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 use the information we have at t = 0 seconds and t = 2 seconds.
First, let's understand the motion of the object. At t = 2 seconds, the object is 3 feet to the right of the origin, so it has moved to the right. However, we also know that the object is moving to the leftward at a speed of 8 feet per second. Therefore, between t = 0 seconds and t = 2 seconds, the object must have started at some point to the left of the origin, moved to the right, and then started moving to the left.
Next, let's find out how far the object moved to the right between t = 0 seconds and t = 2 seconds. Since we know the object's speed at t = 2 seconds is 8 feet per second, and it has been moving for 2 seconds, we can multiply the speed and time to get the distance. Thus, the object traveled 8 feet/s * 2 s = 16 feet to the right during this time interval.
Now, we need to determine the initial position of the object. Since it was at the origin at t = 0 seconds, and it moved 16 feet to the right between t = 0 seconds and t = 2 seconds, the initial position must be -16 feet.
Finally, we can estimate the position of the object after 2.5 seconds. Since the object started at -16 feet, moved 16 feet to the right, and then started moving to the left, it will continue moving to the leftward. Therefore, after 2.5 seconds, the object will be to the left of its initial position.
In summary, our estimate for the position of the object after 2.5 seconds is somewhere between -16 feet and 0 feet, to the left of the origin.
(2) With calculus using linearization:
Linearization is a method that approximates the value of a function near a specific point by considering its linear equation. In this case, we want to estimate the position of the object after 2.5 seconds, which means we want to approximate x(2.5).
The linear equation relating position with time can be expressed as: x(t) = x(0) + v(0) * t, where x(t) is the position at time t, x(0) is the initial position, and v(0) is the initial velocity. However, we don't have the value of v(0).
To find v(0), we can use the fact that at t = 2 seconds, the object has a velocity of -8 feet per second. Since x(t) is the position function, the derivative of x(t) with respect to t gives us the velocity function: v(t) = x'(t).
Therefore, we have v(2) = -8, which means the velocity at t = 2 seconds is -8 feet per second. This gives us the following equation: v(0) = v(2) + a * (0 - 2), where a is the acceleration.
Since we don't know the acceleration, we can use the average acceleration over the time interval from t = 0 to t = 2: a = (v(2) - v(0)) / (2 - 0) = (-8 - v(0)) / 2.
Substituting this back into our original equation, we can solve for v(0): v(0) = v(2) + [(-8 - v(0)) / 2] * (-2).
Simplifying, we get: v(0) = -8 + (16 + 2v(0)) / 2.
Solving for v(0), we find v(0) = -6 feet per second.
Now we can use the linear equation to estimate x(2.5): x(2.5) = x(0) + v(0) * (2.5 - 0).
Since x(0) is 0 feet (object is at rest initially) and v(0) is -6 feet per second (negative because the object is moving to the left), we have: x(2.5) = 0 + (-6) * 2.5.
Calculating, we find x(2.5) = -15 feet.
Therefore, our estimate for the position of the object after 2.5 seconds is -15 feet, to the left of the origin.