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) Without calculus:
To estimate the position of the object after 2.5 seconds, we can make use of the information we have at times t = 0 seconds and t = 2 seconds.
At t = 0 seconds, the object is at position x = 0 feet. After exactly 2 seconds, we measure the position of the object to be 3 feet to the right of the origin. We also know that the object is moving to the leftward at a speed of 8 feet per second.
From this information, we can infer that the object was initially moving to the right, starting from a position that is less than 3 feet to the left of the origin.
Since the object is moving to the leftward at a speed of 8 feet per second, we can estimate that after 2.5 seconds, the object would have moved further to the left.
However, since we do not know the exact motion of the object between t = 0 seconds and t = 2 seconds, we can only make an estimate. We can assume that the object's motion during this time period is roughly constant.
Based on this assumption, we can estimate that the object would have moved approximately half the distance it covered between t = 0 seconds and t = 2 seconds. This means that after 2.5 seconds, the object would be approximately 2.5 feet to the left of its position at t = 2 seconds. Therefore, its estimated position after 2.5 seconds would be around 0.5 feet to the left of the origin.
(2) With calculus using the ideas of linearization:
To estimate the position of the object after 2.5 seconds using calculus, we can use linearization.
First, let's find the object's average velocity between t = 0 seconds and t = 2 seconds. The average velocity is given by the change in position divided by the change in time:
Average velocity = (Change in position) / (Change in time)
= (3 ft - 0 ft) / (2 s - 0 s)
= 1.5 ft/s
Now, let's find the object's instantaneous velocity at t = 2 seconds. We know that the object is moving to the leftward at a speed of 8 feet per second. Therefore, at t = 2 seconds, the object's instantaneous velocity is -8 ft/s.
Since the object's motion is not constant, we can use linearization to estimate its position at t = 2.5 seconds. Linearization involves approximating a nonlinear function with a straight line.
We can start by finding the equation of the line that represents the object's motion between t = 0 seconds and t = 2 seconds. This line would have a slope equal to the object's average velocity and pass through the point (0, 0). The equation of the line can be written as:
y = mx + b
where y represents the position of the object at a given time, x represents the time, m represents the slope (average velocity), and b represents the y-intercept.
Using the information we have, the equation becomes:
y = 1.5x + 0
Now, we can approximate the position of the object at t = 2.5 seconds by plugging in x = 2.5 into the equation:
y = 1.5(2.5)
= 3.75 ft
Therefore, using linearization, we can estimate that the object's position at t = 2.5 seconds is approximately 3.75 feet to the right of the origin.
Note: The estimate obtained using linearization is more accurate than the estimate obtained without calculus, as it takes into account the non-constant motion of the object.