the acceleration of the car (in feet per seconds squared) at time t . Let v ( t ) be the velocity of the car (in feet per second) at time t . Let s ( t ) 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 88ft/sec. Therefore, the initial velocity is v (0) =88 ft/sec. Since the car is decelerating, the acceleration is a ( t ) =−15ft/s 2 .
The acceleration is the derivative of the velocity,
v ′( t ) =−15.
Therefore, we have an initial-value problem to solve:
v ′( t ) =−15, v (0) =88.
Integrating, we find that
v ( t ) =−15 t + C . Since v (0) =88, C =88. Thus, the velocity function is v ( t ) =−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 −15 t +88=0, we obtain t = 88 15 sec. b. To find how far the car travels during this time, we need to find the position of the car after 88 15 sec.We know the velocity v ( t ) is the derivative of the position s ( t ). Consider the initial position to be s (0) =0. Therefore, we need to solve the initial-value problem s ′( t ) =−15 t +88, s (0) =0.
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