An object at rest begins moving
horizontally on the x-axis, starting at position x(t) = 0 feet at time t = 0 seconds.Suppose, after exactly 2 seconds we measure the position of the
object to be 3 feet to the right of the origin AND measure the object to be moving to the leftward at a speed of 8 feet per second.
Caution: We do NOT know anything about the position or motion of the object, other than what we know at times t = 0
seconds and t = 2 seconds. So do NOT simply assume the object is moving at constant speed the whole time.
Goal of the Problem: Estimate the position of the object after 2.5 seconds:
(1) by writing a solution as if you were carefully and clearly
explaining it (without calculus) to someone who has NOT had
calculus AND
(2) by writing a solution (with calculus) using the ideas of
linearization.
(1) To estimate the position of the object after 2.5 seconds without using calculus, we can make use of the information we have at times t = 0 seconds and t = 2 seconds.
From the given information, we know that at t = 0 seconds, the object is at position x(t) = 0 feet. At t = 2 seconds, the object is 3 feet to the right of the origin and moving to the left at a speed of 8 feet per second.
Since we want to estimate the position at t = 2.5 seconds, we can assume that the object continues moving with the same speed and direction. Therefore, we can estimate that at t = 2.5 seconds, the object will have moved half of the distance it traveled between t = 0 seconds and t = 2 seconds.
The distance between t = 0 seconds and t = 2 seconds is 3 feet (since the object is 3 feet to the right of the origin at t = 2 seconds). Therefore, the object will have moved 1.5 feet at t = 2.5 seconds.
Since at t = 2 seconds, the object is moving to the left at a speed of 8 feet per second, we can estimate that at t = 2.5 seconds, the object will have moved another 0.5 seconds in the leftward direction. This means the object will have moved 0.5 * (-8) = -4 feet at t = 2.5 seconds.
Therefore, our estimate for the position of the object at t = 2.5 seconds is 1.5 feet to the left of the origin, which is at position x(t) = -1.5 feet.
(2) To solve the problem using calculus and the idea of linearization, we can start by finding the average rate of change of position between t = 0 seconds and t = 2 seconds.
The average rate of change of position is given by the formula:
average rate of change = (change in position) / (change in time)
Between t = 0 seconds and t = 2 seconds, the change in position is 3 feet (since the object is 3 feet to the right of the origin at t = 2 seconds). The change in time is 2 seconds.
Therefore, the average rate of change of position is 3 feet / 2 seconds = 1.5 feet per second.
Next, we can use this average rate of change to estimate the instantaneous rate of change of position at t = 2 seconds.
Since the object is moving to the left at a speed of 8 feet per second at t = 2 seconds, we can estimate the instantaneous rate of change of position as -8 feet per second.
Using the idea of linearization, we can approximate the change in position between t = 2 seconds and t = 2.5 seconds by multiplying the instantaneous rate of change of position (-8 feet per second) by the change in time (0.5 seconds):
change in position = (-8 feet per second) * (0.5 seconds) = -4 feet
Therefore, our estimate for the position of the object at t = 2.5 seconds is obtained by adding this change in position to the position at t = 2 seconds:
position at t = 2 seconds + change in position = 3 feet - 4 feet = -1 feet.
So, our estimate for the position of the object at t = 2.5 seconds is x(t) = -1 feet.