The velocity v(t) of a vehicle is given as a function of time t by v(t) = 24 + 5t − t^2.

Question

The velocity v(t) of a vehicle is given as a function of time t by v(t) = 24 + 5t − t^2. You are required to find the displacement of the vehicle after 5 seconds by assuming that the velocity of the vehicle is constant over each one second time interval.

(a) Find the overestimate U5 for the displacement of the vehicle after 5 seconds using the maximum velocity of the vehicle on each one second time interval.

v(0) = 24 (at t = 0)

v(1) = 24 + 5(1) - (1)^2 = 24 + 5 - 1 = 28 (at t = 1)

v(2) = 24 + 5(2) - (2)^2 = 24 + 10 - 4 = 30 (at t = 2)

v(3) = 24 + 5(3) - (3)^2 = 24 + 15 - 9 = 30 (at t = 3)

v(4) = 24 + 5(4) - (4)^2 = 24 + 20 - 16 = 28 (at t = 4)

v(5) = 24 + 5(5) - (5)^2 = 24 + 25 - 25 = 24 (at t = 5)

U1 = v(1) × 1 = 28 × 1 = 28 meters

U2 = v(2) × 1 = 30 × 1 = 30 meters

U3 = v(3) × 1 = 30 × 1 = 30 meters

U4 = v(4) × 1 = 28 × 1 = 28 meters

U5 = v(5) × 1 = 24 × 1 = 24 meters

U1 + U2 + U3 + U4 + U5 = 28 + 30 + 30 + 28 + 24 = 140 meters

the overestimate for the displacement of the vehicle after 5 seconds is 140 meters.

(b) Find the underestimate L5 for the displacement of the vehicle after 5 seconds using the minimum velocity of the vehicle on each one second time interval.

Minimum velocity is v(0) = 24 m/s

Minimum velocity is v(1) = 28 m/s

Minimum velocity is v(2) = 30 m/s

Minimum velocity is v(3) = 30 m/s

Minimum velocity is v(5) = 24 m/s

L1 = v(0) × 1 = 24 × 1 = 24 meters

L2 = v(1) × 1 = 28 × 1 = 28 meters

L3 = v(2) × 1 = 30 × 1 = 30 meters

L4 = v(3) × 1 = 30 × 1 = 30 meters

L5 = v(5) × 1 = 24 × 1 = 24 meters

L1 + L2 + L3 + L4 + L5 = 24 + 28 + 30 + 30 + 24 = 136 meters

Am I doing this question correctly?

Answer 1

Your reasoning is sound for a function that continuously increases or decreases: then, your minimum or maximum value would occur at the beginning or end of a second, because all the values in-between have velocities that are between the velocities for the endpoints. See below, y=x, as an example of that being true.

But, let's take a look at the function we have here:

This makes it a little harder: the maximum or minimum in the 2nd second (between x=2 seconds and x=3 seconds) is in the middle.

An easy way to see where maxima or minima occur on our interval is to use the first derivative test: wherever the slope of this function is 0 (called "critical points"), we could have a minimum or a maximum if the signs of the slope are opposite on both sides. (This is also true if the slope is undefined; for now, let's focus on slope equaling 0.) That can be hard to get your head around without seeing it; the visual below illustrates this idea nicely: see that if the slopes are opposite, you get a max or min, and if not, they're not max or min.

So, let's do it. Take the derivative of our function here to find the slope function, set it to zero to find where the slope might be zero, and find the x-coordinates of the points that could be or maxima/minima.

f(x) = 24 + 5x - x2 -> f'(x) = 5-2x = 0 -> x = 2.5

Ok, now we know x = 2.5 is our critical point. Let's look at the slope of either side of the function by choosing one random point less than 2.5 and one greater.

f'(0) = 5-2(0) = 5

f'(3) = 5-2(3) = 5-6 = -1

So our function is increasing to the left and decreasing to the right, and we have a maximum point at 2.5.

Since we know that points less than 2.5 are increasing and points more than 2.5 are decreasing, we can identify the maxima and minima for the first, second, fourth, and fifth second as you had figured out:

f(0) = 24 (minimum for the 1st second)

f(1) = 28 (maximum for the 1st second, minimum for the 2nd second)

f(2) = 30 (maximum for the 2nd second)

f(3) = 30 (maximum for the 4th second)

f(4) = 28 (minimum for the 4th second, maximum for the 5th second)

f(5) = 24 (minimum for the 5th second)

But now we have to find the maximum and minimum for the 3rd second. So, we plug in x=2.5, our time at which the function is at a maximum, to find the maximum in the 3rd second:

f(2.5) = 24 + 5(2.5) - (2.5)2 = 30.25 (maximum for the 3rd second)

Since we know the function increases to the left and decreases to the right, we also know that at our minimum time on the 3rd second interval, x = 2, we have our minimum in the 3rd second: f(2) = 30.

And from here, we can find the overestimate and underestimate as you did in your answer.

U5 = 28(1) + 30(1) + 30.25(1) + 30(1) + 28(1) = 146.25 meters

L5 = 24(1) + 28(1) + 30(1) + 28(1) + 24(1) = 134 meters